Ray Optics and Optical Instruments – Complete Notes, Formula Sheet, PYQs and Numericals

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KUMAR PHYSICS CLASSES • MASTER CHAPTER

Ray Optics and Optical Instruments

Complete Notes, Formula Sheet, Derivations, NCERT Exercises, Exam-Oriented Question Banks and Numericals for CBSE, NEET, JEE, IB, IGCSE and A-Level Physics.

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

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

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

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

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

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

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

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 = 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

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

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

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

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

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

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

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: 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

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

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

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

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∞=(L/f₀)(D/fₑ); Mᴅ=(L/f₀)(1+D/fₑ)

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

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

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

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

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

|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: 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

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

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.

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

Short solution: The eye changes lens power during accommodation. The retina distance is nearly fixed.

Final answer: The eye changes lens power during accommodation. The retina distance is nearly fixed.

75A student claims “Astigmatism needs cylindrical power.” Justify or correct the claim.NEET

Answer and short solution

Answer and short solution

Answer and short solution

Answer and short solution

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:

24. Important Numericals

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:

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:

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:

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:

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

View detailed solution

Concept used: Successive mirror imaging.

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

Step-by-step solution:

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.

If Ray Optics formulas, PYQs or numericals are not clear, contact Kumar Sir for one-to-one online Physics classes.+91-9958461445 | kumarsirphysics@gmail.com | kumarphysicsclasses.com