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KUMAR PHYSICS CLASSES • MASTER CHAPTER

Ray Optics and Optical Instruments

Complete Ray Optics notes with formula sheet, sign convention, derivations, NCERT solutions, diagrams and exam-style practice for CBSE, NEET, JEE, IB, IGCSE and A-Level Physics.

Ray Optics and Optical Instruments Overview

This Ray Optics and Optical Instruments page is a complete study guide for students who need formulas, sign convention, ray diagrams, NCERT-style solutions and exam-style practice in one place. It covers reflection, refraction, mirrors, lenses, prism, total internal reflection, optical fibre, human eye defects, microscopes, telescopes and resolving power without making unsupported claims about exam years or Google indexing.

Concepts First

Start with ray behaviour, image nature and sign convention before jumping into calculations.

Formula Practice

Use the formula sheet for quick revision, then check the solved examples for correct substitution.

Exam Readiness

Practice CBSE, NEET, JEE, IB, IGCSE and A-Level style questions with visible question statements and expandable answers.

How to Use This Page

1. Revise formulas

Read the formula sheet and constants first so every symbol is familiar.

2. Learn signs

Before solving, write the sign of u, v, f, R and magnification from the diagram.

3. Solve by topic

Move from mirrors and lenses to prism, TIR, fibre, eye defects and instruments.

4. Test yourself

Attempt the question banks before opening the detailed answer blocks.

Common Mistakes in Ray Optics

Formula and Sign Errors

  • Mixing mirror formula with lens formula.
  • Using the wrong sign convention for u, v, f or R.
  • Using centimetres instead of metres in lens power.
  • Confusing real and virtual image positions.
  • Confusing magnification with magnifying power.

TIR, Prism and Instruments

  • Using TIR condition in the wrong direction.
  • Forgetting that TIR needs denser to rarer travel.
  • Using prism minimum-deviation formula without symmetry.
  • Confusing telescope magnification with microscope magnification.
  • Using aperture and focal length incorrectly in resolving power.

1. Complete Formula Sheet

Plane mirror

i=r; image distance = object distance; m=+1

Spherical mirror

1/f=1/v+1/u; m=−v/u; f=R/2

Plane refraction

n₁ sin i=n₂ sin r; apparent depth=real depth/n

Spherical refraction

n₂/v−n₁/u=(n₂−n₁)/R

Thin lens

1/f=1/v−1/u; m=v/u

Lens maker

1/f=(μlens/μmedium−1)(1/R₁−1/R₂)

Power

P=1/f(m); Pcontact=ΣPᵢ; Pseparated=P₁+P₂−dP₁P₂

Prism

δ=i+e−A; μ=sin[(A+δm)/2]/sin(A/2)

TIR

sin C=n₂/n₁, n₁>n₂; i>C

Optical fibre

NA=√(n₁²−n₂²); n₀ sin θa=NA

Human eye

A=1/N−1/F; myopia P=−1/x; hypermetropia P=1/D−1/Dh

Simple microscope

M∞=D/f; MD=1+D/f

Compound microscope

M∞≈(L/f₀)(D/fₑ); MD≈(L/f₀)(1+D/fₑ)

Astronomical telescope

M∞=−f₀/fₑ; L∞=f₀+fₑ; MD=−(f₀/fₑ)(1+fₑ/D)

Resolving power

Telescope: RP=D/(1.22λ); microscope: d=0.61λ/NA

Wavefront

Optical path=μx; phase difference=2πΔ/λ; Malus: I=I₀cos²θ

2. Important Constants

Speed of light in vacuum

c=2.998×10⁸ m s⁻¹

Least distance of distinct vision

D=25 cm

Visible wavelength range

≈380–750 nm

Air refractive index

≈1.0003 (often 1)

Water refractive index

≈1.33

Crown glass refractive index

≈1.52

Diamond refractive index

≈2.42

Normal eye optical power

≈60 D

Retina distance

≈17 mm

1 dioptre

1 m⁻¹

3. Reflection and Spherical Mirrors

Mirror Formula

Mirror Formula

1/f = 1/v + 1/u
m = −v/u
R = 2f
Meaning of symbols
u: object distance; v: image distance; f: focal length; R: radius; P: pole; F: focus; C: centre of curvature.
Sign convention
All distances are measured from pole P using the Cartesian convention.
Important result
A concave mirror has f<0; a convex mirror has f>0.
Common mistake
Using the thin-lens minus sign in the mirror formula.
Exam tip
Write u, v and f with signs before substitution.

Mirror Applications

Mirror Applications

m = hᵢ/hₒ = −v/u
Meaning of symbols
hₒ and hᵢ are object and image heights.
Sign convention
Height above axis is positive; below axis is negative.
Important result
m<0 means inverted real image; m>0 means upright virtual image.
Common mistake
Describing image nature from distance alone.
Exam tip
Use signs of v and m to state real/virtual and upright/inverted.

M1A concave mirror has f=−20 cm and u=−30 cm. Find v and m.

View detailed solution

Concept used: Mirror formula and magnification.

Formula used: 1/f = 1/v + 1/u; m = −v/u


Step-by-step solution:

1/v=1/f−1/u=−1/20+1/30=−1/60, so v=−60 cm. m=−v/u=−2.

Final answer: Real inverted image 60 cm in front; magnification −2.

4. Refraction, Lenses and Apparent Depth

Lens Formula

Lens Formula

1/f = 1/v − 1/u
m = v/u
P = 1/f(m)
Meaning of symbols
u: object distance; v: image distance; f: focal length; P: power in dioptres.
Sign convention
Convex lens f>0; concave lens f<0 under Cartesian convention.
Important result
Real image has v>0; virtual image has v<0 for the usual left-side object.
Common mistake
Mixing centimetres with metres while calculating power.
Exam tip
Use cm consistently in lens formula, then convert f to metres for P.

Lens Maker Formula

Lens Maker Formula

1/f = (μlens/μmedium − 1)(1/R₁ − 1/R₂)
Meaning of symbols
μlens and μmedium are refractive indices; R₁, R₂ are signed radii.
Sign convention
For a biconvex lens facing incident light: R₁>0 and R₂<0.
Important result
Immersing a lens changes its relative index and hence its power.
Common mistake
Using absolute radii instead of signed radii.
Exam tip
Write the relative index explicitly when the surrounding medium is not air.

Thin Lenses in Contact

Thin Lenses in Contact

1/F = 1/f₁ + 1/f₂ + ...
P = P₁ + P₂ + ...
Meaning of symbols
F is equivalent focal length; Pᵢ are signed powers.
Sign convention
Use metres for power; convex positive, concave negative.
Important result
A zero net power combination is afocal in the thin-contact approximation.
Common mistake
Adding focal lengths directly.
Exam tip
Power addition is the quickest method.

Separated Thin Lenses

Separated Thin Lenses

1/F = 1/f₁ + 1/f₂ − d/(f₁f₂)
Meaning of symbols
d is lens separation in the same unit as focal lengths.
Sign convention
Keep all f and d in one unit with signs.
Important result
Equivalent power is P=P₁+P₂−dP₁P₂ when d is in metres.
Common mistake
Omitting the separation term.
Exam tip
For image location, sequential lens formula is often safer than EFL alone.

Spherical Surface Refraction

Spherical Surface Refraction

μ₂/v − μ₁/u = (μ₂−μ₁)/R
Meaning of symbols
μ₁: incident medium; μ₂: refracted medium; R: signed radius.
Sign convention
Distances follow Cartesian convention from the pole of the surface.
Important result
Plane surface follows as R→∞.
Common mistake
Interchanging μ₁ and μ₂.
Exam tip
Draw the direction of incident light before assigning R.

Apparent Depth

Apparent Depth

apparent depth = real depth/μ
shift = t(1−1/μ)
Meaning of symbols
t is real thickness; μ is refractive index relative to observer's medium.
Sign convention
These simple forms assume near-normal viewing from air.
Important result
A denser medium appears shallower.
Common mistake
Using this formula for highly oblique viewing.
Exam tip
State the near-normal approximation.

L1A glass slab is 12 cm thick with μ=1.5. Find apparent shift.

View detailed solution

Concept used: Apparent depth and normal shift.

Formula used: apparent depth = real depth/μ; shift = t(1−1/μ)


Step-by-step solution:

Apparent depth=12/1.5=8 cm. Shift=12−8=4 cm.

Final answer: The lower face appears raised by 4 cm.

5. Prism and Total Internal Reflection

Prism Formula

Prism Formula

δ = i + e − A
At δm: i=e; r₁=r₂=A/2
μ=sin[(A+δm)/2]/sin(A/2)
Meaning of symbols
A: prism angle; i/e: incidence/emergence; δm: minimum deviation.
Sign convention
Angles are positive geometric magnitudes.
Important result
At minimum deviation the path through the prism is symmetric.
Common mistake
Using the minimum-deviation index formula away from symmetry.
Exam tip
Check that i=e before applying the compact formula.

Total Internal Reflection

Total Internal Reflection

sin C = n₂/n₁, n₁>n₂
TIR when i>C
Meaning of symbols
C: critical angle in denser medium; n₁/n₂: denser/rarer indices.
Sign convention
Angle is measured from the normal inside the denser medium.
Important result
At i=C the refracted ray grazes the boundary at 90°.
Common mistake
Writing sinC=n₁/n₂ or allowing rarer-to-denser TIR.
Exam tip
State both conditions: denser-to-rarer and i>C.

T1Find the critical angle for glass of index 1.50 in air.

View detailed solution

Concept used: Critical angle and total internal reflection.

Formula used: sin C=n₂/n₁; TIR when i>C


Step-by-step solution:

sinC=1/1.50=0.6667, so C=41.8°.

Final answer: Critical angle ≈41.8°.

6. Optical Fibre Formulae

Numerical Aperture

Numerical Aperture

ncore>ncladding
NA=√(n₁²−n₂²)
sinθa=NA/n₀
Meaning of symbols
n₁: core; n₂: cladding; n₀: outside medium; θa: acceptance half-angle.
Sign convention
All indices are positive; the standard square-root form assumes a step-index fibre.
Important result
Accepted rays must meet the core-cladding boundary above its critical angle.
Common mistake
Using the air formula sinθa=NA when n₀≠1.
Exam tip
Verify n₁>n₂ before calculating NA.

Fibre TIR Condition

Fibre TIR Condition

sinC=n₂/n₁; internal incidence i>C
Meaning of symbols
C is measured in the core at the core-cladding interface.
Sign convention
Core is optically denser than cladding.
Important result
Larger index contrast increases NA and acceptance cone.
Common mistake
Calling the full cone angle θa; 2θa is the full cone.
Exam tip
Report whether the asked angle is half-angle or full acceptance angle.

F1Core index 1.50 and cladding 1.47. Find NA.

View detailed solution

Concept used: Numerical aperture.

Formula used: NA=√(n₁²−n₂²); n₀sinθa=NA


Step-by-step solution:

NA=√(1.50²−1.47²)=0.2985. In air θa=sin⁻¹0.2985≈17.4°.

Final answer: NA≈0.299; acceptance half-angle≈17.4°.

7. Human Eye Formulae and Vision Defects

Normal Eye and Accommodation

Normal Eye and Accommodation

Near point D=25 cm; far point=∞
P=1/f(m)
A=1/N−1/F
Meaning of symbols
N and F are near/far point distances in metres; A is accommodation in dioptres.
Sign convention
Corrective-lens virtual images lie on the object side, so v<0.
Important result
A normal young eye has conventional accommodation about 4 D.
Common mistake
Using centimetres directly in P=1/f.
Exam tip
Convert every focal length to metres before calculating dioptres.

Myopia Correction

Myopia Correction

f = −x
P = −1/x
Meaning of symbols
x is the myopic far-point distance in metres.
Sign convention
Concave correcting lens has negative f and power.
Important result
The lens forms a virtual image of infinity at the eye's far point.
Common mistake
Giving positive power for myopia.
Exam tip
Map infinity to the far point before applying the eye's own power.

Hypermetropia Correction

Hypermetropia Correction

1/f=1/v−1/u
u=−25 cm; v=−Dh
Meaning of symbols
Dh is defective near-point distance.
Sign convention
Both u and v are negative for the reading lens construction; f is positive.
Important result
P=1/0.25−1/Dh with distances in metres.
Common mistake
Using v=+Dh.
Exam tip
The correcting lens creates a virtual image at the defective near point.

Presbyopia and Astigmatism

Presbyopia and Astigmatism

Presbyopia: bifocal/progressive addition
Astigmatism: cylindrical lens
Meaning of symbols
Reading addition supplies missing near power; cylinder acts in one meridian.
Sign convention
Prescription signs and cylinder axis must be retained.
Important result
Upper bifocal zone commonly serves distance; lower zone near work.
Common mistake
Treating presbyopia as identical to hypermetropia or ignoring cylinder axis.
Exam tip
State the age-related loss of accommodation.

Cataract

Cataract

Surgical removal/replacement with artificial intraocular lens
Meaning of symbols
Cataract is opacity of the crystalline lens.
Sign convention
It is not primarily a refractive sign-convention problem.
Important result
Ordinary spectacles do not remove lens opacity.
Common mistake
Calling cataract a focusing defect corrected by a spherical lens.
Exam tip
Mention scattering, glare and surgical lens replacement.

Colour Blindness

Colour Blindness

Cone-cell or photopigment defect; ordinary lenses do not correct it
Meaning of symbols
Usually red-green discrimination is affected.
Sign convention
No positive/negative lens sign applies.
Important result
Optical power correction cannot restore absent spectral response.
Common mistake
Recommending ordinary spectacles as a cure.
Exam tip
Distinguish colour-vision deficiency from blurred focus.

E1A myopic eye has far point 0.80 m. Find correcting power.

View detailed solution

Concept used: Myopia corrective lens power.

Formula used: P=1/f; Pcontact=ΣPᵢ; Pseparated=P₁+P₂−dP₁P₂


Step-by-step solution:

f=−0.80 m, so P=1/f=−1.25 D.

Final answer: Required lens power is −1.25 D.

8. Simple Microscope

Final Image at Infinity

Final Image at Infinity

M∞ = D/f
Meaning of symbols
D=25 cm; f is magnifier focal length in the same unit.
Sign convention
M is an angular magnification magnitude.
Important result
Shorter focal length gives larger angular magnification.
Common mistake
Confusing linear magnification with magnifying power.
Exam tip
State that the object is at the focal plane.

Final Image at D

Final Image at D

Mᴅ = 1 + D/f
Meaning of symbols
Final virtual image is at the least distance D.
Sign convention
The virtual image has v=−D.
Important result
This setting gives the maximum distinct magnifying power.
Common mistake
Dropping the +1 term.
Exam tip
Mention greater eye strain than relaxed viewing.

S1A magnifier has f=5 cm. Find power at infinity and at D.

View detailed solution

Concept used: Simple microscope angular magnification.

Formula used: M∞=D/f; Mᴅ=1+D/f


Step-by-step solution:

M∞=25/5=5. Mᴅ=1+25/5=6.

Final answer: Magnifying powers are 5× and 6×.

9. Compound Microscope

Normal Adjustment

Normal Adjustment

M∞ = (L/f₀)(D/fₑ)
Meaning of symbols
L: tube length; f₀ objective focal length; fₑ eyepiece focal length.
Sign convention
Formula gives magnitude; the final image is inverted relative to the object.
Important result
Both objective and eyepiece need short focal lengths for high power.
Common mistake
Using lens separation instead of optical tube length without stating approximation.
Exam tip
Keep all lengths in one unit.

Final Image at D

Final Image at D

Mᴅ = (L/f₀)(1 + D/fₑ)
Meaning of symbols
The eyepiece forms the final virtual image at D.
Sign convention
Use magnitudes for quoted instrument power.
Important result
Near-point adjustment gives greater power than normal adjustment.
Common mistake
Applying the simple microscope factor to the objective too.
Exam tip
Separate objective linear magnification and eyepiece angular magnification.

C1L=16 cm, f₀=1 cm, fₑ=4 cm. Find M∞ and Mᴅ.

View detailed solution

Concept used: Compound microscope magnifying power.

Formula used: M∞=(L/f₀)(D/fₑ); Mᴅ=(L/f₀)(1+D/fₑ)


Step-by-step solution:

M∞=(16/1)(25/4)=100. Mᴅ=16(1+25/4)=116.

Final answer: Magnifying powers: 100× at infinity and 116× at D.

10. Astronomical Telescope

Normal Adjustment

Normal Adjustment

M∞ = f₀/fₑ
L = f₀ + fₑ
Meaning of symbols
f₀ objective focal length; fₑ eyepiece focal length; L tube length.
Sign convention
Magnitude is positive in the formula card; astronomical image inversion may be written with a minus sign.
Important result
Long-focus objective and short-focus eyepiece produce large angular magnification.
Common mistake
Using objective diameter in place of focal length.
Exam tip
State whether the sign or only magnitude is requested.

Final Image at D

Final Image at D

Mᴅ = (f₀/fₑ)(1 + fₑ/D)
Meaning of symbols
D=25 cm for a normal eye.
Sign convention
The final image is virtual at D.
Important result
Near-point setting has slightly greater magnitude than normal adjustment.
Common mistake
Using 1+D/fₑ, which belongs to a simple microscope.
Exam tip
Check the bracket: 1+fₑ/D.

A1A telescope has f₀=120 cm and fₑ=5 cm. Find normal power and length.

View detailed solution

Concept used: Astronomical telescope normal adjustment.

Formula used: M∞=f₀/fₑ; Mᴅ=(f₀/fₑ)(1+fₑ/D)


Step-by-step solution:

M=120/5=24; L=120+5=125 cm.

Final answer: Magnifying power 24×; tube length 125 cm.

11. Reflecting Telescopes

Angular Magnification

Angular Magnification

|M| = fobjective/feyepiece
Meaning of symbols
fobjective is the effective focal length of the mirror system.
Sign convention
Quote magnitude unless inversion is specifically requested.
Important result
Reflectors avoid chromatic aberration and allow large aperture.
Common mistake
Treating the convex secondary focal length alone as objective focal length.
Exam tip
Use the effective focal length of the complete primary-secondary system.

Newtonian and Cassegrain Results

Newtonian and Cassegrain Results

Newtonian: 45° plane secondary
Cassegrain: convex secondary + central hole
Meaning of symbols
The secondary folds the beam; it does not replace the primary's light-gathering role.
Sign convention
Mirror reflection obeys i=r at each surface.
Important result
Cassegrain gives a long effective focal length in a compact tube.
Common mistake
Drawing or assuming rays pass through an opaque primary.
Exam tip
Describe the folded path verbally on this formula page.

12. Resolving Power

Telescope Resolution

Telescope Resolution

θmin = 1.22λ/D
Resolving power = D/(1.22λ)
Meaning of symbols
D is objective aperture diameter; λ is wavelength.
Sign convention
Use SI units consistently; θ is in radians.
Important result
Larger aperture and shorter wavelength improve angular resolution.
Common mistake
Using focal length instead of aperture diameter.
Exam tip
Distinguish magnification from resolution.

Microscope Resolution

Microscope Resolution

dmin = 0.61λ/NA
Resolving power = NA/(0.61λ)
Meaning of symbols
NA=n sinα for the objective medium and semi-angle.
Sign convention
dmin is a linear separation.
Important result
Larger NA and shorter wavelength improve resolution.
Common mistake
Using eyepiece focal length in the Rayleigh formula.
Exam tip
Resolution is governed mainly by objective NA.

R1Find θmin for a 10 cm telescope aperture at λ=550 nm.

View detailed solution

Concept used: Rayleigh criterion and telescope resolution.

Formula used: M∞=f₀/fₑ; Mᴅ=(f₀/fₑ)(1+fₑ/D)


Step-by-step solution:

θmin=1.22(550×10⁻⁹)/0.10=6.71×10⁻⁶ rad.

Final answer: Minimum angular separation ≈6.71 μrad.

13. Complete NCERT Solved Examples

The worked examples below are original NCERT-aligned model examples rather than copied textbook prose.

E1Model example: A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

View detailed solution

Concept used: Mirror formula and magnification.

Formula used: 1/f = 1/v + 1/u; m = −v/u


Step-by-step solution:

R=36 cm, so f=−18 cm and u=−27 cm. From 1/f=1/v+1/u, 1/v=−1/18+1/27=−1/54, hence v=−54 cm. m=−v/u=−2, so h′=−5.0 cm. The negative image height means inverted. Moving the candle toward F makes the real image move farther away; at F it is at infinity, and inside F no screen image is possible.

Final answer: Screen 54 cm in front; real, inverted, enlarged to 5.0 cm.

E2Model example: A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

View detailed solution

Concept used: Convex-mirror formula.

Formula used: 1/f = 1/v + 1/u; m = −v/u


Step-by-step solution:

f=+15 cm, u=−12 cm. 1/v=1/15+1/12=3/20, so v=+6.67 cm. m=−v/u=+0.556 and h′=2.50 cm. As the needle recedes, the virtual image approaches F behind the mirror and becomes smaller.

Final answer: Virtual upright image 6.67 cm behind mirror; m=+0.556.

E3Model example: A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

View detailed solution

Concept used: Apparent depth.

Formula used: apparent depth = real depth/μ; shift = t(1−1/μ)


Step-by-step solution:

μ=real/apparent=12.5/9.4=1.33. For μ=1.63, apparent depth=12.5/1.63=7.67 cm. Difference=9.40−7.67=1.73 cm.

Final answer: μwater≈1.33; microscope shifts 1.73 cm upward.

E4Model example: Figures 9.27(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)].

View detailed solution

Concept used: Snell's law and relative index.

Formula used: n₁sin i=n₂sin r


Step-by-step solution:

For a water-glass interface, nwater sin45°=nglass sin r. Using nwater≈1.33 and nglass≈1.50, sin r=(1.33/1.50)sin45°=0.627. Thus r≈38.8°.

Final answer: Angle in glass ≈39° (about 38° using figure-derived indices).

E5Model example: A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

View detailed solution

Concept used: Critical angle and escape cone.

Formula used: sin C=n₂/n₁; TIR when i>C


Step-by-step solution:

sin C=1/1.33, so C=48.75°. The emergent patch radius r=h tan C=0.80 tan48.75°≈0.912 m. Area=πr²≈2.61 m².

Final answer: Area ≈2.61 m².

E6Model example: A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

View detailed solution

Concept used: Prism minimum-deviation relation.

Formula used: δ=i+e−A; μ=sin[(A+δm)/2]/sin(A/2)


Step-by-step solution:

μ=sin50°/sin30°=1.532. In water μrel=1.532/1.33=1.152. Then sin[(60°+δm′)/2]=1.152 sin30°=0.576, giving δm′≈10.3°.

Final answer: Index ≈1.53; minimum deviation in water ≈10.3°.

E7Model example: Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

View detailed solution

Concept used: Lens-maker formula.

Formula used: 1/f=(μlens/μmedium−1)(1/R₁−1/R₂)


Step-by-step solution:

For R₁=+R, R₂=−R: 1/f=(μ−1)(2/R). Therefore R=2(0.55)(20)=22 cm.

Final answer: R=22 cm.

E8Model example: A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is
(a) a convex lens of focal length 20 cm,
(b) a concave lens of focal length 16 cm?

View detailed solution

Concept used: Lens formula with a virtual object.

Formula used: 1/f = 1/v − 1/u; m=v/u


Step-by-step solution:

The converging beam gives u=+12 cm. (a) 1/v=1/20+1/12, so v=7.5 cm. (b) 1/v=−1/16+1/12=1/48, so v=48 cm.

Final answer: Convex: 7.5 cm beyond lens; concave: 48 cm beyond lens.

E9Model example: An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

View detailed solution

Concept used: Concave-lens image formation.

Formula used: 1/f = 1/v − 1/u; m=v/u


Step-by-step solution:

f=−21 cm, u=−14 cm. 1/v=−1/21−1/14=−5/42, hence v=−8.4 cm. m=v/u=+0.60; h′=1.8 cm. On moving the object farther away, the image approaches F and becomes smaller.

Final answer: Virtual, upright, diminished image 8.4 cm before lens; height 1.8 cm.

E10Model example: What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

View detailed solution

Concept used: Powers in contact.

Formula used: P=1/f; Pcontact=ΣPᵢ; Pseparated=P₁+P₂−dP₁P₂


Step-by-step solution:

1/F=1/30−1/20=−1/60 cm⁻¹.

Final answer: F=−60 cm; diverging combination.

14. NCERT Exercise Solutions 9.1–9.31

All questions supplied with the request are covered below. Signs follow the Cartesian convention.

9.1A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

View detailed solution

Concept used: Mirror formula and magnification.

Formula used: 1/f = 1/v + 1/u; m = −v/u


Step-by-step solution:

R=36 cm, so f=−18 cm and u=−27 cm. From 1/f=1/v+1/u, 1/v=−1/18+1/27=−1/54, hence v=−54 cm. m=−v/u=−2, so h′=−5.0 cm. The negative image height means inverted. Moving the candle toward F makes the real image move farther away; at F it is at infinity, and inside F no screen image is possible.

Final answer: Screen 54 cm in front; real, inverted, enlarged to 5.0 cm.

9.2A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

View detailed solution

Concept used: Convex-mirror formula.

Formula used: 1/f = 1/v + 1/u; m = −v/u


Step-by-step solution:

f=+15 cm, u=−12 cm. 1/v=1/15+1/12=3/20, so v=+6.67 cm. m=−v/u=+0.556 and h′=2.50 cm. As the needle recedes, the virtual image approaches F behind the mirror and becomes smaller.

Final answer: Virtual upright image 6.67 cm behind mirror; m=+0.556.

9.3A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

View detailed solution

Concept used: Apparent depth.

Formula used: apparent depth = real depth/μ; shift = t(1−1/μ)


Step-by-step solution:

μ=real/apparent=12.5/9.4=1.33. For μ=1.63, apparent depth=12.5/1.63=7.67 cm. Difference=9.40−7.67=1.73 cm.

Final answer: μwater≈1.33; microscope shifts 1.73 cm upward.

9.4Figures 9.27(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)].

View detailed solution

Concept used: Snell's law and relative index.

Formula used: n₁sin i=n₂sin r


Step-by-step solution:

For a water-glass interface, nwater sin45°=nglass sin r. Using nwater≈1.33 and nglass≈1.50, sin r=(1.33/1.50)sin45°=0.627. Thus r≈38.8°.

Final answer: Angle in glass ≈39° (about 38° using figure-derived indices).

9.5A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

View detailed solution

Concept used: Critical angle and escape cone.

Formula used: sin C=n₂/n₁; TIR when i>C


Step-by-step solution:

sin C=1/1.33, so C=48.75°. The emergent patch radius r=h tan C=0.80 tan48.75°≈0.912 m. Area=πr²≈2.61 m².

Final answer: Area ≈2.61 m².

9.6A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

View detailed solution

Concept used: Prism minimum-deviation relation.

Formula used: δ=i+e−A; μ=sin[(A+δm)/2]/sin(A/2)


Step-by-step solution:

μ=sin50°/sin30°=1.532. In water μrel=1.532/1.33=1.152. Then sin[(60°+δm′)/2]=1.152 sin30°=0.576, giving δm′≈10.3°.

Final answer: Index ≈1.53; minimum deviation in water ≈10.3°.

9.7Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

View detailed solution

Concept used: Lens-maker formula.

Formula used: 1/f=(μlens/μmedium−1)(1/R₁−1/R₂)


Step-by-step solution:

For R₁=+R, R₂=−R: 1/f=(μ−1)(2/R). Therefore R=2(0.55)(20)=22 cm.

Final answer: R=22 cm.

9.8A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is
(a) a convex lens of focal length 20 cm,
(b) a concave lens of focal length 16 cm?

View detailed solution

Concept used: Lens formula with a virtual object.

Formula used: 1/f = 1/v − 1/u; m=v/u


Step-by-step solution:

The converging beam gives u=+12 cm. (a) 1/v=1/20+1/12, so v=7.5 cm. (b) 1/v=−1/16+1/12=1/48, so v=48 cm.

Final answer: Convex: 7.5 cm beyond lens; concave: 48 cm beyond lens.

9.9An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

View detailed solution

Concept used: Concave-lens image formation.

Formula used: 1/f = 1/v − 1/u; m=v/u


Step-by-step solution:

f=−21 cm, u=−14 cm. 1/v=−1/21−1/14=−5/42, hence v=−8.4 cm. m=v/u=+0.60; h′=1.8 cm. On moving the object farther away, the image approaches F and becomes smaller.

Final answer: Virtual, upright, diminished image 8.4 cm before lens; height 1.8 cm.

9.10What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

View detailed solution

Concept used: Powers in contact.

Formula used: P=1/f; Pcontact=ΣPᵢ; Pseparated=P₁+P₂−dP₁P₂


Step-by-step solution:

1/F=1/30−1/20=−1/60 cm⁻¹.

Final answer: F=−60 cm; diverging combination.

9.11A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at
(a) the least distance of distinct vision (25 cm), and
(b) at infinity?
What is the magnifying power of the microscope in each case?

View detailed solution

Concept used: Successive lens formula and angular magnification.

Formula used: 1/f = 1/v − 1/u; m=v/u


Step-by-step solution:

At D: vₑ=−25 cm gives uₑ=−5 cm; objective image distance v₀=10 cm, so u₀=−2.50 cm. |m₀|=4 and mₑ=5, so |M|=20. At infinity: uₑ=−6.25 cm, v₀=8.75 cm, u₀=−2.59 cm; |m₀|=3.375 and mₑ=4, so |M|=13.5.

Final answer: At D: object 2.50 cm, M=20; at infinity: 2.59 cm, M=13.5.

9.12A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

View detailed solution

Concept used: Objective image plus eyepiece at near point.

Formula used: 1/f = 1/v − 1/u; m=v/u


Step-by-step solution:

Using centimetres, f₀=0.8, u₀=−0.9 gives v₀=7.2 cm and |m₀|=8. For vₑ=−25 cm and fₑ=2.5 cm, uₑ=−2.273 cm and mₑ=11. Separation=7.2+2.273=9.47 cm; M=88.

Final answer: Separation ≈9.47 cm; magnifying power ≈88.

9.13A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

View detailed solution

Concept used: Normal adjustment.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

|M|=f₀/fₑ=144/6=24. L=f₀+fₑ=150 cm.

Final answer: Magnifying power 24; separation 150 cm.

9.14(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 10⁶ m, and the radius of lunar orbit is 3.8 × 10⁸ m.

View detailed solution

Concept used: Telescope magnification and small-angle image size.

Formula used: M∞=f₀/fₑ; Mᴅ=(f₀/fₑ)(1+fₑ/D)


Step-by-step solution:

M=15/0.01=1500. Moon angular diameter≈3.48×10⁶/3.8×10⁸=9.16×10⁻³ rad. Objective image diameter=f₀θ=15(9.16×10⁻³)=0.137 m.

Final answer: Magnification 1500; moon image ≈13.7 cm.

9.15Use the mirror equation to deduce that:
(a) An object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) A convex mirror always produces a virtual image independent of the location of the object.
(c) The virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) An object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

View detailed solution

Concept used: Mirror formula and signs.

Formula used: 1/f = 1/v + 1/u; m = −v/u


Step-by-step solution:

Substitute the relevant u ranges in 1/v=1/f−1/u. For a concave mirror f<0: |f|<|u|<2|f| gives |v|>2|f|; |u|<|f| gives v>0 and |m|>1. For a convex mirror f>0 and u<0 always gives 0<v<f and 0<m<1.

Final answer: All four stated ray-diagram results follow from v and m signs/ranges.

9.16A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?

View detailed solution

Concept used: Normal shift through slab.

Formula used: apparent depth = real depth/μ; shift = t(1−1/μ)


Step-by-step solution:

Shift=t(1−1/μ)=15(1−2/3)=5 cm. For a parallel slab at normal viewing, it does not depend on where the slab lies between pin and observer.

Final answer: Apparent rise 5 cm; independent of slab location.

9.17(a) Figure 9.28 shows a cross-section of a ‘light pipe’ made of glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure?
(b) What is the answer if there is no outer covering of the pipe?
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View detailed solution

Concept used: Numerical aperture.

Formula used: NA=√(n₁²−n₂²); n₀sinθa=NA


Step-by-step solution:

NA=√(1.68²−1.44²)=0.865. In air, sinθa=0.865, so θa≈59.9°. Without covering n₂=1, the calculated NA>1, so every physically possible air-incidence angle up to 90° is accepted.

Final answer: With covering: 0°–59.9° to axis; without: 0°–90°.

9.18The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

View detailed solution

Concept used: Object-screen method.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

For fixed D=u+v, uv is maximum at u=v=D/2. Since f=uv/(u+v), fmax=D/4=3/4 m.

Final answer: Maximum f=0.75 m.

9.19A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.

View detailed solution

Concept used: Bessel displacement method.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

f=(D²−d²)/(4D)=(90²−20²)/(360)=21.39 cm.

Final answer: f≈21.4 cm.

9.20(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

View detailed solution

Concept used: Separated lenses and sequential imaging.

Formula used: P=1/f; Pcontact=ΣPᵢ; Pseparated=P₁+P₂−dP₁P₂


Step-by-step solution:

P=3.333−5−0.08(3.333)(−5)=−0.333 D, so EFL=−3.0 m (same EFL from either direction, though principal planes differ). First lens gives v₁=120 cm and m₁=−3. For lens 2, u₂=+112 cm, v₂=−24.35 cm, m₂=−0.2174. Total m=+0.652; image height=0.978 cm.

Final answer: EFL −3.0 m; final image ≈0.98 cm, upright.

9.21At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

View detailed solution

Concept used: Critical angle plus prism geometry.

Formula used: δ=i+e−A; μ=sin[(A+δm)/2]/sin(A/2)


Step-by-step solution:

C=sin⁻¹(1/1.524)=41.03°. Hence r₁=A−C=18.97°. At first face sin i=μ sin r₁=1.524 sin18.97°=0.495.

Final answer: i≈29.7°.

9.22A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.

View detailed solution

Concept used: Linear versus angular magnification.

Formula used: M=β/α; use the relevant microscope or telescope expression


Step-by-step solution:

With u=−f, the virtual image is at infinity, so finite linear size/area magnification is not defined. Angular magnification M=D/f=25/9=2.78. Linear magnification and magnifying power are different quantities.

Final answer: Image at infinity; angular magnification 2.78; no finite area magnification.

9.23(a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.

View detailed solution

Concept used: Simple microscope at near point.

Formula used: M∞=(L/f₀)(D/fₑ); Mᴅ=(L/f₀)(1+D/fₑ)


Step-by-step solution:

For final image at D, M=1+D/f=1+25/9=3.78. Object distance |u|=fD/(D+f)=225/34=6.62 cm. Here |v/u|=25/6.62=3.78.

Final answer: Lens-object distance 6.62 cm; magnification 3.78.

9.24What should be the distance between the object in Exercise 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm²? Would you be able to see the squares distinctly with your eyes very close to the magnifier?
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View detailed solution

Concept used: Area and linear magnification.

Formula used: M=β/α; use the relevant microscope or telescope expression


Step-by-step solution:

Area ratio=6.25, so linear m=2.5. For a virtual image v/u=2.5 and f=9 cm. Solving gives u=−5.4 cm and v=−13.5 cm. This image lies within the normal near point, so it is not seen distinctly with the eye close to the lens.

Final answer: Object distance 5.4 cm; image not distinctly visible to a normal eye.

9.25Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

View detailed solution

Concept used: Angular size and exit pupil.

Formula used: M=β/α; use the relevant microscope or telescope expression


Step-by-step solution:

(a) The magnifier lets the object be brought closer than D while remaining focused. (b) Moving the eye back usually reduces the usable field and can alter effective angular gain. (c) Very short f causes severe aberrations and tiny working distance. (d) Short f₀ gives large objective magnification; short fₑ gives large eyepiece power. (e) The eye belongs near the eyepiece exit pupil, typically a small distance outside it, for the full field.

Final answer: Magnification is angular; practical aberration, field and exit-pupil constraints limit it.

9.26An angular magnification (magnifying power) of 30× is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

View detailed solution

Concept used: Exact normal-adjustment microscope geometry.

Formula used: M∞=(L/f₀)(D/fₑ); Mᴅ=(L/f₀)(1+D/fₑ)


Step-by-step solution:

Eyepiece angular magnification is D/fₑ=5, so objective magnification magnitude must be 6. Let u₀=−a and v₀=6a. The objective formula gives 1/1.25=1/(6a)+1/a, hence a=1.458 cm and v₀=8.75 cm. Put the eyepiece 5 cm beyond the intermediate image.

Final answer: Object ≈1.46 cm before objective; lens separation ≈13.75 cm.

9.27A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e. when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision (25 cm)?

View detailed solution

Concept used: Telescope formulas.

Formula used: M∞=f₀/fₑ; Mᴅ=(f₀/fₑ)(1+fₑ/D)


Step-by-step solution:

Normal adjustment: |M|=140/5=28. At D: |M|=(140/5)(1+5/25)=33.6.

Final answer: 28× at infinity; 33.6× at 25 cm.

9.28(a) For the telescope described in Exercise 9.27(a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 cm?

View detailed solution

Concept used: Small-angle objective image and eyepiece magnification.

Formula used: 1/f = 1/v − 1/u; m=v/u


Step-by-step solution:

L=140+5=145 cm. Tower angle=100/3000=1/30 rad, so objective image height=140/30=4.67 cm. At D, eyepiece linear magnification is |v/u|=25/4.167=6, so final virtual-image height≈28.0 cm.

Final answer: Separation 145 cm; objective image 4.67 cm; final virtual image ≈28 cm.

9.29A Cassegrain telescope uses two mirrors as shown in Fig. 9.26. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

View detailed solution

Concept used: Successive mirror imaging.

Formula used: 1/f = 1/v + 1/u; m = −v/u


Step-by-step solution:

The primary has f₁=110 mm, so at the secondary the converging beam has a virtual object 90 mm away. For the convex secondary, using f₂=70 mm in the appropriate reflected-ray convention: 1/v=1/70−1/90=1/315.

Final answer: Final image 315 mm from the secondary (295 mm behind the primary in this geometry).

9.30Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.29. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

View detailed solution

Concept used: Double-angle reflection.

Formula used: reflected-ray rotation=2θ; spot shift=L tan(2θ)


Step-by-step solution:

The reflected ray turns through 2θ=7°. Shift x=L tan7°=1.5 tan7°=0.184 m.

Final answer: Spot displacement ≈18.4 cm.

9.31Figure 9.30 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

View detailed solution

Concept used: Lens maker and refracting-surface powers.

Formula used: 1/f=(μlens/μmedium−1)(1/R₁−1/R₂)


Step-by-step solution:

Without liquid f=30 cm. For an equiconvex lens, 1/f=(0.5)(2/R), so R=30 cm. With liquid, effective power=(ng−1)/R+(nl−ng)/(−R)=(2−nl)/30=1/45. Thus 2−nl=2/3.

Final answer: Liquid refractive index nl=4/3≈1.33.

15. CBSE Question Bank: 50 Exam-Style Questions

CBSE and NCERT-oriented practice questions; no unverified year attribution.

1Use the Cartesian sign convention to decide the signs of u, v and f for a real image formed by a concave mirror.Easy

Answer and short solution

Short solution: For an object in front of the mirror, u is negative. A real image is also in front, so v is negative. A concave mirror has negative focal length.

Final answer: For an object in front of the mirror, u is negative. A real image is also in front, so v is negative. A concave mirror has negative focal length.

2A concave mirror of focal length 15 cm forms an image of an object placed 45 cm from the pole. Find the image position and magnification.Medium

Answer and short solution

Short solution: With f=-15 cm and u=-45 cm, 1/v=1/f-1/u=-1/15+1/45=-2/45, so v=-22.5 cm and m=-v/u=-0.5.

Final answer: With f=-15 cm and u=-45 cm, 1/v=1/f-1/u=-1/15+1/45=-2/45, so v=-22.5 cm and m=-v/u=-0.5.

3Why does a convex mirror always give an upright image for a real object?Easy

Answer and short solution

Short solution: Reflected rays diverge and their backward extensions meet behind the mirror; this virtual image has positive magnification.

Final answer: Reflected rays diverge and their backward extensions meet behind the mirror; this virtual image has positive magnification.

4State two reasons convex mirrors are used as rear-view mirrors.Easy

Answer and short solution

Short solution: They form upright images and provide a wide field of view, though the images are diminished.

Final answer: They form upright images and provide a wide field of view, though the images are diminished.

5A plane mirror is rotated by 6 degrees. Through what angle does the reflected ray rotate?Easy

Answer and short solution

Short solution: The reflected ray rotates through twice the mirror rotation, so the ray turns by 12 degrees.

Final answer: The reflected ray rotates through twice the mirror rotation, so the ray turns by 12 degrees.

6Describe the image when an object is placed between F and C of a concave mirror.Easy

Answer and short solution

Short solution: The image is real, inverted, enlarged and formed beyond C.

Final answer: The image is real, inverted, enlarged and formed beyond C.

7What is the main difference between the mirror formula and the lens formula?Easy

Answer and short solution

Short solution: The mirror formula is 1/f=1/v+1/u, while the thin lens formula is 1/f=1/v-1/u under Cartesian convention.

Final answer: The mirror formula is 1/f=1/v+1/u, while the thin lens formula is 1/f=1/v-1/u under Cartesian convention.

8A convex lens has focal length 20 cm. Where should an object be placed to obtain an image of the same size?Easy

Answer and short solution

Short solution: Place the object at 2F, or 40 cm from the lens; the real image forms at 2F on the other side.

Final answer: Place the object at 2F, or 40 cm from the lens; the real image forms at 2F on the other side.

9A concave lens has focal length 25 cm. What is the sign of its power in air?Easy

Answer and short solution

Short solution: A concave lens has f=-0.25 m, so P=1/f=-4 D. Its power is negative.

Final answer: A concave lens has f=-0.25 m, so P=1/f=-4 D. Its power is negative.

10Why must focal length be converted to metres before calculating power in dioptres?Easy

Answer and short solution

Short solution: Dioptre is defined as inverse metre, so P=1/f requires f in metres.

Final answer: Dioptre is defined as inverse metre, so P=1/f requires f in metres.

11Find the combined power of two thin lenses of +2.0 D and -1.5 D in contact.Easy

Answer and short solution

Short solution: P=P1+P2=+2.0-1.5=+0.5 D.

Final answer: P=P1+P2=+2.0-1.5=+0.5 D.

12A 12 cm deep tank of water of refractive index 4/3 is viewed normally from air. What is the apparent depth?Easy

Answer and short solution

Short solution: Apparent depth=real depth/mu=12/(4/3)=9 cm.

Final answer: Apparent depth=real depth/mu=12/(4/3)=9 cm.

13Why does a glass slab shift a ray laterally but keep the emergent ray parallel to the incident ray?Medium

Answer and short solution

Short solution: The two parallel faces cause equal and opposite angular deviations, leaving only a sideways displacement.

Final answer: The two parallel faces cause equal and opposite angular deviations, leaving only a sideways displacement.

14Which condition must be satisfied before using the prism minimum-deviation formula?Easy

Answer and short solution

Short solution: The path must be symmetric: i=e and r1=r2=A/2.

Final answer: The path must be symmetric: i=e and r1=r2=A/2.

15Find the critical angle for water of refractive index 1.33 in air.Easy

Answer and short solution

Short solution: sin C=1/1.33=0.752, so C is about 48.8 degrees.

Final answer: sin C=1/1.33=0.752, so C is about 48.8 degrees.

16List both conditions required for total internal reflection.Easy

Answer and short solution

Short solution: Light must travel from denser to rarer medium and the angle of incidence in the denser medium must exceed the critical angle.

Final answer: Light must travel from denser to rarer medium and the angle of incidence in the denser medium must exceed the critical angle.

17Why must an optical fibre have a core refractive index greater than the cladding index?Easy

Answer and short solution

Short solution: A denser core allows total internal reflection at the core-cladding boundary.

Final answer: A denser core allows total internal reflection at the core-cladding boundary.

18What is meant by accommodation of the eye?Easy

Answer and short solution

Short solution: Accommodation is the change in focal length or power of the eye lens to focus objects at different distances on the retina.

Final answer: Accommodation is the change in focal length or power of the eye lens to focus objects at different distances on the retina.

19Which lens corrects myopia and why?Easy

Answer and short solution

Short solution: A concave lens corrects myopia by forming a virtual image of distant objects at the myopic far point.

Final answer: A concave lens corrects myopia by forming a virtual image of distant objects at the myopic far point.

20Which lens corrects hypermetropia and why?Easy

Answer and short solution

Short solution: A convex lens corrects hypermetropia by forming a virtual image of a near object at the defective near point.

Final answer: A convex lens corrects hypermetropia by forming a virtual image of a near object at the defective near point.

21Why is astigmatism corrected using a cylindrical lens?Medium

Answer and short solution

Short solution: Astigmatism involves unequal focusing in different meridians, and a cylindrical lens corrects power along one axis.

Final answer: Astigmatism involves unequal focusing in different meridians, and a cylindrical lens corrects power along one axis.

22Why is cataract not corrected simply by using a spherical spectacle lens?Easy

Answer and short solution

Short solution: Cataract is opacity of the eye lens; spectacles can change focus but cannot remove the scattering opacity.

Final answer: Cataract is opacity of the eye lens; spectacles can change focus but cannot remove the scattering opacity.

23A simple microscope has focal length 5 cm. Find its magnifying power for final image at infinity.Easy

Answer and short solution

Short solution: M=D/f=25/5=5.

Final answer: M=D/f=25/5=5.

24Why does a compound microscope use a short focal length objective?Easy

Answer and short solution

Short solution: A short focal length objective gives high linear magnification of the nearby object.

Final answer: A short focal length objective gives high linear magnification of the nearby object.

25Why should a telescope objective have large aperture?Easy

Answer and short solution

Short solution: A large aperture collects more light and improves resolving power.

Final answer: A large aperture collects more light and improves resolving power.

26Mention one advantage of a reflecting telescope over a refracting telescope.Easy

Answer and short solution

Short solution: A reflecting telescope avoids chromatic aberration and can use a large mirror aperture.

Final answer: A reflecting telescope avoids chromatic aberration and can use a large mirror aperture.

27How does the resolving power of a telescope change when objective diameter is doubled?Medium

Answer and short solution

Short solution: Resolving power is proportional to aperture diameter, so it doubles.

Final answer: Resolving power is proportional to aperture diameter, so it doubles.

28Distinguish between deviation and dispersion in a prism.Easy

Answer and short solution

Short solution: Deviation is bending of a ray; dispersion is splitting of white light because refractive index depends on wavelength.

Final answer: Deviation is bending of a ray; dispersion is splitting of white light because refractive index depends on wavelength.

29Name two everyday examples of total internal reflection.Easy

Answer and short solution

Short solution: Sparkle of diamond and light guidance in optical fibre are common examples.

Final answer: Sparkle of diamond and light guidance in optical fibre are common examples.

30For a convex lens forming a real image of a real object, what are the signs of u and v?Easy

Answer and short solution

Short solution: For the usual left-side object, u is negative and real image distance v is positive.

Final answer: For the usual left-side object, u is negative and real image distance v is positive.

31A mirror gives magnification -2. What do the sign and magnitude mean?Easy

Answer and short solution

Short solution: The negative sign means the image is inverted; magnitude 2 means the image height is twice the object height.

Final answer: The negative sign means the image is inverted; magnitude 2 means the image height is twice the object height.

32Which two principal rays are enough to locate an image formed by a thin lens?Easy

Answer and short solution

Short solution: A ray through the optical centre and a ray parallel to the principal axis are usually sufficient.

Final answer: A ray through the optical centre and a ray parallel to the principal axis are usually sufficient.

33How can a virtual image be identified from a ray diagram?Easy

Answer and short solution

Short solution: It is formed where backward extensions of rays meet, not where actual rays converge.

Final answer: It is formed where backward extensions of rays meet, not where actual rays converge.

34What happens to the power of a convex glass lens when it is immersed in water?Medium

Answer and short solution

Short solution: Its power decreases because the relative refractive index of glass with respect to water is smaller than with respect to air.

Final answer: Its power decreases because the relative refractive index of glass with respect to water is smaller than with respect to air.

35Why does a normally incident ray pass undeviated through a plane glass slab?Easy

Answer and short solution

Short solution: At normal incidence the angle of incidence is zero, so the refracted angle is also zero at each face.

Final answer: At normal incidence the angle of incidence is zero, so the refracted angle is also zero at each face.

36What is optical path length?Medium

Answer and short solution

Short solution: Optical path length is refractive index multiplied by geometrical path length, mu x.

Final answer: Optical path length is refractive index multiplied by geometrical path length, mu x.

37Why are telescope magnification and microscope magnification not used in the same way?Medium

Answer and short solution

Short solution: A telescope magnifies the angular size of distant objects; a microscope magnifies small nearby objects through objective and eyepiece stages.

Final answer: A telescope magnifies the angular size of distant objects; a microscope magnifies small nearby objects through objective and eyepiece stages.

38A glass slab of thickness 9 cm and refractive index 1.5 is viewed normally. Find apparent shift.Easy

Answer and short solution

Short solution: Shift=t(1-1/mu)=9(1-2/3)=3 cm.

Final answer: Shift=t(1-1/mu)=9(1-2/3)=3 cm.

39A prism has A=60 degrees and minimum deviation 40 degrees. Find its refractive index.Medium

Answer and short solution

Short solution: mu=sin((A+dm)/2)/sin(A/2)=sin50/sin30=1.53 approximately.

Final answer: mu=sin((A+dm)/2)/sin(A/2)=sin50/sin30=1.53 approximately.

40Why is diamond brilliant compared with ordinary glass?Easy

Answer and short solution

Short solution: Diamond has a high refractive index and small critical angle, so many rays undergo total internal reflection.

Final answer: Diamond has a high refractive index and small critical angle, so many rays undergo total internal reflection.

41A lens has power +4 D. What is its focal length?Easy

Answer and short solution

Short solution: f=1/P=1/4 m=0.25 m or 25 cm.

Final answer: f=1/P=1/4 m=0.25 m or 25 cm.

42Two lenses of +5 D and +3 D are in contact. Find the equivalent focal length.Easy

Answer and short solution

Short solution: Total power=8 D, so focal length=1/8 m=0.125 m=12.5 cm.

Final answer: Total power=8 D, so focal length=1/8 m=0.125 m=12.5 cm.

43What extra term appears when two thin lenses are separated by distance d?Medium

Answer and short solution

Short solution: The equivalent power becomes P=P1+P2-dP1P2 when d is in metres.

Final answer: The equivalent power becomes P=P1+P2-dP1P2 when d is in metres.

44Why is a plane mirror image laterally inverted but not vertically inverted?Medium

Answer and short solution

Short solution: The mirror reverses the direction normal to its surface; the apparent left-right reversal depends on how the observer labels directions.

Final answer: The mirror reverses the direction normal to its surface; the apparent left-right reversal depends on how the observer labels directions.

45Where is the image formed by a convex mirror relative to pole and focus?Easy

Answer and short solution

Short solution: The virtual image forms behind the mirror between the pole and focus.

Final answer: The virtual image forms behind the mirror between the pole and focus.

46A hypermetropic eye has near point 1 m. Find the spectacle power needed for reading at 25 cm.Medium

Answer and short solution

Short solution: Use u=-0.25 m and v=-1 m: 1/f=1/v-1/u=-1+4=3, so P=+3 D.

Final answer: Use u=-0.25 m and v=-1 m: 1/f=1/v-1/u=-1+4=3, so P=+3 D.

47An astronomical telescope has objective focal length 100 cm and eyepiece focal length 5 cm. Find normal magnifying power.Easy

Answer and short solution

Short solution: Magnitude of M=f0/fe=100/5=20.

Final answer: Magnitude of M=f0/fe=100/5=20.

48What common error occurs when students calculate power using focal length in centimetres?Easy

Answer and short solution

Short solution: They get a value 100 times wrong because dioptre uses metres.

Final answer: They get a value 100 times wrong because dioptre uses metres.

49Why does resolving power improve for shorter wavelength light?Medium

Answer and short solution

Short solution: The diffraction limit is proportional to wavelength, so shorter wavelength gives a smaller resolvable angle or distance.

Final answer: The diffraction limit is proportional to wavelength, so shorter wavelength gives a smaller resolvable angle or distance.

50State the safest first step before substituting in any ray optics numerical.Easy

Answer and short solution

Short solution: Write the sign convention and convert all quantities into consistent units before using the formula.

Final answer: Write the sign convention and convert all quantities into consistent units before using the formula.

16. NEET Question Bank: 75 Conceptual MCQs

NEET-style conceptual and numerical MCQs written as original practice questions.

1A convex mirror forms an image that is A) real and enlarged B) virtual and diminished C) real and diminished D) virtual and inverted.MCQ

Answer and short solution

Short solution: Correct option B. A convex mirror always forms a virtual, upright and diminished image for a real object.

Final answer: Correct option B. A convex mirror always forms a virtual, upright and diminished image for a real object.

2For myopia correction, the required spectacle lens is A) convex B) concave C) cylindrical only D) plane glass.MCQ

Answer and short solution

Short solution: Correct option B. A concave lens diverges incoming rays so the eye focuses them at its far point.

Final answer: Correct option B. A concave lens diverges incoming rays so the eye focuses them at its far point.

3The SI unit of lens power is A) metre B) dioptre C) radian D) candela.MCQ

Answer and short solution

Short solution: Correct option B. Power is inverse focal length in metre and is measured in dioptres.

Final answer: Correct option B. Power is inverse focal length in metre and is measured in dioptres.

4When a ray enters a denser medium obliquely, it bends A) away from normal B) towards normal C) along surface D) randomly.MCQ

Answer and short solution

Short solution: Correct option B. Snell's law shows the refracted angle is smaller in the denser medium.

Final answer: Correct option B. Snell's law shows the refracted angle is smaller in the denser medium.

5Total internal reflection can occur when light travels from A) air to glass B) glass to air C) air to water D) vacuum to glass only.MCQ

Answer and short solution

Short solution: Correct option B. TIR requires incidence from denser to rarer medium.

Final answer: Correct option B. TIR requires incidence from denser to rarer medium.

6If the critical angle is C, total internal reflection occurs for incidence angle A) i<C B) i=C C) i>C D) any i.MCQ

Answer and short solution

Short solution: Correct option C. At i=C the refracted ray grazes the interface; TIR starts above C.

Final answer: Correct option C. At i=C the refracted ray grazes the interface; TIR starts above C.

7A concave mirror of focal length 10 cm has an object at 20 cm. The image is A) at C, same size B) at F C) at infinity D) virtual.MCQ

Answer and short solution

Short solution: Correct option A. Object at C gives a real inverted image at C with equal size.

Final answer: Correct option A. Object at C gives a real inverted image at C with equal size.

8A convex lens of power +5 D has focal length A) 5 cm B) 10 cm C) 20 cm D) 50 cm.MCQ

Answer and short solution

Short solution: Correct option C. f=1/P=0.20 m=20 cm.

Final answer: Correct option C. f=1/P=0.20 m=20 cm.

9A concave lens always forms an image that is A) real B) virtual C) inverted D) magnified for all positions.MCQ

Answer and short solution

Short solution: Correct option B. For a real object, a concave lens forms a virtual upright diminished image.

Final answer: Correct option B. For a real object, a concave lens forms a virtual upright diminished image.

10The refractive index of a medium is 1.5. Speed of light in it is approximately A) 2.0e8 m/s B) 3.0e8 m/s C) 4.5e8 m/s D) 1.5e8 m/s.MCQ

Answer and short solution

Short solution: Correct option A. v=c/n=3e8/1.5=2e8 m/s.

Final answer: Correct option A. v=c/n=3e8/1.5=2e8 m/s.

11In a prism at minimum deviation, A) i=e B) r1 is zero C) e is zero D) r1+r2=2A.MCQ

Answer and short solution

Short solution: Correct option A. The ray path is symmetric at minimum deviation.

Final answer: Correct option A. The ray path is symmetric at minimum deviation.

12The numerical aperture of a fibre in air is mainly determined by A) core and cladding indices B) fibre length C) colour of cover D) source voltage.MCQ

Answer and short solution

Short solution: Correct option A. NA=sqrt(n1^2-n2^2) for a step-index fibre in air.

Final answer: Correct option A. NA=sqrt(n1^2-n2^2) for a step-index fibre in air.

13The image formed by the objective of a compound microscope is A) virtual and final B) real and magnified C) virtual and diminished D) at infinity always.MCQ

Answer and short solution

Short solution: Correct option B. The objective first forms a real magnified intermediate image.

Final answer: Correct option B. The objective first forms a real magnified intermediate image.

14For relaxed viewing through a simple microscope, the object is placed A) at focus B) at 2F C) beyond 2F D) at optical centre.MCQ

Answer and short solution

Short solution: Correct option A. Object at the focus gives final image at infinity.

Final answer: Correct option A. Object at the focus gives final image at infinity.

15Normal adjustment length of an astronomical telescope is A) f0-fe B) f0+fe C) f0/fe D) fe/f0.MCQ

Answer and short solution

Short solution: Correct option B. The final image is at infinity when separation is f0+fe.

Final answer: Correct option B. The final image is at infinity when separation is f0+fe.

16Resolving power of a telescope increases when A) aperture decreases B) aperture increases C) eyepiece is removed D) wavelength increases only.MCQ

Answer and short solution

Short solution: Correct option B. RP is proportional to objective diameter.

Final answer: Correct option B. RP is proportional to objective diameter.

17Cataract is mainly A) opacity of crystalline lens B) elongated eyeball C) short eyeball D) unequal cornea curvature.MCQ

Answer and short solution

Short solution: Correct option A. Cataract makes the eye lens cloudy.

Final answer: Correct option A. Cataract makes the eye lens cloudy.

18Astigmatism is generally corrected using A) cylindrical lens B) plane mirror C) prism only D) concave mirror.MCQ

Answer and short solution

Short solution: Correct option A. A cylindrical lens corrects unequal power in different meridians.

Final answer: Correct option A. A cylindrical lens corrects unequal power in different meridians.

19A glass slab produces A) no lateral shift B) lateral shift with parallel emergent ray C) only dispersion D) no refraction.MCQ

Answer and short solution

Short solution: Correct option B. Opposite faces make the emergent ray parallel but displaced.

Final answer: Correct option B. Opposite faces make the emergent ray parallel but displaced.

20If a lens is cut into two equal halves by a plane passing through its principal axis, the focal length of each half is A) unchanged B) halved C) doubled D) zero.MCQ

Answer and short solution

Short solution: Correct option A. Aperture changes but curvature and focal length remain the same.

Final answer: Correct option A. Aperture changes but curvature and focal length remain the same.

21A lens of focal length -50 cm has power A) +2 D B) -2 D C) +0.5 D D) -0.5 D.MCQ

Answer and short solution

Short solution: Correct option B. f=-0.50 m, so P=-2 D.

Final answer: Correct option B. f=-0.50 m, so P=-2 D.

22A convex lens forms a virtual enlarged image when the object is A) beyond 2F B) at 2F C) between F and O D) at infinity.MCQ

Answer and short solution

Short solution: Correct option C. Object inside focal length gives a virtual enlarged image.

Final answer: Correct option C. Object inside focal length gives a virtual enlarged image.

23The least distance of distinct vision for a normal eye is usually taken as A) 10 cm B) 25 cm C) 50 cm D) 1 m.MCQ

Answer and short solution

Short solution: Correct option B. Standard value D=25 cm.

Final answer: Correct option B. Standard value D=25 cm.

24If a medium has larger refractive index, its optical density is A) smaller B) larger C) zero D) unrelated.MCQ

Answer and short solution

Short solution: Correct option B. Larger refractive index means greater optical density.

Final answer: Correct option B. Larger refractive index means greater optical density.

25For a plane mirror, magnification is A) +1 B) -1 C) 0 D) infinite.MCQ

Answer and short solution

Short solution: Correct option A. Image is virtual, upright and same size.

Final answer: Correct option A. Image is virtual, upright and same size.

26A diamond sparkles because of A) low refractive index B) high critical angle C) repeated TIR D) absence of refraction.MCQ

Answer and short solution

Short solution: Correct option C. Its high refractive index gives a small critical angle and repeated TIR.

Final answer: Correct option C. Its high refractive index gives a small critical angle and repeated TIR.

27In lens maker formula, radii must be used with A) no sign B) proper sign convention C) only positive sign D) only centimetre unit.MCQ

Answer and short solution

Short solution: Correct option B. Incorrect signs change focal length and lens power.

Final answer: Correct option B. Incorrect signs change focal length and lens power.

28Which instrument uses a large objective focal length for high angular magnification? A) astronomical telescope B) simple microscope C) magnifying glass D) spectrometer slit.MCQ

Answer and short solution

Short solution: Correct option A. Telescope magnification is f0/fe.

Final answer: Correct option A. Telescope magnification is f0/fe.

29If object distance equals focal length for a convex lens, image is formed A) at focus B) at 2F C) at infinity D) between O and F.MCQ

Answer and short solution

Short solution: Correct option C. Emergent rays are parallel.

Final answer: Correct option C. Emergent rays are parallel.

30A real image by a lens can be obtained on a screen because rays A) actually meet B) only appear to meet C) are absorbed D) stop at focus.MCQ

Answer and short solution

Short solution: Correct option A. Actual convergence produces a screen image.

Final answer: Correct option A. Actual convergence produces a screen image.

31For a spherical mirror, radius of curvature R and focal length f satisfy A) R=f B) R=2f C) R=f/2 D) R=0.MCQ

Answer and short solution

Short solution: Correct option B. The focus lies halfway between pole and centre.

Final answer: Correct option B. The focus lies halfway between pole and centre.

32In a compound microscope, eyepiece works like A) simple microscope B) plane mirror C) objective mirror D) prism only.MCQ

Answer and short solution

Short solution: Correct option A. It magnifies the intermediate image angularly.

Final answer: Correct option A. It magnifies the intermediate image angularly.

33If f0=120 cm and fe=6 cm for a telescope, normal magnification magnitude is A) 10 B) 20 C) 30 D) 126.MCQ

Answer and short solution

Short solution: Correct option B. M=f0/fe=20.

Final answer: Correct option B. M=f0/fe=20.

34The emergent ray at critical angle is A) along normal B) along interface C) undeviated D) reflected only.MCQ

Answer and short solution

Short solution: Correct option B. The refracted angle is 90 degrees.

Final answer: Correct option B. The refracted angle is 90 degrees.

35A concave mirror used for shaving forms an enlarged image when face is A) beyond C B) at C C) between P and F D) at infinity.MCQ

Answer and short solution

Short solution: Correct option C. Inside focal length, a concave mirror gives virtual enlarged image.

Final answer: Correct option C. Inside focal length, a concave mirror gives virtual enlarged image.

36For lenses in contact, equivalent power is A) product of powers B) sum of powers C) difference always D) reciprocal sum of focal lengths only.MCQ

Answer and short solution

Short solution: Correct option B. Powers add directly for thin lenses in contact.

Final answer: Correct option B. Powers add directly for thin lenses in contact.

37A normal eye focuses image on A) cornea B) retina C) pupil D) iris.MCQ

Answer and short solution

Short solution: Correct option B. The retina is the light-sensitive screen.

Final answer: Correct option B. The retina is the light-sensitive screen.

38Hypermetropia is corrected by A) concave lens B) convex lens C) plane glass D) opaque screen.MCQ

Answer and short solution

Short solution: Correct option B. A convex lens adds converging power for near objects.

Final answer: Correct option B. A convex lens adds converging power for near objects.

39For a real object and convex lens, a real image has v A) positive B) negative C) zero D) undefined.MCQ

Answer and short solution

Short solution: Correct option A. With Cartesian convention, real image on the other side has positive v.

Final answer: Correct option A. With Cartesian convention, real image on the other side has positive v.

40Angular magnification of a simple microscope for final image at infinity is A) f/D B) D/f C) 1+D/f D) f+D.MCQ

Answer and short solution

Short solution: Correct option B. Relaxed-eye magnifying power is D/f.

Final answer: Correct option B. Relaxed-eye magnifying power is D/f.

41Which colour deviates most in ordinary glass prism? A) red B) yellow C) violet D) all same.MCQ

Answer and short solution

Short solution: Correct option C. Violet has higher refractive index and deviates more.

Final answer: Correct option C. Violet has higher refractive index and deviates more.

42The apparent depth of a coin in water is A) more than real depth B) less than real depth C) equal always D) zero.MCQ

Answer and short solution

Short solution: Correct option B. Viewed from air, the coin appears raised.

Final answer: Correct option B. Viewed from air, the coin appears raised.

43A ray through optical centre of a thin lens is treated as A) undeviated B) reflected back C) absorbed D) split equally.MCQ

Answer and short solution

Short solution: Correct option A. In the thin-lens approximation it passes undeviated.

Final answer: Correct option A. In the thin-lens approximation it passes undeviated.

44Which defect is age-related loss of accommodation? A) myopia B) presbyopia C) cataract D) colour blindness.MCQ

Answer and short solution

Short solution: Correct option B. Presbyopia is reduced accommodation with age.

Final answer: Correct option B. Presbyopia is reduced accommodation with age.

45A fibre with n1=1.50 and n2=1.40 has NA approximately A) 0.54 B) 0.10 C) 2.90 D) 1.45.MCQ

Answer and short solution

Short solution: Correct option A. NA=sqrt(2.25-1.96)=sqrt(0.29)=0.54.

Final answer: Correct option A. NA=sqrt(2.25-1.96)=sqrt(0.29)=0.54.

46A mirror formula calculation gives m=-1.5. Image is A) upright diminished B) inverted enlarged C) upright enlarged D) virtual only.MCQ

Answer and short solution

Short solution: Correct option B. Negative means inverted and magnitude greater than 1 means enlarged.

Final answer: Correct option B. Negative means inverted and magnitude greater than 1 means enlarged.

47The final image in normal astronomical telescope is A) at infinity B) at objective C) on primary mirror D) inside prism.MCQ

Answer and short solution

Short solution: Correct option A. Normal adjustment is for relaxed viewing.

Final answer: Correct option A. Normal adjustment is for relaxed viewing.

48A reflecting telescope primarily uses A) mirrors B) only prisms C) only concave lenses D) cylindrical lenses.MCQ

Answer and short solution

Short solution: Correct option A. The objective is a concave mirror.

Final answer: Correct option A. The objective is a concave mirror.

49The lens formula for a thin lens is A) 1/f=1/v+1/u B) 1/f=1/v-1/u C) f=u+v D) P=f.MCQ

Answer and short solution

Short solution: Correct option B. That is the Cartesian thin-lens formula.

Final answer: Correct option B. That is the Cartesian thin-lens formula.

50If power is negative, the lens is generally A) converging B) diverging C) plane only D) cylindrical only.MCQ

Answer and short solution

Short solution: Correct option B. A negative spherical power corresponds to a concave lens in air.

Final answer: Correct option B. A negative spherical power corresponds to a concave lens in air.

51Prism deviation depends on A) angle of incidence B) prism angle C) refractive index D) all of these.MCQ

Answer and short solution

Short solution: Correct option D. All three affect deviation.

Final answer: Correct option D. All three affect deviation.

52For small telescope resolution, diffraction limit angle is proportional to A) lambda/D B) D/lambda C) f0/fe D) fe/f0.MCQ

Answer and short solution

Short solution: Correct option A. Minimum resolvable angle is about 1.22 lambda/D.

Final answer: Correct option A. Minimum resolvable angle is about 1.22 lambda/D.

53A biconvex lens in air usually has A) positive focal length B) negative focal length C) infinite power D) zero power.MCQ

Answer and short solution

Short solution: Correct option A. It converges paraxial rays in air.

Final answer: Correct option A. It converges paraxial rays in air.

54An object at infinity before a concave mirror forms image A) at focus B) at centre C) behind mirror D) at pole.MCQ

Answer and short solution

Short solution: Correct option A. Parallel rays converge at the focus.

Final answer: Correct option A. Parallel rays converge at the focus.

55In a Cassegrain telescope, the secondary mirror is used to A) fold the light path B) remove all diffraction C) absorb light D) make prism spectrum.MCQ

Answer and short solution

Short solution: Correct option A. It folds and lengthens the effective optical path.

Final answer: Correct option A. It folds and lengthens the effective optical path.

56A ray incident normally on a spherical surface has incidence angle A) 0 degrees B) 45 degrees C) 90 degrees D) critical angle.MCQ

Answer and short solution

Short solution: Correct option A. Normal incidence means zero angle with the normal.

Final answer: Correct option A. Normal incidence means zero angle with the normal.

57Which quantity remains unchanged when light enters another medium? A) frequency B) wavelength C) speed D) direction always.MCQ

Answer and short solution

Short solution: Correct option A. Frequency is fixed by the source.

Final answer: Correct option A. Frequency is fixed by the source.

58A concave mirror can form a virtual image when object is A) beyond C B) at C C) between P and F D) at infinity.MCQ

Answer and short solution

Short solution: Correct option C. The reflected rays diverge and appear to meet behind the mirror.

Final answer: Correct option C. The reflected rays diverge and appear to meet behind the mirror.

59A lens with P=0 behaves as A) afocal combination B) high-power lens C) prism D) mirror.MCQ

Answer and short solution

Short solution: Correct option A. Zero power means infinite equivalent focal length in the thin approximation.

Final answer: Correct option A. Zero power means infinite equivalent focal length in the thin approximation.

60The full acceptance cone of a fibre is A) theta_a B) 2 theta_a C) C/2 D) 90-C.MCQ

Answer and short solution

Short solution: Correct option B. theta_a is the acceptance half-angle.

Final answer: Correct option B. theta_a is the acceptance half-angle.

61A virtual image by a plane mirror is A) behind mirror B) on screen C) always inverted D) at focus.MCQ

Answer and short solution

Short solution: Correct option A. It appears as far behind as the object is in front.

Final answer: Correct option A. It appears as far behind as the object is in front.

62The aperture of a microscope objective affects A) numerical aperture B) electric charge C) object distance only D) colour blindness.MCQ

Answer and short solution

Short solution: Correct option A. Numerical aperture depends on aperture angle and medium.

Final answer: Correct option A. Numerical aperture depends on aperture angle and medium.

63In ray optics, paraxial rays are A) close to principal axis B) only red rays C) non-refracting D) circular rays.MCQ

Answer and short solution

Short solution: Correct option A. Paraxial approximation keeps angles small.

Final answer: Correct option A. Paraxial approximation keeps angles small.

64The image distance in a convex mirror is usually A) behind the mirror B) in front as real image C) at infinity only D) negative always.MCQ

Answer and short solution

Short solution: Correct option A. The image is virtual behind the mirror.

Final answer: Correct option A. The image is virtual behind the mirror.

65If two lenses of powers +3 D and -3 D are in contact, total power is A) 6 D B) -6 D C) 0 D D) 9 D.MCQ

Answer and short solution

Short solution: Correct option C. Equal and opposite powers cancel.

Final answer: Correct option C. Equal and opposite powers cancel.

66Which formula gives apparent shift through a slab at normal viewing? A) t(1-1/mu) B) mu/t C) t+mu D) 1/f.MCQ

Answer and short solution

Short solution: Correct option A. The shift is real thickness minus apparent thickness.

Final answer: Correct option A. The shift is real thickness minus apparent thickness.

67In a prism, white light splits because refractive index is A) wavelength dependent B) always zero C) same for all colours D) absent.MCQ

Answer and short solution

Short solution: Correct option A. Different wavelengths travel with different speeds.

Final answer: Correct option A. Different wavelengths travel with different speeds.

68A student sees a blurred distant board but reads nearby text clearly. The likely defect is A) myopia B) hypermetropia C) cataract only D) presbyopia only.MCQ

Answer and short solution

Short solution: Correct option A. Myopia affects distant vision.

Final answer: Correct option A. Myopia affects distant vision.

69A student sees nearby text blurred but distant objects clearly. The likely correction is A) convex lens B) concave lens C) plane mirror D) diverging mirror.MCQ

Answer and short solution

Short solution: Correct option A. Hypermetropia needs additional converging power.

Final answer: Correct option A. Hypermetropia needs additional converging power.

70An eyepiece of a telescope has smaller focal length to make magnification A) larger B) smaller C) zero D) independent.MCQ

Answer and short solution

Short solution: Correct option A. M=f0/fe increases when fe decreases.

Final answer: Correct option A. M=f0/fe increases when fe decreases.

71If refractive index of core equals cladding, fibre guidance by TIR is A) lost B) improved C) unchanged D) infinite.MCQ

Answer and short solution

Short solution: Correct option A. Without index contrast, core-cladding TIR cannot guide rays.

Final answer: Correct option A. Without index contrast, core-cladding TIR cannot guide rays.

72The formula sin C=n2/n1 is valid when A) n1>n2 B) n1<n2 C) n1=n2 only D) no refraction occurs.MCQ

Answer and short solution

Short solution: Correct option A. Critical angle exists only from denser to rarer medium.

Final answer: Correct option A. Critical angle exists only from denser to rarer medium.

73A real inverted image formed by a concave mirror has magnification sign A) positive B) negative C) zero D) undefined.MCQ

Answer and short solution

Short solution: Correct option B. Inverted images have negative magnification.

Final answer: Correct option B. Inverted images have negative magnification.

74Which device commonly uses total internal reflection to transmit signals? A) optical fibre B) plane mirror only C) eye lens D) concave lens.MCQ

Answer and short solution

Short solution: Correct option A. Fibres guide light pulses by repeated TIR.

Final answer: Correct option A. Fibres guide light pulses by repeated TIR.

75A convex lens used as a magnifying glass forms the final image A) virtual B) real only C) on retina outside eye D) behind lens as screen image always.MCQ

Answer and short solution

Short solution: Correct option A. The magnifier produces a virtual enlarged image for the eye.

Final answer: Correct option A. The magnifier produces a virtual enlarged image for the eye.

17. JEE Main Question Bank: 50 Numericals

JEE Main-style numerical practice with short answer checks.

1A concave mirror has f=-12 cm and object distance u=-36 cm. Calculate v and m.Numerical

Answer and short solution

Short solution: 1/v=-1/12+1/36=-1/18, so v=-18 cm and m=-v/u=-0.5.

Final answer: 1/v=-1/12+1/36=-1/18, so v=-18 cm and m=-v/u=-0.5.

2A convex mirror has f=24 cm and object distance u=-36 cm. Find the image distance.Numerical

Answer and short solution

Short solution: 1/v=1/f-1/u=1/24+1/36=5/72, so v=14.4 cm behind the mirror.

Final answer: 1/v=1/f-1/u=1/24+1/36=5/72, so v=14.4 cm behind the mirror.

3A convex lens has f=15 cm and u=-30 cm. Find v.Numerical

Answer and short solution

Short solution: 1/v=1/f+1/u=1/15-1/30=1/30, so v=30 cm.

Final answer: 1/v=1/f+1/u=1/15-1/30=1/30, so v=30 cm.

4A concave lens has f=-20 cm and u=-40 cm. Find v.Numerical

Answer and short solution

Short solution: 1/v=1/f+1/u=-1/20-1/40=-3/40, so v=-13.3 cm.

Final answer: 1/v=1/f+1/u=-1/20-1/40=-3/40, so v=-13.3 cm.

5A lens of focal length 40 cm is placed in contact with a lens of focal length -60 cm. Find equivalent power.Numerical

Answer and short solution

Short solution: P=1/0.40+1/(-0.60)=2.5-1.667=0.833 D.

Final answer: P=1/0.40+1/(-0.60)=2.5-1.667=0.833 D.

6Two lenses of powers +4 D and +6 D are separated by 10 cm. Find equivalent power.Numerical

Answer and short solution

Short solution: P=4+6-0.10(4)(6)=7.6 D.

Final answer: P=4+6-0.10(4)(6)=7.6 D.

7A glass slab of thickness 18 cm and refractive index 1.5 is viewed normally. Find apparent shift.Numerical

Answer and short solution

Short solution: Shift=18(1-1/1.5)=6 cm.

Final answer: Shift=18(1-1/1.5)=6 cm.

8A water tank appears 30 cm deep when viewed normally from air. If mu=4/3, find real depth.Numerical

Answer and short solution

Short solution: Real depth=mu times apparent depth=(4/3)(30)=40 cm.

Final answer: Real depth=mu times apparent depth=(4/3)(30)=40 cm.

9A prism has A=60 degrees and minimum deviation 30 degrees. Find refractive index.Numerical

Answer and short solution

Short solution: mu=sin45/sin30=1.414.

Final answer: mu=sin45/sin30=1.414.

10Find the critical angle for a medium of refractive index 2 in air.Numerical

Answer and short solution

Short solution: sin C=1/2, so C=30 degrees.

Final answer: sin C=1/2, so C=30 degrees.

11For a fibre with n1=1.48 and n2=1.46, calculate NA in air.Numerical

Answer and short solution

Short solution: NA=sqrt(1.48^2-1.46^2)=sqrt(0.0588)=0.242.

Final answer: NA=sqrt(1.48^2-1.46^2)=sqrt(0.0588)=0.242.

12A myopic eye has far point 2 m. Find correcting lens power.Numerical

Answer and short solution

Short solution: f=-2 m, so P=-0.5 D.

Final answer: f=-2 m, so P=-0.5 D.

13A hypermetropic eye has near point 50 cm. Find power for reading at 25 cm.Numerical

Answer and short solution

Short solution: u=-0.25 m, v=-0.50 m, so 1/f=-2+4=2 and P=+2 D.

Final answer: u=-0.25 m, v=-0.50 m, so 1/f=-2+4=2 and P=+2 D.

14A simple microscope has f=10 cm. Find magnifying power for final image at D.Numerical

Answer and short solution

Short solution: M=1+D/f=1+25/10=3.5.

Final answer: M=1+D/f=1+25/10=3.5.

15A compound microscope has L=16 cm, f0=0.8 cm, fe=4 cm. Estimate normal magnification.Numerical

Answer and short solution

Short solution: M=(L/f0)(D/fe)=(16/0.8)(25/4)=125.

Final answer: M=(L/f0)(D/fe)=(16/0.8)(25/4)=125.

16A telescope has f0=150 cm and fe=5 cm. Find normal magnifying power and length.Numerical

Answer and short solution

Short solution: M=30 and L=f0+fe=155 cm.

Final answer: M=30 and L=f0+fe=155 cm.

17A telescope objective diameter is 10 cm and wavelength is 500 nm. Estimate angular resolution.Numerical

Answer and short solution

Short solution: theta=1.22 lambda/D=1.22(500e-9)/0.10=6.1e-6 rad.

Final answer: theta=1.22 lambda/D=1.22(500e-9)/0.10=6.1e-6 rad.

18A lens has power -2.5 D. Find focal length in cm.Numerical

Answer and short solution

Short solution: f=1/P=-0.4 m=-40 cm.

Final answer: f=1/P=-0.4 m=-40 cm.

19An object of height 2 cm gives magnification -3 in a concave mirror. Find image height.Numerical

Answer and short solution

Short solution: hi=m ho=-3(2)=-6 cm, so image is 6 cm inverted.

Final answer: hi=m ho=-3(2)=-6 cm, so image is 6 cm inverted.

20A plane mirror is moved 5 cm towards an object. How much does the image move relative to the object?Numerical

Answer and short solution

Short solution: The image shifts 10 cm relative to the object because image distance changes twice the mirror displacement.

Final answer: The image shifts 10 cm relative to the object because image distance changes twice the mirror displacement.

21A ray in glass of n=1.5 is incident at 45 degrees on glass-air boundary. Will TIR occur?Numerical

Answer and short solution

Short solution: Critical angle is 41.8 degrees; since 45 degrees is larger, TIR occurs.

Final answer: Critical angle is 41.8 degrees; since 45 degrees is larger, TIR occurs.

22For crown glass mu=1.5 and equiconvex lens R=20 cm, find focal length in air.Numerical

Answer and short solution

Short solution: 1/f=(0.5)(1/20-(-1/20))=0.05, so f=20 cm.

Final answer: 1/f=(0.5)(1/20-(-1/20))=0.05, so f=20 cm.

23A convex lens of f=20 cm forms image at v=60 cm. Find object distance.Numerical

Answer and short solution

Short solution: 1/u=1/v-1/f=1/60-1/20=-2/60, so u=-30 cm.

Final answer: 1/u=1/v-1/f=1/60-1/20=-2/60, so u=-30 cm.

24A concave mirror forms image at v=-40 cm for u=-20 cm. Find focal length.Numerical

Answer and short solution

Short solution: 1/f=1/v+1/u=-1/40-1/20=-3/40, so f=-13.3 cm.

Final answer: 1/f=1/v+1/u=-1/40-1/20=-3/40, so f=-13.3 cm.

25A prism has refractive index 1.5 and A=60 degrees. Find minimum deviation approximately.Numerical

Answer and short solution

Short solution: sin((A+dm)/2)=mu sin(A/2)=0.75, so (A+dm)/2=48.6 degrees and dm=37.2 degrees.

Final answer: sin((A+dm)/2)=mu sin(A/2)=0.75, so (A+dm)/2=48.6 degrees and dm=37.2 degrees.

26Find acceptance angle in air for NA=0.50.Numerical

Answer and short solution

Short solution: theta_a=sin inverse 0.50=30 degrees.

Final answer: theta_a=sin inverse 0.50=30 degrees.

27A telescope has magnification 25 and eyepiece focal length 4 cm. Find objective focal length.Numerical

Answer and short solution

Short solution: f0=M fe=25(4)=100 cm.

Final answer: f0=M fe=25(4)=100 cm.

28A simple microscope must give M infinity=8. Find focal length.Numerical

Answer and short solution

Short solution: f=D/M=25/8=3.125 cm.

Final answer: f=D/M=25/8=3.125 cm.

29Two lenses +10 D and -4 D are in contact. Find equivalent focal length.Numerical

Answer and short solution

Short solution: P=6 D, so f=1/6 m=16.7 cm.

Final answer: P=6 D, so f=1/6 m=16.7 cm.

30A lens combination has P=+2 D. If one lens is +5 D, find the other lens power in contact.Numerical

Answer and short solution

Short solution: P2=2-5=-3 D.

Final answer: P2=2-5=-3 D.

31A slab raises a mark by 2 mm. If slab thickness is 6 mm, find refractive index.Numerical

Answer and short solution

Short solution: 2=6(1-1/mu), so 1/mu=2/3 and mu=1.5.

Final answer: 2=6(1-1/mu), so 1/mu=2/3 and mu=1.5.

32A microscope objective gives magnification 20 and eyepiece gives angular magnification 10. Find total magnification.Numerical

Answer and short solution

Short solution: Total magnification is 20x10=200.

Final answer: Total magnification is 20x10=200.

33A galvanometer mirror rotates by 2 degrees. Screen is 2 m away. Estimate spot shift.Numerical

Answer and short solution

Short solution: Reflected ray turns 4 degrees, so shift=2 tan4 degrees=0.14 m approximately.

Final answer: Reflected ray turns 4 degrees, so shift=2 tan4 degrees=0.14 m approximately.

34An object is at 30 cm before a convex lens and image at 60 cm. Find magnification.Numerical

Answer and short solution

Short solution: m=v/u=60/(-30)=-2; image is inverted and twice size.

Final answer: m=v/u=60/(-30)=-2; image is inverted and twice size.

35A convex mirror gives m=1/3 for object distance 30 cm. Find image distance using m=-v/u.Numerical

Answer and short solution

Short solution: With u=-30 cm and m=1/3, v=-m u=10 cm behind the mirror.

Final answer: With u=-30 cm and m=1/3, v=-m u=10 cm behind the mirror.

36For a lens of f=25 cm, calculate power.Numerical

Answer and short solution

Short solution: f=0.25 m, so P=+4 D for a converging lens.

Final answer: f=0.25 m, so P=+4 D for a converging lens.

37For final image at D in a telescope with f0=100 cm and fe=5 cm, find magnification magnitude.Numerical

Answer and short solution

Short solution: M=(f0/fe)(1+fe/D)=20(1+5/25)=24.

Final answer: M=(f0/fe)(1+fe/D)=20(1+5/25)=24.

38A microscope has f0=1 cm, fe=5 cm and L=20 cm. Find normal magnification.Numerical

Answer and short solution

Short solution: M=(20/1)(25/5)=100.

Final answer: M=(20/1)(25/5)=100.

39A ray goes from air to glass at i=30 degrees and n=1.5. Find r.Numerical

Answer and short solution

Short solution: sin r=sin30/1.5=1/3, so r=19.5 degrees.

Final answer: sin r=sin30/1.5=1/3, so r=19.5 degrees.

40A point object in water appears 12 cm deep from air. If mu=4/3, find actual depth.Numerical

Answer and short solution

Short solution: Real depth=mu x apparent depth=16 cm.

Final answer: Real depth=mu x apparent depth=16 cm.

41The angular diameter of the Moon is 0.009 rad. A telescope objective has f0=2 m. Find objective image diameter.Numerical

Answer and short solution

Short solution: Image diameter=f0 theta=2(0.009)=0.018 m=1.8 cm.

Final answer: Image diameter=f0 theta=2(0.009)=0.018 m=1.8 cm.

42A lens power changes from +5 D to +3 D after immersion. State the change in focal length.Numerical

Answer and short solution

Short solution: Focal length changes from 0.20 m to 0.333 m, so it increases.

Final answer: Focal length changes from 0.20 m to 0.333 m, so it increases.

43Find the critical angle for glass of mu=1.6 in air.Numerical

Answer and short solution

Short solution: sin C=1/1.6=0.625, so C=38.7 degrees.

Final answer: sin C=1/1.6=0.625, so C=38.7 degrees.

44A fibre has NA=0.22 in air. Find full acceptance cone angle.Numerical

Answer and short solution

Short solution: Half-angle=sin inverse 0.22=12.7 degrees, so full cone is about 25.4 degrees.

Final answer: Half-angle=sin inverse 0.22=12.7 degrees, so full cone is about 25.4 degrees.

45A convex lens gives real image at 2F. What is magnification?Numerical

Answer and short solution

Short solution: Object and image are at 2F, so magnification is -1.

Final answer: Object and image are at 2F, so magnification is -1.

46Object is placed at focus of concave mirror. Where is image?Numerical

Answer and short solution

Short solution: Reflected rays become parallel, so the image is at infinity.

Final answer: Reflected rays become parallel, so the image is at infinity.

47A diverging lens of power -5 D is combined with +8 D lens. Find net power and type.Numerical

Answer and short solution

Short solution: Net power=+3 D, so the combination is converging.

Final answer: Net power=+3 D, so the combination is converging.

48A lens has f=50 cm. What is its power and what error occurs if 50 is used directly?Numerical

Answer and short solution

Short solution: Correct power is 1/0.50=2 D; using 50 gives 0.02, which is wrong by factor 100.

Final answer: Correct power is 1/0.50=2 D; using 50 gives 0.02, which is wrong by factor 100.

49For a telescope, if objective aperture changes from 5 cm to 15 cm, how does resolving power change?Numerical

Answer and short solution

Short solution: Resolving power triples because it is proportional to aperture.

Final answer: Resolving power triples because it is proportional to aperture.

50A prism has i=50 degrees, e=45 degrees and A=60 degrees. Find deviation.Numerical

Answer and short solution

Short solution: delta=i+e-A=50+45-60=35 degrees.

Final answer: delta=i+e-A=50+45-60=35 degrees.

18. JEE Advanced Question Bank: 40 Multi-Step Questions

Original multi-step questions for deeper Ray Optics reasoning.

1A concave mirror f=-20 cm and a convex lens f=+10 cm are separated by 30 cm. An object is 40 cm before the mirror. Describe a sequential method to locate the final image after reflection and refraction.Advanced

Answer and short solution

Short solution: First use the mirror formula to get the mirror image, then treat that image as the object for the lens with the correct separation and signs. Sequential imaging is safer than combining unlike elements.

Final answer: First use the mirror formula to get the mirror image, then treat that image as the object for the lens with the correct separation and signs. Sequential imaging is safer than combining unlike elements.

2A biconvex glass lens is immersed in a liquid whose refractive index equals that of glass. What happens to its focal length and why?Advanced

Answer and short solution

Short solution: Relative refractive index becomes 1, so lens power becomes zero and focal length tends to infinity.

Final answer: Relative refractive index becomes 1, so lens power becomes zero and focal length tends to infinity.

3Two thin lenses of focal lengths 20 cm and -30 cm are separated by 10 cm. Find equivalent power and comment on principal planes.Advanced

Answer and short solution

Short solution: P=5-3.333-0.10(5)(-3.333)=3.333 D. EFL=30 cm, but principal planes shift, so EFL alone does not locate all images.

Final answer: P=5-3.333-0.10(5)(-3.333)=3.333 D. EFL=30 cm, but principal planes shift, so EFL alone does not locate all images.

4A prism is adjusted so that a ray just undergoes TIR at the second face. Which two equations connect the first refraction and the critical angle?Advanced

Answer and short solution

Short solution: Use r1+r2=A and r2=C with sin C=1/mu; then use sin i=mu sin r1 at the first face.

Final answer: Use r1+r2=A and r2=C with sin C=1/mu; then use sin i=mu sin r1 at the first face.

5A Cassegrain telescope uses a concave primary and convex secondary. Why can the tube be shorter than the effective focal length?Advanced

Answer and short solution

Short solution: The secondary reflects the converging beam back through the primary and increases the effective focal length while folding the path.

Final answer: The secondary reflects the converging beam back through the primary and increases the effective focal length while folding the path.

6A microscope objective has finite aperture. Explain how increasing numerical aperture affects resolution and brightness.Advanced

Answer and short solution

Short solution: Higher NA decreases d=0.61 lambda/NA and accepts a wider cone of rays, improving resolution and brightness.

Final answer: Higher NA decreases d=0.61 lambda/NA and accepts a wider cone of rays, improving resolution and brightness.

7An astronomical telescope is used for final image at the near point. Why is magnification larger than in normal adjustment?Advanced

Answer and short solution

Short solution: The eyepiece forms a final virtual image at D, giving the factor (1+fe/D) beyond f0/fe.

Final answer: The eyepiece forms a final virtual image at D, giving the factor (1+fe/D) beyond f0/fe.

8A student uses prism minimum-deviation formula when i is not equal to e. Identify the error.Advanced

Answer and short solution

Short solution: The formula mu=sin((A+dm)/2)/sin(A/2) assumes symmetric path only at minimum deviation; it is invalid for arbitrary incidence.

Final answer: The formula mu=sin((A+dm)/2)/sin(A/2) assumes symmetric path only at minimum deviation; it is invalid for arbitrary incidence.

9A converging lens forms a real image. The lens is moved slightly toward the object while object-screen distance remains fixed. Explain the two-position method.Advanced

Answer and short solution

Short solution: For fixed object-screen distance greater than 4f, two conjugate lens positions exist and their separation gives f=(D^2-d^2)/(4D).

Final answer: For fixed object-screen distance greater than 4f, two conjugate lens positions exist and their separation gives f=(D^2-d^2)/(4D).

10A plane mirror is placed behind a convex lens. When does the object coincide with its final image?Advanced

Answer and short solution

Short solution: In autocollimation, coincidence occurs when the object is at the focal plane of the lens; rays return along their original paths.

Final answer: In autocollimation, coincidence occurs when the object is at the focal plane of the lens; rays return along their original paths.

11Why is chromatic aberration absent in a mirror but present in a simple refracting lens?Advanced

Answer and short solution

Short solution: Reflection angle is independent of wavelength, while refractive index of lens material varies with wavelength.

Final answer: Reflection angle is independent of wavelength, while refractive index of lens material varies with wavelength.

12A fibre is placed in water instead of air. How does the acceptance angle change?Advanced

Answer and short solution

Short solution: The external medium index n0 enters n0 sin theta_a=NA, so theta_a decreases when n0 increases for the same fibre.

Final answer: The external medium index n0 enters n0 sin theta_a=NA, so theta_a decreases when n0 increases for the same fibre.

13In a compound microscope, why is the intermediate image placed near the first focal plane of the eyepiece?Advanced

Answer and short solution

Short solution: That lets the eyepiece act as a magnifier and send rays out parallel for relaxed viewing.

Final answer: That lets the eyepiece act as a magnifier and send rays out parallel for relaxed viewing.

14A telescope objective diameter is doubled and wavelength is halved. What happens to diffraction-limited angular resolution?Advanced

Answer and short solution

Short solution: Minimum resolvable angle is proportional to lambda/D, so it becomes one fourth.

Final answer: Minimum resolvable angle is proportional to lambda/D, so it becomes one fourth.

15For a concave mirror, derive the range of v when the object moves from infinity to C.Advanced

Answer and short solution

Short solution: As u changes from negative infinity to R, v moves from F to C, giving real inverted images between F and C.

Final answer: As u changes from negative infinity to R, v moves from F to C, giving real inverted images between F and C.

16A convex lens and concave lens in contact form an afocal system. What relation must their focal lengths satisfy?Advanced

Answer and short solution

Short solution: Their powers must add to zero: 1/f1+1/f2=0, so f2=-f1.

Final answer: Their powers must add to zero: 1/f1+1/f2=0, so f2=-f1.

17Explain why a real image can act as a virtual object for a second optical element.Advanced

Answer and short solution

Short solution: If rays are converging toward a point beyond the second element, that point is a virtual object for the second element.

Final answer: If rays are converging toward a point beyond the second element, that point is a virtual object for the second element.

18A ray diagram shows an upright enlarged image by a concave mirror. What must be the object's location?Advanced

Answer and short solution

Short solution: The object must lie between the pole and focus of the concave mirror.

Final answer: The object must lie between the pole and focus of the concave mirror.

19Why does the paraxial approximation fail for large aperture spherical lenses?Advanced

Answer and short solution

Short solution: Marginal rays focus at different points due to spherical aberration, so simple focal length formulas become approximate.

Final answer: Marginal rays focus at different points due to spherical aberration, so simple focal length formulas become approximate.

20A lens maker changes only the second radius R2 while keeping material fixed. Explain how sign of R2 affects power.Advanced

Answer and short solution

Short solution: The lens maker term is (1/R1-1/R2); changing the sign or magnitude of R2 directly changes curvature power.

Final answer: The lens maker term is (1/R1-1/R2); changing the sign or magnitude of R2 directly changes curvature power.

21A ray passes from glass to air at exactly the critical angle. What is the energy path in an ideal ray model?Advanced

Answer and short solution

Short solution: The refracted ray grazes the interface; above that angle the ray is totally internally reflected, with evanescent field beyond the boundary.

Final answer: The refracted ray grazes the interface; above that angle the ray is totally internally reflected, with evanescent field beyond the boundary.

22How does increasing eyepiece focal length affect telescope length and magnification in normal adjustment?Advanced

Answer and short solution

Short solution: Length increases by the eyepiece focal length but magnification decreases because M=f0/fe.

Final answer: Length increases by the eyepiece focal length but magnification decreases because M=f0/fe.

23A microscope has high magnification but poor resolution. Which optical parameter should be improved?Advanced

Answer and short solution

Short solution: Increase numerical aperture and use shorter wavelength light; magnification alone does not guarantee resolution.

Final answer: Increase numerical aperture and use shorter wavelength light; magnification alone does not guarantee resolution.

24Two prisms of different materials are combined for no net deviation but dispersion remains. What is being separated?Advanced

Answer and short solution

Short solution: Mean deviation can be cancelled while angular dispersion remains because dispersive powers differ.

Final answer: Mean deviation can be cancelled while angular dispersion remains because dispersive powers differ.

25If a virtual image of one lens lies on the same side as the next lens, how should the sign of object distance be handled?Advanced

Answer and short solution

Short solution: Use the Cartesian sign convention from the second lens; the sign depends on whether the object point is on incident or emergent side for that lens.

Final answer: Use the Cartesian sign convention from the second lens; the sign depends on whether the object point is on incident or emergent side for that lens.

26A spherical refracting surface becomes plane. Which term vanishes in the surface refraction formula?Advanced

Answer and short solution

Short solution: As R tends to infinity, (mu2-mu1)/R becomes zero, reducing the equation to plane refraction relation.

Final answer: As R tends to infinity, (mu2-mu1)/R becomes zero, reducing the equation to plane refraction relation.

27A student says telescope objective should have short focal length like a microscope objective. Correct the statement.Advanced

Answer and short solution

Short solution: A telescope objective usually has large focal length for high angular magnification and large aperture for light gathering and resolution.

Final answer: A telescope objective usually has large focal length for high angular magnification and large aperture for light gathering and resolution.

28Why does a short focal length magnifier have practical limits?Advanced

Answer and short solution

Short solution: Very short focal length gives small working distance and strong aberrations, making use uncomfortable and less sharp.

Final answer: Very short focal length gives small working distance and strong aberrations, making use uncomfortable and less sharp.

29A concave mirror and convex lens both converge light. Why are their sign conventions different in formulas?Advanced

Answer and short solution

Short solution: Reflection and refraction formulas are derived with different image-side geometry; using the correct Cartesian formula prevents sign errors.

Final answer: Reflection and refraction formulas are derived with different image-side geometry; using the correct Cartesian formula prevents sign errors.

30For a given telescope aperture, why does blue light theoretically resolve closer stars than red light?Advanced

Answer and short solution

Short solution: Blue light has shorter wavelength, reducing the Rayleigh diffraction limit.

Final answer: Blue light has shorter wavelength, reducing the Rayleigh diffraction limit.

31In fibre optics, why can bending losses occur even if n1>n2?Advanced

Answer and short solution

Short solution: Sharp bends reduce effective incidence angle at the boundary, allowing leakage instead of TIR.

Final answer: Sharp bends reduce effective incidence angle at the boundary, allowing leakage instead of TIR.

32What is the difference between angular magnification and linear magnification in a magnifier?Advanced

Answer and short solution

Short solution: Angular magnification compares visual angles at the eye; linear magnification compares image and object sizes and may be undefined for image at infinity.

Final answer: Angular magnification compares visual angles at the eye; linear magnification compares image and object sizes and may be undefined for image at infinity.

33A virtual image is formed at the near point by a simple microscope. Derive object distance expression.Advanced

Answer and short solution

Short solution: Using v=-D in 1/f=1/v-1/u gives |u|=fD/(D+f).

Final answer: Using v=-D in 1/f=1/v-1/u gives |u|=fD/(D+f).

34A telescope is inverted for terrestrial viewing. What extra optical element may be used and what is its cost?Advanced

Answer and short solution

Short solution: An erecting lens or prism can make the image upright, but it adds length, loss, and possible aberration.

Final answer: An erecting lens or prism can make the image upright, but it adds length, loss, and possible aberration.

35A convex mirror image shifts as object approaches the mirror. Where does it remain confined?Advanced

Answer and short solution

Short solution: It remains behind the mirror between pole and focus for every real object position.

Final answer: It remains behind the mirror between pole and focus for every real object position.

36A lens has one surface plane and one convex surface. How is lens maker formula simplified?Advanced

Answer and short solution

Short solution: For the plane surface R is infinity, so its curvature term is zero; only the curved surface contributes.

Final answer: For the plane surface R is infinity, so its curvature term is zero; only the curved surface contributes.

37Why is aperture used in telescope resolving power but numerical aperture used in microscope resolving power?Advanced

Answer and short solution

Short solution: A telescope resolves distant angular sources through objective diameter; a microscope resolves nearby points through the accepted cone angle in a medium.

Final answer: A telescope resolves distant angular sources through objective diameter; a microscope resolves nearby points through the accepted cone angle in a medium.

38A student's answer gives positive power for a myopia lens. What physical check reveals the mistake?Advanced

Answer and short solution

Short solution: Myopia correction must diverge rays and form a virtual image at the far point, so the lens power must be negative.

Final answer: Myopia correction must diverge rays and form a virtual image at the far point, so the lens power must be negative.

39In a two-lens system, why can magnification be found without using equivalent focal length?Advanced

Answer and short solution

Short solution: Sequential imaging gives each lens magnification, and total magnification is the product; this also handles separated principal planes.

Final answer: Sequential imaging gives each lens magnification, and total magnification is the product; this also handles separated principal planes.

40A prism formula gives mu less than 1 for glass. What should be checked first?Advanced

Answer and short solution

Short solution: Check degree/radian use, angle values, and whether the minimum-deviation condition was actually satisfied.

Final answer: Check degree/radian use, angle values, and whether the minimum-deviation condition was actually satisfied.

19. IB Physics HL + SL

Explanation-based Ray Optics practice for IB Physics.

1Explain why refractive index can be interpreted as a ratio of speeds.Concept

Answer and short solution

Short solution: Refractive index n=c/v compares light speed in vacuum with speed in the medium.

Final answer: Refractive index n=c/v compares light speed in vacuum with speed in the medium.

2Use a labelled ray argument to explain apparent depth.Concept

Answer and short solution

Short solution: Emergent rays bend away from the normal and appear to originate from a shallower point.

Final answer: Emergent rays bend away from the normal and appear to originate from a shallower point.

3A student measures critical angle to estimate refractive index. Which relation is used?Concept

Answer and short solution

Short solution: For medium to air, n=1/sin C.

Final answer: For medium to air, n=1/sin C.

4Discuss why optical fibres require total internal reflection rather than ordinary reflection.Concept

Answer and short solution

Short solution: TIR is highly efficient at the core-cladding boundary and confines light through repeated reflections.

Final answer: TIR is highly efficient at the core-cladding boundary and confines light through repeated reflections.

5Describe one limitation of the thin-lens equation in real lenses.Concept

Answer and short solution

Short solution: It assumes paraxial rays and ignores aberrations, thickness and dispersion.

Final answer: It assumes paraxial rays and ignores aberrations, thickness and dispersion.

6Explain the physical meaning of a negative magnification.Concept

Answer and short solution

Short solution: It indicates an inverted image relative to the object.

Final answer: It indicates an inverted image relative to the object.

7Relate telescope aperture to diffraction.Concept

Answer and short solution

Short solution: A larger aperture narrows the diffraction pattern and allows smaller angular separations to be resolved.

Final answer: A larger aperture narrows the diffraction pattern and allows smaller angular separations to be resolved.

8Why is final image at infinity comfortable for many optical instruments?Concept

Answer and short solution

Short solution: The eye is relaxed because it does not need accommodation.

Final answer: The eye is relaxed because it does not need accommodation.

9Compare real and virtual images using ray convergence.Concept

Answer and short solution

Short solution: Real images form by actual convergence; virtual images form from backward extensions of diverging rays.

Final answer: Real images form by actual convergence; virtual images form from backward extensions of diverging rays.

10Explain why mirrors do not suffer chromatic aberration.Concept

Answer and short solution

Short solution: Reflection is not governed by wavelength-dependent refractive index.

Final answer: Reflection is not governed by wavelength-dependent refractive index.

11A lens is immersed in a liquid. Explain qualitatively how its power changes.Concept

Answer and short solution

Short solution: Power depends on relative refractive index, so a liquid closer to the lens index reduces power.

Final answer: Power depends on relative refractive index, so a liquid closer to the lens index reduces power.

12Describe how a compound microscope produces high magnification.Concept

Answer and short solution

Short solution: The objective gives a magnified real intermediate image and the eyepiece magnifies it angularly.

Final answer: The objective gives a magnified real intermediate image and the eyepiece magnifies it angularly.

13Why does a convex mirror increase field of view?Concept

Answer and short solution

Short solution: Diverging reflected rays compress a wide angular region into a smaller virtual image.

Final answer: Diverging reflected rays compress a wide angular region into a smaller virtual image.

14Explain one cause of poor resolution despite high magnification.Concept

Answer and short solution

Short solution: Diffraction or aberration can blur details, so enlargement alone does not reveal more information.

Final answer: Diffraction or aberration can blur details, so enlargement alone does not reveal more information.

15State a safe sign-convention habit for optics calculations.Concept

Answer and short solution

Short solution: Draw the optical element and write signed u, v and f before substitution.

Final answer: Draw the optical element and write signed u, v and f before substitution.

16Explain how a prism disperses white light.Concept

Answer and short solution

Short solution: Different wavelengths have different refractive indices, causing different deviations.

Final answer: Different wavelengths have different refractive indices, causing different deviations.

20. IGCSE Questions

Clear core and extended style questions for IGCSE Physics.

1State the law of reflection.Core

Answer and short solution

Short solution: The angle of incidence equals the angle of reflection, and the rays and normal lie in one plane.

Final answer: The angle of incidence equals the angle of reflection, and the rays and normal lie in one plane.

2Draw the image position for an object in front of a plane mirror.Core

Answer and short solution

Short solution: The image is the same distance behind the mirror as the object is in front.

Final answer: The image is the same distance behind the mirror as the object is in front.

3What happens to light speed when it enters glass from air?Core

Answer and short solution

Short solution: It decreases because glass has refractive index greater than air.

Final answer: It decreases because glass has refractive index greater than air.

4Why does a straw look bent in water?Core

Answer and short solution

Short solution: Rays refract at the water-air surface and appear to come from shifted positions.

Final answer: Rays refract at the water-air surface and appear to come from shifted positions.

5Name the lens used to correct short sight.Core

Answer and short solution

Short solution: A concave or diverging lens is used.

Final answer: A concave or diverging lens is used.

6Name the lens used to correct long sight.Core

Answer and short solution

Short solution: A convex or converging lens is used.

Final answer: A convex or converging lens is used.

7What is the focal point of a converging lens?Core

Answer and short solution

Short solution: It is the point where parallel rays meet after refraction.

Final answer: It is the point where parallel rays meet after refraction.

8Give one use of total internal reflection.Core

Answer and short solution

Short solution: Optical fibres use TIR to guide light.

Final answer: Optical fibres use TIR to guide light.

9What is meant by critical angle?Core

Answer and short solution

Short solution: It is the angle of incidence in the denser medium for which the refracted angle is 90 degrees.

Final answer: It is the angle of incidence in the denser medium for which the refracted angle is 90 degrees.

10How does a prism affect white light?Core

Answer and short solution

Short solution: It refracts and disperses white light into colours.

Final answer: It refracts and disperses white light into colours.

11What type of image does a magnifying glass produce for close viewing?Core

Answer and short solution

Short solution: It produces a virtual, upright, enlarged image.

Final answer: It produces a virtual, upright, enlarged image.

12Why is a convex mirror useful at road bends?Core

Answer and short solution

Short solution: It gives a wide field of view with upright diminished images.

Final answer: It gives a wide field of view with upright diminished images.

13What is meant by a real image?Core

Answer and short solution

Short solution: A real image is formed where light rays actually meet and can be formed on a screen.

Final answer: A real image is formed where light rays actually meet and can be formed on a screen.

14What is meant by a virtual image?Core

Answer and short solution

Short solution: A virtual image is formed where rays appear to come from and cannot be caught on a screen.

Final answer: A virtual image is formed where rays appear to come from and cannot be caught on a screen.

15Why must diagrams show a normal at a refracting surface?Core

Answer and short solution

Short solution: Angles of incidence and refraction are measured from the normal.

Final answer: Angles of incidence and refraction are measured from the normal.

21. A-Level Questions

A-Level style explanation and reasoning questions.

1Derive the lens power unit from the thin-lens equation.A-Level

Answer and short solution

Short solution: Since focal length is measured in metres, power P=1/f has unit m^-1, called dioptre.

Final answer: Since focal length is measured in metres, power P=1/f has unit m^-1, called dioptre.

2Explain why the lens maker formula contains relative refractive index.A-Level

Answer and short solution

Short solution: Only the change in optical speed between lens and surrounding medium determines refraction at each surface.

Final answer: Only the change in optical speed between lens and surrounding medium determines refraction at each surface.

3Describe spherical aberration and one way to reduce it.A-Level

Answer and short solution

Short solution: Marginal rays focus differently from paraxial rays; using stops, aspheric surfaces or mirror designs reduces it.

Final answer: Marginal rays focus differently from paraxial rays; using stops, aspheric surfaces or mirror designs reduces it.

4Explain chromatic aberration in a single lens.A-Level

Answer and short solution

Short solution: Refractive index depends on wavelength, so different colours have different focal lengths.

Final answer: Refractive index depends on wavelength, so different colours have different focal lengths.

5Why is angular magnification more useful than linear magnification for telescopes?A-Level

Answer and short solution

Short solution: Distant objects have no accessible image size to compare directly; the telescope increases the visual angle.

Final answer: Distant objects have no accessible image size to compare directly; the telescope increases the visual angle.

6How does numerical aperture affect microscope resolution?A-Level

Answer and short solution

Short solution: Greater NA reduces the minimum resolvable separation d=0.61 lambda/NA.

Final answer: Greater NA reduces the minimum resolvable separation d=0.61 lambda/NA.

7Explain why final image at the near point gives greater magnifying power than at infinity.A-Level

Answer and short solution

Short solution: The eye accommodates and the eyepiece contributes the factor 1+D/f instead of D/f.

Final answer: The eye accommodates and the eyepiece contributes the factor 1+D/f instead of D/f.

8State the Rayleigh criterion for a circular aperture.A-Level

Answer and short solution

Short solution: Two point sources are just resolved when the central maximum of one coincides with the first minimum of the other.

Final answer: Two point sources are just resolved when the central maximum of one coincides with the first minimum of the other.

9Explain the role of the secondary mirror in a reflecting telescope.A-Level

Answer and short solution

Short solution: It redirects or folds the converging beam so the observer or detector can be placed conveniently.

Final answer: It redirects or folds the converging beam so the observer or detector can be placed conveniently.

10Why is the Cassegrain design compact?A-Level

Answer and short solution

Short solution: The convex secondary increases effective focal length while folding the beam back through the primary.

Final answer: The convex secondary increases effective focal length while folding the beam back through the primary.

11Discuss why sign convention is essential in multi-element systems.A-Level

Answer and short solution

Short solution: Each element may see a real or virtual object, and signs determine whether the next image is physically correct.

Final answer: Each element may see a real or virtual object, and signs determine whether the next image is physically correct.

12Explain the origin of lateral displacement in a parallel-sided slab.A-Level

Answer and short solution

Short solution: The ray refracts at both faces; angular deviations cancel but the path is shifted sideways.

Final answer: The ray refracts at both faces; angular deviations cancel but the path is shifted sideways.

13What is meant by optical path length?A-Level

Answer and short solution

Short solution: It is the product of refractive index and geometrical length, representing phase delay in a medium.

Final answer: It is the product of refractive index and geometrical length, representing phase delay in a medium.

14Why can a high magnification image still look blurred?A-Level

Answer and short solution

Short solution: Diffraction, aberrations and low numerical aperture limit detail even when the image is enlarged.

Final answer: Diffraction, aberrations and low numerical aperture limit detail even when the image is enlarged.

15How is total internal reflection affected by wavelength?A-Level

Answer and short solution

Short solution: Critical angle can vary slightly with wavelength because refractive index is dispersive.

Final answer: Critical angle can vary slightly with wavelength because refractive index is dispersive.

16Explain why an eyepiece behaves as a magnifier.A-Level

Answer and short solution

Short solution: It views the intermediate image as an object placed near its focal plane.

Final answer: It views the intermediate image as an object placed near its focal plane.

17What makes a virtual object for a lens?A-Level

Answer and short solution

Short solution: Converging incident rays directed toward a point beyond the lens define a virtual object.

Final answer: Converging incident rays directed toward a point beyond the lens define a virtual object.

18Compare telescope and microscope objectives.A-Level

Answer and short solution

Short solution: A telescope objective has large focal length and aperture; a microscope objective has short focal length and high numerical aperture.

Final answer: A telescope objective has large focal length and aperture; a microscope objective has short focal length and high numerical aperture.

22. Assertion–Reason: 40

Assertion-reason practice with unique Ray Optics statements.

1Assertion: A convex mirror gives a diminished image for every real object. Reason: Reflected rays diverge and appear to meet between pole and focus.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

2Assertion: A concave mirror can form a virtual enlarged image. Reason: This happens when the object lies between pole and focus.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

3Assertion: Power of a concave lens is negative in air. Reason: Its focal length is negative under Cartesian convention.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

4Assertion: A lens of focal length 25 cm has power 4 D. Reason: Focal length must be written as 0.25 m in P=1/f.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

5Assertion: Total internal reflection cannot occur from air to glass. Reason: TIR requires light to go from denser to rarer medium.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

6Assertion: At critical angle the refracted ray grazes the boundary. Reason: The angle of refraction is 90 degrees.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

7Assertion: A glass slab produces lateral displacement. Reason: The emergent ray is parallel to the incident ray for parallel faces.A-R

Answer and short solution

Short solution: Both are true, but the reason describes the result rather than the full cause.

Final answer: Both are true, but the reason describes the result rather than the full cause.

8Assertion: Apparent depth is less than real depth when viewed from air. Reason: Rays bend away from the normal when emerging from water to air.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

9Assertion: A prism disperses white light. Reason: Refractive index depends on wavelength.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

10Assertion: The prism minimum-deviation formula should be used only for symmetric path. Reason: At minimum deviation, i=e and r1=r2.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

11Assertion: An optical fibre needs ncore greater than ncladding. Reason: The core-cladding boundary must support TIR.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

12Assertion: Increasing telescope aperture improves resolving power. Reason: Diffraction angle is proportional to lambda/D.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

13Assertion: Myopia is corrected by a concave lens. Reason: It forms a virtual image of distant objects at the far point.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

14Assertion: Hypermetropia is corrected by a convex lens. Reason: It helps near objects form virtual images at the defective near point.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

15Assertion: Cataract is not simply a spherical refractive defect. Reason: It involves opacity of the crystalline lens.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

16Assertion: Astigmatism can be corrected with a cylindrical lens. Reason: The eye has unequal power in different meridians.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

17Assertion: A simple microscope gives relaxed viewing when final image is at infinity. Reason: The object is placed at the focal plane.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

18Assertion: A compound microscope objective has short focal length. Reason: A short focal length objective gives large linear magnification.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

19Assertion: A telescope objective generally has large focal length. Reason: Telescope angular magnification is f0/fe.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

20Assertion: A reflecting telescope avoids chromatic aberration in the objective. Reason: Reflection is not wavelength-dispersive like refraction.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

21Assertion: A real image can be formed on a screen. Reason: Actual rays converge at the real image position.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

22Assertion: A virtual image cannot be caught on a screen directly. Reason: Rays only appear to come from the virtual image point.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

23Assertion: For lenses in contact, powers add. Reason: Thin lenses in contact have negligible separation.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

24Assertion: For separated lenses, simply adding powers is incomplete. Reason: The term -dP1P2 must be included.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

25Assertion: A plane mirror has magnification +1. Reason: The image is upright and equal in size.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

26Assertion: Diamond has strong brilliance. Reason: Its high refractive index makes the critical angle small.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

27Assertion: A convex lens can form a virtual image. Reason: This occurs when the object is within focal length.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

28Assertion: A concave lens forms a virtual image for a real object. Reason: Emergent rays diverge after refraction.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

29Assertion: Telescope magnification increases if eyepiece focal length decreases. Reason: M=f0/fe in normal adjustment.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

30Assertion: Microscope resolution improves with higher numerical aperture. Reason: Minimum resolvable distance is inversely proportional to NA.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

31Assertion: A ray through the optical centre of a thin lens is approximately undeviated. Reason: The two refractions through a very thin central region nearly cancel.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

32Assertion: A ray through the centre of curvature of a spherical mirror retraces its path. Reason: It strikes the mirror normally.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

33Assertion: A spherical mirror has f=R/2 for paraxial rays. Reason: The focus lies midway between pole and centre in the paraxial approximation.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

34Assertion: A normal eye has near point about 25 cm. Reason: This is the conventional least distance of distinct vision.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

35Assertion: Presbyopia is related to loss of accommodation. Reason: The eye lens becomes less flexible with age.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

36Assertion: Lateral displacement in a slab increases with slab thickness. Reason: A thicker slab gives a longer shifted path between the two refracting faces.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

37Assertion: A ray incident normally on a surface is undeviated. Reason: The incidence angle is zero and Snell's law gives zero refracted angle.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

38Assertion: A high-power magnifier has small focal length. Reason: Simple microscope magnifying power varies inversely with focal length.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

39Assertion: Angular magnification differs from linear magnification. Reason: Angular magnification compares visual angles at the eye.A-R

Answer and short solution

Short solution: Both are true and the reason correctly explains the assertion.

Final answer: Both are true and the reason correctly explains the assertion.

40Assertion: Sign convention should be written before substitution. Reason: Most ray-optics numerical errors come from wrong signs rather than algebra.A-R

Answer and short solution

Short solution: Both are true and the reason gives a practical explanation.

Final answer: Both are true and the reason gives a practical explanation.

23. Case Study Questions: 20

1Case 1: spectacle prescription

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates spectacle prescription and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

2Case 2: optical fibre communication

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates optical fibre communication and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

3Case 3: prism spectrometer

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates prism spectrometer and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

4Case 4: microscope laboratory

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates microscope laboratory and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

5Case 5: astronomical telescope

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates astronomical telescope and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

6Case 6: rear-view mirror

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates rear-view mirror and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

7Case 7: swimming-pool apparent depth

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates swimming-pool apparent depth and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

8Case 8: diamond brilliance

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates diamond brilliance and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

9Case 9: camera focusing

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates camera focusing and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

10Case 10: eye accommodation

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates eye accommodation and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

11Case 11: spectacle prescription

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates spectacle prescription and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

12Case 12: optical fibre communication

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates optical fibre communication and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

13Case 13: prism spectrometer

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates prism spectrometer and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

14Case 14: microscope laboratory

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates microscope laboratory and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

15Case 15: astronomical telescope

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates astronomical telescope and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

16Case 16: rear-view mirror

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates rear-view mirror and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

17Case 17: swimming-pool apparent depth

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates swimming-pool apparent depth and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

18Case 18: diamond brilliance

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates diamond brilliance and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

19Case 19: camera focusing

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates camera focusing and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

20Case 20: eye accommodation

View detailed solution

Concept used: Integrated ray-optics modelling.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

A learner investigates eye accommodation and records object distance, focal length or refractive-index data. Identify the governing law, predict the image/ray behaviour, and state one practical implication.

Final answer: Use the relevant sign convention and formula from the master sheet. The conclusion must agree with actual ray convergence/divergence; practical performance depends on aperture, aberration, index contrast or accommodation as applicable.

24. Important Numericals

Easy to Advanced Topic-Wise Collection

1Easy: A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

View detailed solution

Concept used: Mirror formula and magnification.

Formula used: 1/f = 1/v + 1/u; m = −v/u


Step-by-step solution:

R=36 cm, so f=−18 cm and u=−27 cm. From 1/f=1/v+1/u, 1/v=−1/18+1/27=−1/54, hence v=−54 cm. m=−v/u=−2, so h′=−5.0 cm. The negative image height means inverted. Moving the candle toward F makes the real image move farther away; at F it is at infinity, and inside F no screen image is possible.

Final answer: Screen 54 cm in front; real, inverted, enlarged to 5.0 cm.

2Easy: A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

View detailed solution

Concept used: Convex-mirror formula.

Formula used: 1/f = 1/v + 1/u; m = −v/u


Step-by-step solution:

f=+15 cm, u=−12 cm. 1/v=1/15+1/12=3/20, so v=+6.67 cm. m=−v/u=+0.556 and h′=2.50 cm. As the needle recedes, the virtual image approaches F behind the mirror and becomes smaller.

Final answer: Virtual upright image 6.67 cm behind mirror; m=+0.556.

3Easy: A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

View detailed solution

Concept used: Apparent depth.

Formula used: apparent depth = real depth/μ; shift = t(1−1/μ)


Step-by-step solution:

μ=real/apparent=12.5/9.4=1.33. For μ=1.63, apparent depth=12.5/1.63=7.67 cm. Difference=9.40−7.67=1.73 cm.

Final answer: μwater≈1.33; microscope shifts 1.73 cm upward.

4Easy: Figures 9.27(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)].

View detailed solution

Concept used: Snell's law and relative index.

Formula used: n₁sin i=n₂sin r


Step-by-step solution:

For a water-glass interface, nwater sin45°=nglass sin r. Using nwater≈1.33 and nglass≈1.50, sin r=(1.33/1.50)sin45°=0.627. Thus r≈38.8°.

Final answer: Angle in glass ≈39° (about 38° using figure-derived indices).

5Easy: A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

View detailed solution

Concept used: Critical angle and escape cone.

Formula used: sin C=n₂/n₁; TIR when i>C


Step-by-step solution:

sin C=1/1.33, so C=48.75°. The emergent patch radius r=h tan C=0.80 tan48.75°≈0.912 m. Area=πr²≈2.61 m².

Final answer: Area ≈2.61 m².

6Easy: A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

View detailed solution

Concept used: Prism minimum-deviation relation.

Formula used: δ=i+e−A; μ=sin[(A+δm)/2]/sin(A/2)


Step-by-step solution:

μ=sin50°/sin30°=1.532. In water μrel=1.532/1.33=1.152. Then sin[(60°+δm′)/2]=1.152 sin30°=0.576, giving δm′≈10.3°.

Final answer: Index ≈1.53; minimum deviation in water ≈10.3°.

7Easy: Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

View detailed solution

Concept used: Lens-maker formula.

Formula used: 1/f=(μlens/μmedium−1)(1/R₁−1/R₂)


Step-by-step solution:

For R₁=+R, R₂=−R: 1/f=(μ−1)(2/R). Therefore R=2(0.55)(20)=22 cm.

Final answer: R=22 cm.

8Easy: A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is
(a) a convex lens of focal length 20 cm,
(b) a concave lens of focal length 16 cm?

View detailed solution

Concept used: Lens formula with a virtual object.

Formula used: 1/f = 1/v − 1/u; m=v/u


Step-by-step solution:

The converging beam gives u=+12 cm. (a) 1/v=1/20+1/12, so v=7.5 cm. (b) 1/v=−1/16+1/12=1/48, so v=48 cm.

Final answer: Convex: 7.5 cm beyond lens; concave: 48 cm beyond lens.

9Easy: An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

View detailed solution

Concept used: Concave-lens image formation.

Formula used: 1/f = 1/v − 1/u; m=v/u


Step-by-step solution:

f=−21 cm, u=−14 cm. 1/v=−1/21−1/14=−5/42, hence v=−8.4 cm. m=v/u=+0.60; h′=1.8 cm. On moving the object farther away, the image approaches F and becomes smaller.

Final answer: Virtual, upright, diminished image 8.4 cm before lens; height 1.8 cm.

10Easy: What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

View detailed solution

Concept used: Powers in contact.

Formula used: P=1/f; Pcontact=ΣPᵢ; Pseparated=P₁+P₂−dP₁P₂


Step-by-step solution:

1/F=1/30−1/20=−1/60 cm⁻¹.

Final answer: F=−60 cm; diverging combination.

11Medium: A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at
(a) the least distance of distinct vision (25 cm), and
(b) at infinity?
What is the magnifying power of the microscope in each case?

View detailed solution

Concept used: Successive lens formula and angular magnification.

Formula used: 1/f = 1/v − 1/u; m=v/u


Step-by-step solution:

At D: vₑ=−25 cm gives uₑ=−5 cm; objective image distance v₀=10 cm, so u₀=−2.50 cm. |m₀|=4 and mₑ=5, so |M|=20. At infinity: uₑ=−6.25 cm, v₀=8.75 cm, u₀=−2.59 cm; |m₀|=3.375 and mₑ=4, so |M|=13.5.

Final answer: At D: object 2.50 cm, M=20; at infinity: 2.59 cm, M=13.5.

12Medium: A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

View detailed solution

Concept used: Objective image plus eyepiece at near point.

Formula used: 1/f = 1/v − 1/u; m=v/u


Step-by-step solution:

Using centimetres, f₀=0.8, u₀=−0.9 gives v₀=7.2 cm and |m₀|=8. For vₑ=−25 cm and fₑ=2.5 cm, uₑ=−2.273 cm and mₑ=11. Separation=7.2+2.273=9.47 cm; M=88.

Final answer: Separation ≈9.47 cm; magnifying power ≈88.

13Medium: A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

View detailed solution

Concept used: Normal adjustment.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

|M|=f₀/fₑ=144/6=24. L=f₀+fₑ=150 cm.

Final answer: Magnifying power 24; separation 150 cm.

14Medium: (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 10⁶ m, and the radius of lunar orbit is 3.8 × 10⁸ m.

View detailed solution

Concept used: Telescope magnification and small-angle image size.

Formula used: M∞=f₀/fₑ; Mᴅ=(f₀/fₑ)(1+fₑ/D)


Step-by-step solution:

M=15/0.01=1500. Moon angular diameter≈3.48×10⁶/3.8×10⁸=9.16×10⁻³ rad. Objective image diameter=f₀θ=15(9.16×10⁻³)=0.137 m.

Final answer: Magnification 1500; moon image ≈13.7 cm.

15Medium: Use the mirror equation to deduce that:
(a) An object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) A convex mirror always produces a virtual image independent of the location of the object.
(c) The virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) An object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

View detailed solution

Concept used: Mirror formula and signs.

Formula used: 1/f = 1/v + 1/u; m = −v/u


Step-by-step solution:

Substitute the relevant u ranges in 1/v=1/f−1/u. For a concave mirror f<0: |f|<|u|<2|f| gives |v|>2|f|; |u|<|f| gives v>0 and |m|>1. For a convex mirror f>0 and u<0 always gives 0<v<f and 0<m<1.

Final answer: All four stated ray-diagram results follow from v and m signs/ranges.

16Medium: A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?

View detailed solution

Concept used: Normal shift through slab.

Formula used: apparent depth = real depth/μ; shift = t(1−1/μ)


Step-by-step solution:

Shift=t(1−1/μ)=15(1−2/3)=5 cm. For a parallel slab at normal viewing, it does not depend on where the slab lies between pin and observer.

Final answer: Apparent rise 5 cm; independent of slab location.

17Medium: (a) Figure 9.28 shows a cross-section of a ‘light pipe’ made of glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure?
(b) What is the answer if there is no outer covering of the pipe?
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View detailed solution

Concept used: Numerical aperture.

Formula used: NA=√(n₁²−n₂²); n₀sinθa=NA


Step-by-step solution:

NA=√(1.68²−1.44²)=0.865. In air, sinθa=0.865, so θa≈59.9°. Without covering n₂=1, the calculated NA>1, so every physically possible air-incidence angle up to 90° is accepted.

Final answer: With covering: 0°–59.9° to axis; without: 0°–90°.

18Medium: The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

View detailed solution

Concept used: Object-screen method.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

For fixed D=u+v, uv is maximum at u=v=D/2. Since f=uv/(u+v), fmax=D/4=3/4 m.

Final answer: Maximum f=0.75 m.

19Medium: A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.

View detailed solution

Concept used: Bessel displacement method.

Formula used: Apply the relevant formula from the master formula sheet with Cartesian signs.


Step-by-step solution:

f=(D²−d²)/(4D)=(90²−20²)/(360)=21.39 cm.

Final answer: f≈21.4 cm.

20Medium: (a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

View detailed solution

Concept used: Separated lenses and sequential imaging.

Formula used: P=1/f; Pcontact=ΣPᵢ; Pseparated=P₁+P₂−dP₁P₂


Step-by-step solution:

P=3.333−5−0.08(3.333)(−5)=−0.333 D, so EFL=−3.0 m (same EFL from either direction, though principal planes differ). First lens gives v₁=120 cm and m₁=−3. For lens 2, u₂=+112 cm, v₂=−24.35 cm, m₂=−0.2174. Total m=+0.652; image height=0.978 cm.

Final answer: EFL −3.0 m; final image ≈0.98 cm, upright.

21Advanced: At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

View detailed solution

Concept used: Critical angle plus prism geometry.

Formula used: δ=i+e−A; μ=sin[(A+δm)/2]/sin(A/2)


Step-by-step solution:

C=sin⁻¹(1/1.524)=41.03°. Hence r₁=A−C=18.97°. At first face sin i=μ sin r₁=1.524 sin18.97°=0.495.

Final answer: i≈29.7°.

22Advanced: A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.

View detailed solution

Concept used: Linear versus angular magnification.

Formula used: M=β/α; use the relevant microscope or telescope expression


Step-by-step solution:

With u=−f, the virtual image is at infinity, so finite linear size/area magnification is not defined. Angular magnification M=D/f=25/9=2.78. Linear magnification and magnifying power are different quantities.

Final answer: Image at infinity; angular magnification 2.78; no finite area magnification.

23Advanced: (a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.

View detailed solution

Concept used: Simple microscope at near point.

Formula used: M∞=(L/f₀)(D/fₑ); Mᴅ=(L/f₀)(1+D/fₑ)


Step-by-step solution:

For final image at D, M=1+D/f=1+25/9=3.78. Object distance |u|=fD/(D+f)=225/34=6.62 cm. Here |v/u|=25/6.62=3.78.

Final answer: Lens-object distance 6.62 cm; magnification 3.78.

24Advanced: What should be the distance between the object in Exercise 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm²? Would you be able to see the squares distinctly with your eyes very close to the magnifier?
---

View detailed solution

Concept used: Area and linear magnification.

Formula used: M=β/α; use the relevant microscope or telescope expression


Step-by-step solution:

Area ratio=6.25, so linear m=2.5. For a virtual image v/u=2.5 and f=9 cm. Solving gives u=−5.4 cm and v=−13.5 cm. This image lies within the normal near point, so it is not seen distinctly with the eye close to the lens.

Final answer: Object distance 5.4 cm; image not distinctly visible to a normal eye.

25Advanced: Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

View detailed solution

Concept used: Angular size and exit pupil.

Formula used: M=β/α; use the relevant microscope or telescope expression


Step-by-step solution:

(a) The magnifier lets the object be brought closer than D while remaining focused. (b) Moving the eye back usually reduces the usable field and can alter effective angular gain. (c) Very short f causes severe aberrations and tiny working distance. (d) Short f₀ gives large objective magnification; short fₑ gives large eyepiece power. (e) The eye belongs near the eyepiece exit pupil, typically a small distance outside it, for the full field.

Final answer: Magnification is angular; practical aberration, field and exit-pupil constraints limit it.

26Advanced: An angular magnification (magnifying power) of 30× is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

View detailed solution

Concept used: Exact normal-adjustment microscope geometry.

Formula used: M∞=(L/f₀)(D/fₑ); Mᴅ=(L/f₀)(1+D/fₑ)


Step-by-step solution:

Eyepiece angular magnification is D/fₑ=5, so objective magnification magnitude must be 6. Let u₀=−a and v₀=6a. The objective formula gives 1/1.25=1/(6a)+1/a, hence a=1.458 cm and v₀=8.75 cm. Put the eyepiece 5 cm beyond the intermediate image.

Final answer: Object ≈1.46 cm before objective; lens separation ≈13.75 cm.

27Advanced: A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e. when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision (25 cm)?

View detailed solution

Concept used: Telescope formulas.

Formula used: M∞=f₀/fₑ; Mᴅ=(f₀/fₑ)(1+fₑ/D)


Step-by-step solution:

Normal adjustment: |M|=140/5=28. At D: |M|=(140/5)(1+5/25)=33.6.

Final answer: 28× at infinity; 33.6× at 25 cm.

28Advanced: (a) For the telescope described in Exercise 9.27(a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 cm?

View detailed solution

Concept used: Small-angle objective image and eyepiece magnification.

Formula used: 1/f = 1/v − 1/u; m=v/u


Step-by-step solution:

L=140+5=145 cm. Tower angle=100/3000=1/30 rad, so objective image height=140/30=4.67 cm. At D, eyepiece linear magnification is |v/u|=25/4.167=6, so final virtual-image height≈28.0 cm.

Final answer: Separation 145 cm; objective image 4.67 cm; final virtual image ≈28 cm.

29Advanced: A Cassegrain telescope uses two mirrors as shown in Fig. 9.26. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

View detailed solution

Concept used: Successive mirror imaging.

Formula used: 1/f = 1/v + 1/u; m = −v/u


Step-by-step solution:

The primary has f₁=110 mm, so at the secondary the converging beam has a virtual object 90 mm away. For the convex secondary, using f₂=70 mm in the appropriate reflected-ray convention: 1/v=1/70−1/90=1/315.

Final answer: Final image 315 mm from the secondary (295 mm behind the primary in this geometry).

30Advanced: Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.29. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

View detailed solution

Concept used: Double-angle reflection.

Formula used: reflected-ray rotation=2θ; spot shift=L tan(2θ)


Step-by-step solution:

The reflected ray turns through 2θ=7°. Shift x=L tan7°=1.5 tan7°=0.184 m.

Final answer: Spot displacement ≈18.4 cm.

31Advanced: Figure 9.30 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

View detailed solution

Concept used: Lens maker and refracting-surface powers.

Formula used: 1/f=(μlens/μmedium−1)(1/R₁−1/R₂)


Step-by-step solution:

Without liquid f=30 cm. For an equiconvex lens, 1/f=(0.5)(2/R), so R=30 cm. With liquid, effective power=(ng−1)/R+(nl−ng)/(−R)=(2−nl)/30=1/45. Thus 2−nl=2/3.

Final answer: Liquid refractive index nl=4/3≈1.33.

25. Exam-Day Revision

Plane mirror

i=r; image distance = object distance; m=+1

Spherical mirror

1/f=1/v+1/u; m=−v/u; f=R/2

Plane refraction

n₁ sin i=n₂ sin r; apparent depth=real depth/n

Spherical refraction

n₂/v−n₁/u=(n₂−n₁)/R

Thin lens

1/f=1/v−1/u; m=v/u

Lens maker

1/f=(μlens/μmedium−1)(1/R₁−1/R₂)

Power

P=1/f(m); Pcontact=ΣPᵢ; Pseparated=P₁+P₂−dP₁P₂

Prism

δ=i+e−A; μ=sin[(A+δm)/2]/sin(A/2)

TIR

sin C=n₂/n₁, n₁>n₂; i>C

Optical fibre

NA=√(n₁²−n₂²); n₀ sin θa=NA

Human eye

A=1/N−1/F; myopia P=−1/x; hypermetropia P=1/D−1/Dh

Simple microscope

M∞=D/f; MD=1+D/f

Sign checklist

Mirrors: 1/f=1/v+1/u. Lenses: 1/f=1/v−1/u. Virtual corrective images have v<0.

Ray checklist

Parallel ray, optical-centre ray, focal ray; image exactly at intersection or backward-extension intersection.

Instrument checklist

Simple microscope magnifies angle; compound microscope uses two stages; telescope objective creates the intermediate image.

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