Optical Instruments Notes for NEET, IIT JEE, CBSE Physics

1. Optical Instruments: Introduction

Optical instruments control light so that the eye receives a larger, clearer or otherwise useful visual angle. They help us observe objects that are too small, too far away, too faint or too closely spaced for unaided vision.

Increase apparent size
Make fine structures easier to inspect.

Increase visual angle
The retinal image size depends mainly on the angle at the eye.

Form a magnified image
Real or virtual images are produced by lenses or mirrors.

Observe extremes
Microscopes study small objects; telescopes study distant objects.

Core definitions

Visual angle (α)
The angle subtended by an object at the eye. For small angles, α≈h/d.

Least distance D
The nearest comfortable viewing distance for a normal eye: D=25 cm.

Angular magnification
M=β/α, where β is the angle with the instrument and α without it.

Magnifying power
The angular magnification under a specified viewing condition.

Objective
The first lens or mirror facing the object. It forms the primary image.

Eyepiece
The lens near the eye; it magnifies the primary or intermediate image.

Key distinction: Magnification makes an image look larger. Resolution determines whether two close details can be distinguished. High magnification without resolution is empty magnification.

2. Simple Microscope / Magnifying Glass

A simple microscope is a single short-focus convex lens. The object AB is placed between the optical centre O and first focus F₁. The lens forms an erect, virtual and enlarged image on the object side.

Case A: Final image at infinity / normal adjustment

AB is at F₁, so the ray through O and the ray refracted through F₂ emerge parallel. The angular inset compares α and β.

Derivation: M = D/f

Without the lens, keep the object at the near point D. For a small object of height h, α ≈ tanα = h/D.
For normal adjustment, the object is at the focal plane, so rays emerge parallel and the virtual image is at infinity.
At the lens, β ≈ tanβ = h/f.
Therefore M = β/α = (h/f)/(h/D).

M_∞ = D/f

Relaxed-eye magnifying power

Case B: Final image at near point

The real rays diverge after refraction; their blue backward extensions meet exactly at the virtual image A′B′ at D=25 cm.

Derivation: M = 1 + D/f

For the virtual image at the near point, use Cartesian signs: v = −D.
Lens formula: 1/f = 1/v − 1/u = −1/D − 1/u.
Hence 1/|u| = 1/f + 1/D.
With small angles, β≈h/|u| and α≈h/D.
Thus M=β/α=D/|u|=D(1/f+1/D).

M_D = 1 + D/f

Maximum comfortable magnifying power

Four solved examples

Simple microscope: normal adjustment

Question: A magnifying glass has focal length 5 cm. Find magnifying power when the final image is at infinity.

Given
D = 25 cm, f = 5 cm

Formula
M = D/f

Substitution
M = 25/5

Calculation
M = 5

Final Answer: 5×

Exam Tip: For a relaxed eye, use M = D/f.

Common Mistake: Adding 1 in the infinity case.

Simple microscope: near point

Question: Find magnifying power of a lens of focal length 10 cm when the image is at the near point.

Given
D = 25 cm, f = 10 cm

Formula
M = 1 + D/f

Substitution
M = 1 + 25/10

Calculation
M = 1 + 2.5 = 3.5

Final Answer: 3.5×

Exam Tip: Near-point viewing gives larger magnification.

Common Mistake: Using D/f for a near-point image.

Find focal length

Question: A simple microscope has magnifying power 6 in normal adjustment. Find its focal length.

Given
M = 6, D = 25 cm

Formula
f = D/M

Substitution
f = 25/6

Calculation
f = 4.167 cm

Final Answer: 4.17 cm

Exam Tip: Rearrange before substituting.

Common Mistake: Using metres for D while keeping f in centimetres.

Compare two settings

Question: A magnifier has f = 8 cm. Find the increase in magnifying power when shifted from infinity to near point.

Given
D = 25 cm, f = 8 cm

Formula
ΔM = (1 + D/f) − D/f

Substitution
ΔM = 1

Calculation
Near-point power exceeds normal power by unity.

Final Answer: Increase = 1×

Exam Tip: This result is general for a simple microscope.

Common Mistake: Calculating a percentage instead of the absolute increase.

3. Compound Microscope

A compound microscope combines a very short-focus objective with an eyepiece. The objective forms a real, inverted and enlarged intermediate image A′B′. The eyepiece then acts as a simple microscope and produces a final virtual image A″B″.

Objective: small f₀ and small apertureEyepiece: moderate short fₑIntermediate image: real, inverted, enlargedFinal image: virtual, inverted

Case A: Final image at infinity

Step-by-step derivation

The objective linear magnification is m₀=v₀/u₀. Since the object is just beyond F₀, |u₀|≈f₀ and v₀≈L, so |m₀|≈L/f₀.
The intermediate image is at the eyepiece focal plane for normal adjustment.
Eyepiece angular magnification is Mₑ=D/fₑ.
Total magnifying power is the product: M=m₀Mₑ.

|M_∞| = (L/f₀)(D/fₑ)

Exact useful form: M=(v₀/u₀)(D/fₑ); negative sign denotes inversion.

Case B: Final image at near point

Objective contribution remains approximately |m₀|≈L/f₀.
The eyepiece behaves like a simple microscope with the final virtual image at D.
Its power is Mₑ=1+D/fₑ.
Multiply the two stages.

|M_D| = (L/f₀)(1 + D/fₑ)

More exact: M=(v₀/u₀)(1+D/fₑ)

Four solved examples

Compound microscope: infinity

Question: A microscope has f₀ = 1 cm, fₑ = 5 cm and tube length 20 cm. Find M in normal adjustment.

Given
L = 20 cm, f₀ = 1 cm, fₑ = 5 cm

Formula
M = (L/f₀)(D/fₑ)

Substitution
M = (20/1)(25/5)

Calculation
M = 20 × 5

Final Answer: 100× (image inverted)

Exam Tip: Magnitude is usually quoted; sign indicates inversion.

Common Mistake: Interchanging objective and eyepiece focal lengths.

Compound microscope: near point

Question: For L = 16 cm, f₀ = 0.8 cm and fₑ = 4 cm, find near-point magnification.

Given
D = 25 cm

Formula
M = (L/f₀)(1 + D/fₑ)

Substitution
M = (16/0.8)(1 + 25/4)

Calculation
20 × 7.25 = 145

Final Answer: 145×

Exam Tip: Keep the bracket intact.

Common Mistake: Writing 1 + D divided by the whole product.

Find objective focal length

Question: A compound microscope gives 200× at infinity. L = 20 cm and fₑ = 5 cm. Find f₀.

Given
M = 200, L = 20 cm, fₑ = 5 cm

Formula
f₀ = LD/(Mfₑ)

Substitution
f₀ = 20×25/(200×5)

Calculation
f₀ = 500/1000

Final Answer: 0.50 cm

Exam Tip: High objective power requires a short focal length.

Common Mistake: Using the near-point formula.

Find eyepiece focal length

Question: A microscope has M = 125 at infinity, L = 25 cm and f₀ = 1 cm. Find fₑ.

Given
D = 25 cm

Formula
fₑ = LD/(Mf₀)

Substitution
fₑ = 25×25/(125×1)

Calculation
fₑ = 625/125

Final Answer: 5 cm

Exam Tip: Check that the eyepiece focal length is larger than f₀.

Common Mistake: Dropping D from the formula.

4. Astronomical Telescope

A refracting astronomical telescope uses a large-aperture, long-focus objective and a short-focus eyepiece. Parallel rays from a distant object form a real inverted image at the objective focal plane. The eyepiece magnifies the angle subtended by this image.

Case A: Normal adjustment / final image at infinity

If the distant object subtends α at the objective, intermediate image height is h≈f₀α.
The same image at the eyepiece focal plane subtends β≈h/fₑ.
Therefore M=β/α=(h/fₑ)/(h/f₀).
The common focal plane makes separation L=f₀+fₑ.

M_∞ = −f₀/fₑ ; L = f₀ + fₑ

The minus sign denotes an inverted final image.

Case B: Final image at near point

For the eyepiece, the virtual image is at vₑ=−D.
From the lens formula, the intermediate image distance from the eyepiece has magnitude |uₑ|=fₑD/(D+fₑ), slightly less than fₑ.
The objective image height is approximately h=f₀α.
β≈h/|uₑ|, giving |M|=f₀/|uₑ|=(f₀/fₑ)(1+fₑ/D).
The tube length is L=f₀+|uₑ|.

M_D = −(f₀/fₑ)(1 + fₑ/D)

L = f₀ + |uₑ|, with uₑ signed negative under Cartesian convention.

Four solved examples

Astronomical telescope: power

Question: An astronomical telescope has f₀ = 120 cm and fₑ = 4 cm. Find normal-adjustment power.

Given
f₀ = 120 cm, fₑ = 4 cm

Formula
|M| = f₀/fₑ

Substitution
|M| = 120/4

Calculation
|M| = 30

Final Answer: 30×, inverted

Exam Tip: The negative sign, if used, denotes inversion.

Common Mistake: Using fₑ/f₀.

Astronomical telescope: length

Question: Find the length of the telescope in the previous problem at normal adjustment.

Given
f₀ = 120 cm, fₑ = 4 cm

Formula
L = f₀ + fₑ

Substitution
L = 120 + 4

Calculation
L = 124 cm

Final Answer: 124 cm

Exam Tip: At normal adjustment the focal planes coincide.

Common Mistake: Subtracting the focal lengths.

Telescope at near point

Question: A telescope has f₀ = 100 cm and fₑ = 5 cm. Find near-point magnifying power.

Given
D = 25 cm

Formula
|M| = (f₀/fₑ)(1 + fₑ/D)

Substitution
|M| = (100/5)(1 + 5/25)

Calculation
20 × 1.2 = 24

Final Answer: 24×, inverted

Exam Tip: Near-point power is slightly greater.

Common Mistake: Using 1 + D/fₑ, which belongs to a microscope eyepiece factor.

Find telescope eyepiece

Question: A telescope has objective focal length 150 cm and magnifying power 50 in normal adjustment. Find fₑ.

Given
f₀ = 150 cm, M = 50

Formula
fₑ = f₀/M

Substitution
fₑ = 150/50

Calculation
fₑ = 3 cm

Final Answer: 3 cm

Exam Tip: A shorter eyepiece increases angular magnification.

Common Mistake: Multiplying f₀ by M.

5. Reflecting Telescope

Reflecting telescopes use a concave primary mirror as the objective. Their angular magnification is still governed by the effective objective focal length and eyepiece focal length: |M|=f₀/fₑ.

Newtonian reflecting telescope

The primary starts a converging cone toward its true focus; the 45° plane secondary intercepts and folds that cone into the side eyepiece.

Cassegrain reflecting telescope

The convex secondary intercepts the primary's converging cone and reflects it through the central hole to the final focus behind the primary.

Why reflectors are preferred

No chromatic aberration in geometric focusing.
Large apertures can be manufactured and supported from behind.
Lower structural weight for a given aperture.
Excellent light-gathering and resolving power.
Parabolic primaries reduce spherical aberration.
Folded designs permit compact astronomical instruments.

Chromatic aberration
Different colours focus at different points in a refracting lens because refractive index depends on wavelength.

Spherical aberration
Marginal and paraxial rays focus at different points when a spherical surface is used. A suitable parabolic primary corrects axial spherical aberration.

Four solved examples

Reflecting telescope power

Question: A Cassegrain telescope has effective objective focal length 2400 mm and eyepiece focal length 20 mm. Find power.

Given
f₀ = 2400 mm, fₑ = 20 mm

Formula
|M| = f₀/fₑ

Substitution
|M| = 2400/20

Calculation
|M| = 120

Final Answer: 120×

Exam Tip: Use effective focal length for folded optical systems.

Common Mistake: Using mirror diameter in place of focal length.

Change of eyepiece

Question: A reflecting telescope of f₀ = 1800 mm uses 30 mm and 10 mm eyepieces. Compare powers.

Given
f₀ = 1800 mm

Formula
M = f₀/fₑ

Substitution
M₁=1800/30; M₂=1800/10

Calculation
M₁=60, M₂=180; ratio=1:3

Final Answer: 60× and 180×

Exam Tip: Smaller fₑ means greater power.

Common Mistake: Assuming aperture determines magnification.

Mirror focal length

Question: A Newtonian telescope gives 80× with a 25 mm eyepiece. Find primary mirror focal length.

Given
M=80, fₑ=25 mm

Formula
f₀ = Mfₑ

Substitution
f₀ = 80×25

Calculation
f₀ = 2000 mm

Final Answer: 2.0 m

Exam Tip: Convert the final unit only after calculation.

Common Mistake: Using mirror radius instead of focal length.

F-ratio

Question: A reflecting telescope has mirror diameter 200 mm and focal length 1200 mm. Find its f-ratio.

Given
f₀=1200 mm, aperture=200 mm

Formula
f-number = f₀/aperture

Substitution
1200/200

Calculation
6

Final Answer: f/6

Exam Tip: F-ratio describes optical speed, not magnification.

Common Mistake: Writing 6×.

6. Terrestrial Telescope

A terrestrial telescope must show distant land objects erect. It therefore uses an objective, an erecting lens or erecting system, and an eyepiece. The objective first forms an inverted image; the erecting stage produces another real image with reversed orientation; the eyepiece gives the final virtual erect view.

|M| = f₀/fₑ

The erecting lens forms I₂ by a second real-image inversion, restoring the original orientation. It ideally changes orientation, not angular magnification. For unit linear magnification, L≈f₀+4fᵣ+fₑ.

Four solved examples

Terrestrial telescope

Question: A terrestrial telescope has f₀ = 80 cm and fₑ = 4 cm. Find angular magnification.

Given
f₀=80 cm, fₑ=4 cm

Formula
M = f₀/fₑ

Substitution
M=80/4

Calculation
M=20

Final Answer: 20×, final image erect

Exam Tip: The erecting lens ideally does not alter M.

Common Mistake: Including erecting-lens focal length in M.

Terrestrial eyepiece

Question: Find eyepiece focal length for an erect telescope of power 25 with f₀ = 100 cm.

Given
M=25, f₀=100 cm

Formula
fₑ=f₀/M

Substitution
fₑ=100/25

Calculation
fₑ=4 cm

Final Answer: 4 cm

Exam Tip: Magnification magnitude follows the astronomical telescope ratio.

Common Mistake: Adding the focal lengths.

Length with erecting lens

Question: A terrestrial telescope has f₀=60 cm, fₑ=5 cm and an erecting lens of f=4 cm arranged for unit linear magnification. Estimate length.

Given
f₀=60, fₑ=5, f=4 cm

Formula
L ≈ f₀ + 4f + fₑ

Substitution
L≈60+16+5

Calculation
L≈81 cm

Final Answer: Approximately 81 cm

Exam Tip: For unit magnification, object and image are at 2f on each side.

Common Mistake: Using only f₀+fₑ.

Orientation

Question: An astronomical telescope forms an inverted view. What does the erecting system do in a terrestrial telescope?

Given
Two additional inversions are arranged by the erecting lens.

Formula
Total orientation = objective inversion × erecting inversion

Substitution
Inverted × inverted

Calculation
Final image becomes erect.

Final Answer: Erect image with essentially unchanged angular power

Exam Tip: Track image orientation stage by stage.

Common Mistake: Claiming the eyepiece itself erects the image.

7. Resolving Power

Resolution is the ability to distinguish two close objects or details as separate. Diffraction creates finite image spots, so every aperture has a fundamental limit.

Overlapping diffraction patterns: unresolved, just resolved by the Rayleigh criterion, and well resolved.

Microscope

d_min = 0.61λ/NA

RP = NA/(0.61λ)

NA=n sinθ; n is medium refractive index and θ is the objective acceptance half-angle.

Telescope

θ_min = 1.22λ/D_a

RP = D_a/(1.22λ)

Dₐ is aperture diameter. θmin is in radians.

Notation alert: This chapter uses D=25 cm for the least distance of distinct vision. In the telescope resolution formula, the aperture is written Dₐ to avoid confusion.

Eight solved resolution examples

Microscope resolution

Question: Find the resolution limit for λ=550 nm and NA=1.25.

Given
λ=550×10⁻⁹ m, NA=1.25

Formula
d=0.61λ/NA

Substitution
d=0.61×550×10⁻⁹/1.25

Calculation
d=2.684×10⁻⁷ m

Final Answer: 0.268 μm

Exam Tip: A smaller d means better resolution.

Common Mistake: Calling d the resolving power.

Microscope resolving power

Question: Find resolving power for λ=500 nm and NA=1.0.

Given
λ=500×10⁻⁹ m, NA=1

Formula
RP=NA/(0.61λ)

Substitution
RP=1/(0.61×500×10⁻⁹)

Calculation
RP=3.279×10⁶ m⁻¹

Final Answer: 3.28×10⁶ m⁻¹

Exam Tip: RP is the reciprocal of the resolution limit.

Common Mistake: Leaving wavelength in nm without conversion.

Numerical aperture

Question: An objective works in oil of n=1.5 with half-angle 60°. Find NA.

Given
n=1.5, θ=60°

Formula
NA=n sinθ

Substitution
NA=1.5 sin60°

Calculation
NA=1.5×0.866

Final Answer: 1.30

Exam Tip: NA can exceed 1 in immersion media.

Common Mistake: Using cosθ.

Effect of wavelength

Question: A microscope changes illumination from 600 nm to 450 nm at the same NA. Find ratio of resolution limits.

Given
d∝λ

Formula
d₂/d₁=λ₂/λ₁

Substitution
d₂/d₁=450/600

Calculation
d₂/d₁=0.75

Final Answer: New limit is 75%; resolution improves by 25%

Exam Tip: Shorter wavelength improves resolution.

Common Mistake: Saying resolving power decreases.

Telescope resolution limit

Question: Find angular resolution for D=0.20 m and λ=550 nm.

Given
D=0.20 m, λ=550×10⁻⁹ m

Formula
θ=1.22λ/D

Substitution
θ=1.22×550×10⁻⁹/0.20

Calculation
θ=3.355×10⁻⁶ rad

Final Answer: 3.36 μrad ≈ 0.692 arcsec

Exam Tip: Multiply radians by 206265 for arcseconds.

Common Mistake: Using focal length instead of aperture.

Required aperture

Question: What aperture resolves 1 arcsecond at λ=500 nm?

Given
θ=1/206265 rad, λ=500×10⁻⁹ m

Formula
D=1.22λ/θ

Substitution
D=1.22×500×10⁻⁹×206265

Calculation
D≈0.1258 m

Final Answer: 12.6 cm

Exam Tip: Convert arcseconds to radians first.

Common Mistake: Treating 1 arcsecond as 1 radian.

Two stars

Question: Two stars are separated by 2 μrad. Can a 0.30 m telescope at 600 nm resolve them?

Given
θsep=2×10⁻⁶ rad

Formula
θmin=1.22λ/D

Substitution
θmin=1.22×600×10⁻⁹/0.30

Calculation
θmin=2.44×10⁻⁶ rad > θsep

Final Answer: No, they are not resolved

Exam Tip: Resolved only if separation ≥ limit.

Common Mistake: Reversing the inequality.

Aperture comparison

Question: Compare resolving powers of 10 cm and 25 cm telescopes at the same wavelength.

Given
RP∝D

Formula
RP₂/RP₁=D₂/D₁

Substitution
25/10

Calculation
2.5

Final Answer: The 25 cm telescope has 2.5 times the resolving power

Exam Tip: Aperture controls both resolution and light gathering.

Common Mistake: Squaring the diameter; that applies to light collection.

8. Magnifying Power Summary

Instrument	Final image	Magnifying power	Length
Simple microscope	Infinity	D/f	—
Simple microscope	Near point	1 + D/f	—
Compound microscope	Infinity	(L/f₀)(D/fₑ)	≈ L
Compound microscope	Near point	(L/f₀)(1 + D/fₑ)	≈ L
Astronomical telescope	Infinity	−f₀/fₑ	f₀ + fₑ
Astronomical telescope	Near point	−(f₀/fₑ)(1 + fₑ/D)	f₀ + \|uₑ\|
Terrestrial telescope	Infinity	f₀/fₑ (erect)	f₀ + 4fᵣ + fₑ for unit erecting stage
Reflecting telescope	Infinity	−f₀/fₑ	Design-dependent folded path

Normal Adjustment, Infinity and Near Point

Normal adjustment
Final image at infinity. Emergent rays are parallel and the eye is relaxed.

Final image at infinity
Most comfortable for prolonged viewing; standard telescope setting.

Final image at near point
Produces somewhat greater angular magnification but requires maximum accommodation.

9. Premium Formula Sheet

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

RPmic = NA/(0.61λ)

RPtel = Dₐ/(1.22λ)

10. Numerical Problems: 40 Fully Solved

The first 28 appeared alongside their concepts. This consolidated bank contains all 40 with question, data, formula, substitution, calculation, answer, tip and common mistake.

Simple microscope: normal adjustment

Question: A magnifying glass has focal length 5 cm. Find magnifying power when the final image is at infinity.

Given
D = 25 cm, f = 5 cm

Formula
M = D/f

Substitution
M = 25/5

Calculation
M = 5

Final Answer: 5×

Exam Tip: For a relaxed eye, use M = D/f.

Common Mistake: Adding 1 in the infinity case.

Simple microscope: near point

Question: Find magnifying power of a lens of focal length 10 cm when the image is at the near point.

Given
D = 25 cm, f = 10 cm

Formula
M = 1 + D/f

Substitution
M = 1 + 25/10

Calculation
M = 1 + 2.5 = 3.5

Final Answer: 3.5×

Exam Tip: Near-point viewing gives larger magnification.

Common Mistake: Using D/f for a near-point image.

Find focal length

Question: A simple microscope has magnifying power 6 in normal adjustment. Find its focal length.

Given
M = 6, D = 25 cm

Formula
f = D/M

Substitution
f = 25/6

Calculation
f = 4.167 cm

Final Answer: 4.17 cm

Exam Tip: Rearrange before substituting.

Common Mistake: Using metres for D while keeping f in centimetres.

Compare two settings

Question: A magnifier has f = 8 cm. Find the increase in magnifying power when shifted from infinity to near point.

Given
D = 25 cm, f = 8 cm

Formula
ΔM = (1 + D/f) − D/f

Substitution
ΔM = 1

Calculation
Near-point power exceeds normal power by unity.

Final Answer: Increase = 1×

Exam Tip: This result is general for a simple microscope.

Common Mistake: Calculating a percentage instead of the absolute increase.

Compound microscope: infinity

Question: A microscope has f₀ = 1 cm, fₑ = 5 cm and tube length 20 cm. Find M in normal adjustment.

Given
L = 20 cm, f₀ = 1 cm, fₑ = 5 cm

Formula
M = (L/f₀)(D/fₑ)

Substitution
M = (20/1)(25/5)

Calculation
M = 20 × 5

Final Answer: 100× (image inverted)

Exam Tip: Magnitude is usually quoted; sign indicates inversion.

Common Mistake: Interchanging objective and eyepiece focal lengths.

Compound microscope: near point

Question: For L = 16 cm, f₀ = 0.8 cm and fₑ = 4 cm, find near-point magnification.

Given
D = 25 cm

Formula
M = (L/f₀)(1 + D/fₑ)

Substitution
M = (16/0.8)(1 + 25/4)

Calculation
20 × 7.25 = 145

Final Answer: 145×

Exam Tip: Keep the bracket intact.

Common Mistake: Writing 1 + D divided by the whole product.

Find objective focal length

Question: A compound microscope gives 200× at infinity. L = 20 cm and fₑ = 5 cm. Find f₀.

Given
M = 200, L = 20 cm, fₑ = 5 cm

Formula
f₀ = LD/(Mfₑ)

Substitution
f₀ = 20×25/(200×5)

Calculation
f₀ = 500/1000

Final Answer: 0.50 cm

Exam Tip: High objective power requires a short focal length.

Common Mistake: Using the near-point formula.

Find eyepiece focal length

Question: A microscope has M = 125 at infinity, L = 25 cm and f₀ = 1 cm. Find fₑ.

Given
D = 25 cm

Formula
fₑ = LD/(Mf₀)

Substitution
fₑ = 25×25/(125×1)

Calculation
fₑ = 625/125

Final Answer: 5 cm

Exam Tip: Check that the eyepiece focal length is larger than f₀.

Common Mistake: Dropping D from the formula.

Astronomical telescope: power

Question: An astronomical telescope has f₀ = 120 cm and fₑ = 4 cm. Find normal-adjustment power.

Given
f₀ = 120 cm, fₑ = 4 cm

Formula
|M| = f₀/fₑ

Substitution
|M| = 120/4

Calculation
|M| = 30

Final Answer: 30×, inverted

Exam Tip: The negative sign, if used, denotes inversion.

Common Mistake: Using fₑ/f₀.

Astronomical telescope: length

Question: Find the length of the telescope in the previous problem at normal adjustment.

Given
f₀ = 120 cm, fₑ = 4 cm

Formula
L = f₀ + fₑ

Substitution
L = 120 + 4

Calculation
L = 124 cm

Final Answer: 124 cm

Exam Tip: At normal adjustment the focal planes coincide.

Common Mistake: Subtracting the focal lengths.

Telescope at near point

Question: A telescope has f₀ = 100 cm and fₑ = 5 cm. Find near-point magnifying power.

Given
D = 25 cm

Formula
|M| = (f₀/fₑ)(1 + fₑ/D)

Substitution
|M| = (100/5)(1 + 5/25)

Calculation
20 × 1.2 = 24

Final Answer: 24×, inverted

Exam Tip: Near-point power is slightly greater.

Common Mistake: Using 1 + D/fₑ, which belongs to a microscope eyepiece factor.

Find telescope eyepiece

Question: A telescope has objective focal length 150 cm and magnifying power 50 in normal adjustment. Find fₑ.

Given
f₀ = 150 cm, M = 50

Formula
fₑ = f₀/M

Substitution
fₑ = 150/50

Calculation
fₑ = 3 cm

Final Answer: 3 cm

Exam Tip: A shorter eyepiece increases angular magnification.

Common Mistake: Multiplying f₀ by M.

Reflecting telescope power

Question: A Cassegrain telescope has effective objective focal length 2400 mm and eyepiece focal length 20 mm. Find power.

Given
f₀ = 2400 mm, fₑ = 20 mm

Formula
|M| = f₀/fₑ

Substitution
|M| = 2400/20

Calculation
|M| = 120

Final Answer: 120×

Exam Tip: Use effective focal length for folded optical systems.

Common Mistake: Using mirror diameter in place of focal length.

Change of eyepiece

Question: A reflecting telescope of f₀ = 1800 mm uses 30 mm and 10 mm eyepieces. Compare powers.

Given
f₀ = 1800 mm

Formula
M = f₀/fₑ

Substitution
M₁=1800/30; M₂=1800/10

Calculation
M₁=60, M₂=180; ratio=1:3

Final Answer: 60× and 180×

Exam Tip: Smaller fₑ means greater power.

Common Mistake: Assuming aperture determines magnification.

Mirror focal length

Question: A Newtonian telescope gives 80× with a 25 mm eyepiece. Find primary mirror focal length.

Given
M=80, fₑ=25 mm

Formula
f₀ = Mfₑ

Substitution
f₀ = 80×25

Calculation
f₀ = 2000 mm

Final Answer: 2.0 m

Exam Tip: Convert the final unit only after calculation.

Common Mistake: Using mirror radius instead of focal length.

F-ratio

Question: A reflecting telescope has mirror diameter 200 mm and focal length 1200 mm. Find its f-ratio.

Given
f₀=1200 mm, aperture=200 mm

Formula
f-number = f₀/aperture

Substitution
1200/200

Calculation
6

Final Answer: f/6

Exam Tip: F-ratio describes optical speed, not magnification.

Common Mistake: Writing 6×.

Terrestrial telescope

Question: A terrestrial telescope has f₀ = 80 cm and fₑ = 4 cm. Find angular magnification.

Given
f₀=80 cm, fₑ=4 cm

Formula
M = f₀/fₑ

Substitution
M=80/4

Calculation
M=20

Final Answer: 20×, final image erect

Exam Tip: The erecting lens ideally does not alter M.

Common Mistake: Including erecting-lens focal length in M.

Terrestrial eyepiece

Question: Find eyepiece focal length for an erect telescope of power 25 with f₀ = 100 cm.

Given
M=25, f₀=100 cm

Formula
fₑ=f₀/M

Substitution
fₑ=100/25

Calculation
fₑ=4 cm

Final Answer: 4 cm

Exam Tip: Magnification magnitude follows the astronomical telescope ratio.

Common Mistake: Adding the focal lengths.

Length with erecting lens

Question: A terrestrial telescope has f₀=60 cm, fₑ=5 cm and an erecting lens of f=4 cm arranged for unit linear magnification. Estimate length.

Given
f₀=60, fₑ=5, f=4 cm

Formula
L ≈ f₀ + 4f + fₑ

Substitution
L≈60+16+5

Calculation
L≈81 cm

Final Answer: Approximately 81 cm

Exam Tip: For unit magnification, object and image are at 2f on each side.

Common Mistake: Using only f₀+fₑ.

Orientation

Question: An astronomical telescope forms an inverted view. What does the erecting system do in a terrestrial telescope?

Given
Two additional inversions are arranged by the erecting lens.

Formula
Total orientation = objective inversion × erecting inversion

Substitution
Inverted × inverted

Calculation
Final image becomes erect.

Final Answer: Erect image with essentially unchanged angular power

Exam Tip: Track image orientation stage by stage.

Common Mistake: Claiming the eyepiece itself erects the image.

Microscope resolution

Question: Find the resolution limit for λ=550 nm and NA=1.25.

Given
λ=550×10⁻⁹ m, NA=1.25

Formula
d=0.61λ/NA

Substitution
d=0.61×550×10⁻⁹/1.25

Calculation
d=2.684×10⁻⁷ m

Final Answer: 0.268 μm

Exam Tip: A smaller d means better resolution.

Common Mistake: Calling d the resolving power.

Microscope resolving power

Question: Find resolving power for λ=500 nm and NA=1.0.

Given
λ=500×10⁻⁹ m, NA=1

Formula
RP=NA/(0.61λ)

Substitution
RP=1/(0.61×500×10⁻⁹)

Calculation
RP=3.279×10⁶ m⁻¹

Final Answer: 3.28×10⁶ m⁻¹

Exam Tip: RP is the reciprocal of the resolution limit.

Common Mistake: Leaving wavelength in nm without conversion.

Numerical aperture

Question: An objective works in oil of n=1.5 with half-angle 60°. Find NA.

Given
n=1.5, θ=60°

Formula
NA=n sinθ

Substitution
NA=1.5 sin60°

Calculation
NA=1.5×0.866

Final Answer: 1.30

Exam Tip: NA can exceed 1 in immersion media.

Common Mistake: Using cosθ.

Effect of wavelength

Question: A microscope changes illumination from 600 nm to 450 nm at the same NA. Find ratio of resolution limits.

Given
d∝λ

Formula
d₂/d₁=λ₂/λ₁

Substitution
d₂/d₁=450/600

Calculation
d₂/d₁=0.75

Final Answer: New limit is 75%; resolution improves by 25%

Exam Tip: Shorter wavelength improves resolution.

Common Mistake: Saying resolving power decreases.

Telescope resolution limit

Question: Find angular resolution for D=0.20 m and λ=550 nm.

Given
D=0.20 m, λ=550×10⁻⁹ m

Formula
θ=1.22λ/D

Substitution
θ=1.22×550×10⁻⁹/0.20

Calculation
θ=3.355×10⁻⁶ rad

Final Answer: 3.36 μrad ≈ 0.692 arcsec

Exam Tip: Multiply radians by 206265 for arcseconds.

Common Mistake: Using focal length instead of aperture.

Required aperture

Question: What aperture resolves 1 arcsecond at λ=500 nm?

Given
θ=1/206265 rad, λ=500×10⁻⁹ m

Formula
D=1.22λ/θ

Substitution
D=1.22×500×10⁻⁹×206265

Calculation
D≈0.1258 m

Final Answer: 12.6 cm

Exam Tip: Convert arcseconds to radians first.

Common Mistake: Treating 1 arcsecond as 1 radian.

Two stars

Question: Two stars are separated by 2 μrad. Can a 0.30 m telescope at 600 nm resolve them?

Given
θsep=2×10⁻⁶ rad

Formula
θmin=1.22λ/D

Substitution
θmin=1.22×600×10⁻⁹/0.30

Calculation
θmin=2.44×10⁻⁶ rad > θsep

Final Answer: No, they are not resolved

Exam Tip: Resolved only if separation ≥ limit.

Common Mistake: Reversing the inequality.

Aperture comparison

Question: Compare resolving powers of 10 cm and 25 cm telescopes at the same wavelength.

Given
RP∝D

Formula
RP₂/RP₁=D₂/D₁

Substitution
25/10

Calculation
2.5

Final Answer: The 25 cm telescope has 2.5 times the resolving power

Exam Tip: Aperture controls both resolution and light gathering.

Common Mistake: Squaring the diameter; that applies to light collection.

Simple microscope design

Question: Design a magnifier of power 11 at the near point. Find f.

Given
M=11, D=25 cm

Formula
f=D/(M−1)

Substitution
f=25/(11−1)

Calculation
f=2.5 cm

Final Answer: 2.5 cm

Exam Tip: Subtract unity before inversion.

Common Mistake: Using f=D/M.

Compound microscope tube length

Question: Find L for M=150 at infinity, f₀=1 cm and fₑ=5 cm.

Given
D=25 cm

Formula
L=Mf₀fₑ/D

Substitution
L=150×1×5/25

Calculation
L=30 cm

Final Answer: 30 cm

Exam Tip: All lengths must use the same unit.

Common Mistake: Putting D in the numerator after rearrangement.

Near vs infinity microscope

Question: For L=20 cm, f₀=1 cm and fₑ=5 cm, compare near-point and infinity powers.

Given
D=25 cm

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

Substitution
M∞=20×5; MD=20×6

Calculation
M∞=100, MD=120

Final Answer: 100× and 120×

Exam Tip: The difference equals objective factor L/f₀.

Common Mistake: Assuming the difference is always 1.

Telescope near-point length

Question: For f₀=100 cm and fₑ=5 cm, find eyepiece object distance and tube length when final image is at D=25 cm.

Given
fₑ=5 cm, vₑ=−25 cm

Formula
1/fₑ=1/vₑ−1/uₑ

Substitution
1/5=−1/25−1/uₑ

Calculation
uₑ=−25/6=−4.167 cm; L=f₀+|uₑ|

Final Answer: uₑ=−4.17 cm, L≈104.17 cm

Exam Tip: Use Cartesian signs for the eyepiece.

Common Mistake: Using uₑ=fₑ exactly in the near-point case.

Telescope power from length

Question: A normal telescope is 105 cm long and has power 20. Find f₀ and fₑ.

Given
f₀+fₑ=105, f₀/fₑ=20

Formula
fₑ=L/(M+1)

Substitution
fₑ=105/21

Calculation
fₑ=5 cm; f₀=100 cm

Final Answer: f₀=100 cm, fₑ=5 cm

Exam Tip: Solve the sum and ratio together.

Common Mistake: Taking f₀=L directly.

Image inversion sign

Question: An astronomical telescope has f₀=90 cm and fₑ=3 cm. Write signed magnification.

Given
f₀=90 cm, fₑ=3 cm

Formula
M=−f₀/fₑ

Substitution
M=−90/3

Calculation
M=−30

Final Answer: −30; magnitude 30× and inverted

Exam Tip: State both sign and physical meaning.

Common Mistake: Interpreting negative magnification as smaller image.

Light gathering

Question: Compare light-gathering powers of 20 cm and 50 cm mirrors.

Given
LGP∝D²

Formula
LGP₂/LGP₁=(D₂/D₁)²

Substitution
(50/20)²

Calculation
2.5²=6.25

Final Answer: The 50 cm mirror gathers 6.25 times more light

Exam Tip: Collection depends on area.

Common Mistake: Using the linear diameter ratio.

Chromatic correction insight

Question: A reflecting telescope observes red and blue light. How do their geometric focal lengths compare?

Given
Reflection law is wavelength independent for the mirror geometry.

Formula
f_red=f_blue

Substitution
Same mirror curvature

Calculation
Both focus at the same geometric focus.

Final Answer: Equal focal lengths; no chromatic aberration

Exam Tip: Diffraction still depends on wavelength.

Common Mistake: Claiming all wavelength effects vanish.

Spherical mirror correction

Question: Why is a parabolic primary preferred for parallel rays?

Given
Parallel rays must meet at one focus.

Formula
Parabola maps axial parallel rays to a common focus

Substitution
Edge and paraxial rays converge together

Calculation
Primary spherical aberration is suppressed.

Final Answer: A sharper axial image

Exam Tip: Distinguish spherical aberration from chromatic aberration.

Common Mistake: Saying a parabola eliminates diffraction.

Resolution with immersion

Question: Compare microscope resolution limits for air NA=0.90 and oil NA=1.35 at the same λ.

Given
d∝1/NA

Formula
d_oil/d_air=NA_air/NA_oil

Substitution
0.90/1.35

Calculation
0.667

Final Answer: Oil limit is two-thirds; resolving power is 1.5×

Exam Tip: State whether comparing d or RP.

Common Mistake: Saying oil increases the minimum resolvable distance.

Mixed instrument choice

Question: A 0.5 mm object must appear 10 times larger in angular size to a relaxed eye. Find magnifier focal length.

Given
M=10, D=25 cm

Formula
f=D/M

Substitution
f=25/10

Calculation
f=2.5 cm

Final Answer: 2.5 cm

Exam Tip: Object size is irrelevant to angular magnification here.

Common Mistake: Trying to use linear magnification.

Telescope field calculation

Question: An eyepiece has apparent field 50° and telescope power 25×. Estimate true field.

Given
AFOV=50°, M=25

Formula
TFOV≈AFOV/M

Substitution
TFOV≈50/25

Calculation
TFOV≈2°

Final Answer: Approximately 2°

Exam Tip: This is a useful observational approximation.

Common Mistake: Multiplying field by power.

Useful magnification vs resolution

Question: A telescope aperture is doubled while eyepiece remains unchanged. State changes in magnification and resolving power.

Given
M=f₀/fₑ; RP∝D

Formula
M unchanged if focal lengths unchanged; RP doubles

Substitution
D₂=2D₁

Calculation
RP₂=2RP₁

Final Answer: Same magnification, twice resolving power

Exam Tip: Magnification and resolution are different properties.

Common Mistake: Assuming a larger aperture automatically magnifies more.

11. PYQs and Exam-Style Questions

Academic integrity note: These are original exam-style questions based on patterns seen across the previous 20 years. No exact year or verbatim past-paper claim is made.

15 CBSE subjective questions with answers

1 Question

Define visual angle and magnifying power.

Answer: Visual angle is the angle subtended by an object or image at the eye. Magnifying power is the ratio β/α of the angle with the instrument to that with the unaided eye at the reference distance D.

2 Question

Derive the magnifying power of a simple microscope in normal adjustment.

Answer: For a small angle, β≈h/f and α≈h/D. Therefore M=β/α=(h/f)/(h/D)=D/f.

3 Question

Why is the final image commonly adjusted at infinity?

Answer: Parallel emergent rays enter a relaxed eye, so accommodation and eye strain are minimum.

4 Question

Derive the near-point magnification of a simple microscope.

Answer: Using lens geometry, angular magnification becomes M=1+D/f when the virtual image is at v=−D.

5 Question

State the functions of objective and eyepiece in a compound microscope.

Answer: The short-focus objective produces a real, inverted, enlarged intermediate image. The eyepiece acts as a simple microscope and produces a virtual enlarged final image.

6 Question

Obtain the normal-adjustment formula of a compound microscope.

Answer: Objective linear magnification is approximately L/f₀ and eyepiece angular power is D/fₑ; hence |M|=(L/f₀)(D/fₑ).

7 Question

Why must a microscope objective have small focal length?

Answer: Because objective magnification is approximately L/f₀, so a smaller f₀ gives greater linear magnification.

8 Question

Draw and describe an astronomical telescope in normal adjustment.

Answer: The objective makes a real image at its focal plane, coincident with the eyepiece focal plane. The eyepiece sends parallel rays; |M|=f₀/fₑ and L=f₀+fₑ.

9 Question

Why does a telescope objective have a large aperture?

Answer: A large aperture gathers more light and improves diffraction-limited resolving power.

10 Question

Differentiate magnifying power and resolving power.

Answer: Magnifying power increases apparent angular size; resolving power distinguishes closely spaced details. Large magnification cannot recover unresolved detail.

11 Question

Define numerical aperture.

Answer: NA=n sinθ, where n is refractive index of the medium and θ is the half-angle of the accepted cone.

12 Question

State Rayleigh criterion for a telescope.

Answer: Two point sources are just resolved when the central maximum of one Airy pattern falls at the first minimum of the other; θmin=1.22λ/D.

13 Question

Why are reflecting telescopes free from chromatic aberration?

Answer: Reflection follows the same geometric law for all visible wavelengths, so the mirror's geometric focal length is not dispersed by refractive index.

14 Question

What is the role of the erecting lens in a terrestrial telescope?

Answer: It re-inverts the objective image so the final view is erect, ideally without changing angular magnification.

15 Question

List two ways to improve microscope resolution.

Answer: Increase NA using a larger acceptance angle or immersion medium, and use shorter-wavelength illumination.

20 NEET-style MCQs with explanations

1 A simple microscope has f=5 cm. Its normal power is

Answer: B. M=25/5=5.

2 Near-point power of a 5 cm magnifier is

Answer: B. M=1+25/5=6.

3 A compound microscope objective generally has

large f and aperture
small f and aperture
large f, small aperture
zero power

Answer: B. A short-focus, small-aperture objective gives high magnification and controls aberrations.

4 Final image in a compound microscope is

real and erect
real and inverted
virtual and inverted
virtual and erect

Answer: C. The objective inverts; the eyepiece preserves that orientation while making a virtual image.

5 Normal telescope length is

f₀−fₑ
f₀+fₑ
f₀fₑ
f₀/fₑ

Answer: B. The focal planes coincide.

6 Telescope magnification magnitude is

fₑ/f₀
f₀/fₑ
D/f₀
D/fₑ

Answer: B. It is the objective-to-eyepiece focal-length ratio.

7 Increasing telescope aperture improves

only magnification
only color
resolution and brightness
focal ratio only

Answer: C. Resolution scales with D and light gathering with D².

8 Microscope resolution limit is proportional to

NA/λ
λ/NA
NAλ
1/λNA

Answer: B. d=0.61λ/NA.

9 Numerical aperture equals

n cosθ
sinθ/n
n sinθ
n/θ

Answer: C. Definition of NA.

10 A reflecting telescope avoids

diffraction
chromatic aberration
all aberrations
inversion

Answer: B. Mirrors do not disperse focus as lenses do.

11 A parabolic mirror reduces

chromatic aberration
spherical aberration
diffraction
dispersion

Answer: B. Parallel axial rays share a focus.

12 If fₑ is halved, telescope power becomes

half
double
four times
unchanged

Answer: B. M∝1/fₑ.

13 If λ decreases, resolving power

decreases
increases
unchanged
becomes zero

Answer: B. RP∝1/λ.

14 The image at normal adjustment is at

D
2f
infinity
objective

Answer: C. Emergent rays are parallel.

15 Least distance of distinct vision for a normal eye is

2.5 cm
25 cm
250 cm
1 m

Answer: B. Standard value D=25 cm.

16 The erecting lens is used in

simple microscope
compound microscope
terrestrial telescope
Newtonian telescope

Answer: C. It makes the terrestrial view erect.

17 For just resolution, separation must be

less than limit
at least the limit
zero
independent of aperture

Answer: B. Rayleigh separation is the threshold.

18 A 20 cm telescope compared with 10 cm has RP

half
same
double
four times

Answer: C. RP∝D.

19 The intermediate image in a microscope is

virtual erect
real inverted
virtual inverted
real erect

Answer: B. It is formed by the objective.

20 At near point, simple microscope power exceeds normal power by

Answer: C. (1+D/f)−D/f=1.

15 JEE Main-style questions

1 Question

A telescope has L=102 cm and M=50 in normal adjustment. Find focal lengths.

Answer: With f₀+fₑ=102 and f₀/fₑ=50, fₑ=2 cm and f₀=100 cm.

2 Question

A microscope has f₀=0.5 cm, fₑ=5 cm, L=20 cm. Find M∞.

Answer: M=(20/0.5)(25/5)=40×5=200.

3 Question

For the previous microscope, find MD.

Answer: M=(20/0.5)(1+25/5)=40×6=240.

4 Question

An objective aperture is doubled. How do telescope resolution limit and light gathering change?

Answer: θmin halves because θmin∝1/D; light gathering becomes four times because area∝D².

5 Question

Find NA for n=4/3 and θ=30°.

Answer: NA=(4/3)(1/2)=2/3≈0.667.

6 Question

A 0.25 m telescope uses λ=500 nm. Find θmin.

Answer: θmin=1.22×500×10⁻⁹/0.25=2.44×10⁻⁶ rad.

7 Question

A magnifier gives powers 5 and 6 in two settings. Identify settings and f.

Answer: 5=D/f is infinity and 6=1+D/f is near point; f=5 cm.

8 Question

Why is a telescope's linear magnification not the relevant quantity?

Answer: The distant object's actual size is inaccessible; the telescope changes the visual angle, so angular magnification is relevant.

9 Question

A reflecting telescope has f/8 and aperture 25 cm. Find focal length.

Answer: f₀=8×25 cm=200 cm=2 m.

10 Question

Find telescope power with the mirror above and a 20 mm eyepiece.

Answer: f₀=2000 mm, so M=2000/20=100.

11 Question

Two wavelengths 400 nm and 600 nm are used in the same microscope. Compare resolution limits.

Answer: d₄₀₀:d₆₀₀=400:600=2:3.

12 Question

A telescope gives 40× with f₀=1.2 m. Find fₑ.

Answer: fₑ=120 cm/40=3 cm.

13 Question

Can magnification be increased indefinitely to see more detail?

Answer: No. Once diffraction and aberrations prevent resolution, extra magnification only enlarges blur (empty magnification).

14 Question

An oil objective has NA=1.4 at 560 nm. Find d.

Answer: d=0.61×560 nm/1.4=244 nm.

15 Question

State the sign of astronomical telescope magnification and its meaning.

Answer: M=−f₀/fₑ; the minus sign means the final image is inverted.

8 JEE Advanced conceptual/numerical questions

1 Question

A telescope has f₀=100 cm and fₑ=5 cm. Derive its near-point tube length.

Answer: For vₑ=−D, 1/fₑ=1/vₑ−1/uₑ gives |uₑ|=fₑD/(D+fₑ)=25/6 cm. Thus L=f₀+|uₑ|=104.17 cm.

2 Question

Show that near-point telescope power is greater than normal power.

Answer: |MD|=(f₀/fₑ)(1+fₑ/D)=|M∞|(1+fₑ/D), and the bracket exceeds unity.

3 Question

A microscope's tube length and both focal lengths are halved. Compare M∞.

Answer: M∝LD/(f₀fₑ). The scale factor is (1/2)/[(1/2)(1/2)]=2, so power doubles.

4 Question

Two telescope designs have equal f₀/fₑ but different apertures. Are they equivalent?

Answer: They have equal angular magnification, but the larger aperture has better resolution and light gathering; they are not observationally equivalent.

5 Question

Derive the reciprocal relation between resolving limit and resolving power.

Answer: If dmin is the smallest distinguishable separation, resolving power is defined as 1/dmin. Thus microscope RP=NA/(0.61λ), and telescope RP=1/θmin=D/(1.22λ).

6 Question

A mirror telescope is achromatic, yet star images have colored diffraction sizes. Explain.

Answer: Geometric focus is achromatic, but Airy angular radius 1.22λ/D depends on wavelength, so diffraction spots are larger for red light.

7 Question

A compound microscope objective forms an image at v₀ while object distance is u₀. Give a more exact power.

Answer: Objective magnification m₀=v₀/u₀ (signed), and eyepiece power is D/fₑ or 1+D/fₑ. Therefore M=m₀Mₑ; L/f₀ is an approximation.

8 Question

Why does increasing magnification not alter Rayleigh resolution?

Answer: Rayleigh resolution is set before the eyepiece by aperture and wavelength. The eyepiece changes angular display size but cannot add missing spatial information.

8 IGCSE / A-Level / British Curriculum questions

1 Question

Explain why a telescope objective is wide.

Answer: A wide objective collects more light from faint objects and reduces the diffraction angle 1.22λ/D.

2 Question

Describe the image formed by the objective of an astronomical telescope.

Answer: It is real, inverted and small, formed in the objective focal plane.

3 Question

What adjustment gives minimum eye strain?

Answer: Normal adjustment, where the final image is at infinity and rays leave the eyepiece parallel.

4 Question

Calculate power for f₀=0.90 m and fₑ=30 mm.

Answer: Convert 0.90 m to 900 mm; M=900/30=30.

5 Question

Define the Rayleigh criterion.

Answer: Two diffraction patterns are just resolved when the central maximum of one coincides with the first minimum of the other.

6 Question

How does immersion oil improve a microscope?

Answer: It raises n and hence NA=n sinθ, reducing the minimum resolvable distance.

7 Question

Compare refracting and reflecting telescopes.

Answer: Reflectors avoid chromatic aberration and permit large supported apertures; refractors use lenses and may suffer dispersion and heavy-edge support problems.

8 Question

Why is an erect image important in a terrestrial telescope?

Answer: For navigation and land observation, scene orientation must match ordinary vision.

8 IB Physics questions

1 Question

Distinguish angular magnification from linear magnification.

Answer: Angular magnification compares visual angles β/α; linear magnification compares image and object heights.

2 Question

Discuss one trade-off in choosing a short focal-length eyepiece.

Answer: It raises magnification but may reduce field of view, eye relief, brightness per apparent area and tolerance to aberrations.

3 Question

A telescope resolves 0.8 arcsec at 550 nm. Estimate aperture.

Answer: D=1.22λ/θ with θ=0.8/206265 rad gives D≈0.173 m.

4 Question

Why is D=25 cm used for a magnifier?

Answer: It is the conventional near point of a normal adult eye and provides a common unaided-view reference angle.

5 Question

Explain why the final microscope image is virtual.

Answer: The intermediate image is placed inside the eyepiece focal length, so rays leave diverging and appear to come from an enlarged image.

6 Question

Evaluate the claim 'higher power always means a better telescope'.

Answer: False. Useful detail is limited by diffraction, seeing, aberrations and brightness; excessive power produces empty magnification.

7 Question

How does wavelength affect a diffraction-limited telescope?

Answer: Shorter wavelength gives a smaller Airy disk and improves angular resolution.

8 Question

State one environmental limitation not included in θ=1.22λ/D.

Answer: Atmospheric seeing, turbulence, thermal currents or imperfect tracking can dominate practical resolution.

8 AP Physics questions

1 Question

A student's eye is relaxed while using a telescope. What is true of outgoing rays?

Answer: They are parallel, so the final image is at infinity.

2 Question

Sketch how the objective changes parallel rays.

Answer: A converging objective bends them toward a real focus in its focal plane.

3 Question

An eyepiece is replaced by one with twice the focal length. What happens to M?

Answer: M is halved because M=f₀/fₑ.

4 Question

Why does stopping down an aperture reduce resolution?

Answer: It increases diffraction angle because θmin=1.22λ/D.

5 Question

A microscope uses λ=480 nm and NA=1.2. Find d.

Answer: d=0.61×480 nm/1.2=244 nm.

6 Question

Which measurement determines light-gathering power?

Answer: The effective aperture area, proportional to diameter squared.

7 Question

Explain image inversion in a Keplerian astronomical telescope.

Answer: The objective forms an inverted real image; the converging eyepiece magnifies its angular size without re-inverting it.

8 Question

Propose one change to improve microscope resolution without changing lenses.

Answer: Use shorter-wavelength illumination, within specimen and detector constraints.

12. Assertion–Reason: 15 Questions

Choose: both true with correct explanation; both true but no correct explanation; Assertion true/Reason false; Assertion false/Reason true; or both false.

1 Assertion–Reason

Assertion: A simple microscope has greater magnifying power at the near point.

Reason: Its near-point formula includes an additional +1 term.

Answer: Both true, and Reason correctly explains Assertion.

2 Assertion–Reason

Assertion: A telescope objective should have a large focal length.

Reason: Magnifying power is proportional to objective focal length.

Answer: Both true, and Reason correctly explains Assertion.

3 Assertion–Reason

Assertion: A telescope objective should have a large aperture.

Reason: Magnifying power is proportional to aperture.

Answer: Assertion true, Reason false; aperture improves brightness and resolution, not the focal-length ratio.

4 Assertion–Reason

Assertion: Reflecting telescopes have no chromatic aberration.

Reason: The law of reflection does not depend on refractive index.

Answer: Both true, and Reason correctly explains Assertion.

5 Assertion–Reason

Assertion: A microscope can resolve finer detail with blue light than red light.

Reason: Resolution limit is proportional to wavelength.

Answer: Both true, and Reason correctly explains Assertion.

6 Assertion–Reason

Assertion: Normal adjustment gives maximum possible magnification.

Reason: The final image is at infinity.

Answer: Assertion false, Reason true; near-point viewing gives slightly larger power.

7 Assertion–Reason

Assertion: The intermediate microscope image is virtual.

Reason: The objective is a converging lens.

Answer: Assertion false, Reason true; the objective forms a real image.

8 Assertion–Reason

Assertion: A smaller telescope eyepiece focal length gives higher power.

Reason: M=f₀/fₑ.

Answer: Both true, and Reason correctly explains Assertion.

9 Assertion–Reason

Assertion: Resolution and magnification are identical concepts.

Reason: Both are measured as ratios.

Answer: Both false in general; resolution is a minimum separation criterion.

10 Assertion–Reason

Assertion: Oil immersion improves microscope resolution.

Reason: Oil increases numerical aperture.

Answer: Both true, and Reason correctly explains Assertion.

11 Assertion–Reason

Assertion: A terrestrial telescope gives an erect image.

Reason: It contains an erecting lens system.

Answer: Both true, and Reason correctly explains Assertion.

12 Assertion–Reason

Assertion: A parabolic mirror is used to reduce spherical aberration.

Reason: All axial parallel rays reflect to a common focus ideally.

Answer: Both true, and Reason correctly explains Assertion.

13 Assertion–Reason

Assertion: A larger telescope aperture gives a smaller Airy disk.

Reason: θmin is inversely proportional to D.

Answer: Both true, and Reason correctly explains Assertion.

14 Assertion–Reason

Assertion: The final image of a compound microscope is real.

Reason: It can be observed by the eye.

Answer: Assertion false, Reason true but not explanatory; the viewed final image is virtual.

15 Assertion–Reason

Assertion: At normal adjustment, telescope length is f₀+fₑ.

Reason: Objective and eyepiece focal planes coincide.

Answer: Both true, and Reason correctly explains Assertion.

13. Case Studies: 10 Sets

1 Biology Lab Microscope

A student uses f₀=1 cm, fₑ=5 cm and L=20 cm.

Subparts: (a) Find normal power. (b) Find near-point power. (c) State image orientation.

Answers: (a) 100×. (b) 120×. (c) Final image is virtual and inverted relative to the original object.

2 School Observatory

A telescope has f₀=120 cm, fₑ=4 cm and aperture 12 cm.

Subparts: (a) Find power. (b) Find normal length. (c) State why aperture matters.

Answers: (a) 30×. (b) 124 cm. (c) It increases light gathering and resolving power.

3 Oil Immersion

A microscope operates at λ=500 nm. NA changes from 0.9 to 1.35.

Subparts: (a) Find both resolution limits. (b) Find improvement factor.

Answers: d₁=339 nm, d₂=226 nm. Resolving power improves by 1.5×.

4 Choosing an Eyepiece

A 1500 mm reflecting telescope has 25 mm and 10 mm eyepieces.

Subparts: (a) Find powers. (b) Which gives wider true field? (c) Which shows a dimmer extended image?

Answers: (a) 60× and 150×. (b) 25 mm usually gives wider field. (c) 10 mm at higher power.

5 Reading Fine Print

A person uses a 10 cm convex lens.

Subparts: (a) Find M∞. (b) Find MD. (c) Which is more comfortable?

Answers: (a) 2.5×. (b) 3.5×. (c) Infinity setting because the eye is relaxed.

6 Resolving Double Stars

Two stars are 1 arcsec apart, λ=550 nm.

Subparts: (a) Find minimum aperture. (b) Would 8 cm work?

Answers: D≈13.8 cm. An 8 cm ideal telescope would not meet the Rayleigh criterion.

7 Newtonian Design

A 200 mm f/5 primary is paired with a 20 mm eyepiece.

Subparts: (a) Find focal length. (b) Find power. (c) Name the secondary.

Answers: (a) 1000 mm. (b) 50×. (c) A small plane mirror near 45°.

8 Cassegrain Path

Light reflects from a concave primary and convex secondary.

Subparts: (a) Where does light finally travel? (b) Why is the tube compact?

Answers: (a) Back through the central hole to an eyepiece behind the primary. (b) The optical path is folded.

9 Terrestrial Survey

An erect telescope has f₀=90 cm and fₑ=4.5 cm.

Subparts: (a) Find power. (b) Why add an erecting lens? (c) Does it ideally change power?

Answers: (a) 20×. (b) To restore upright orientation. (c) No, not ideally.

10 Empty Magnification

A small telescope's eyepiece is changed from 20 mm to 4 mm, but no extra detail appears.

Subparts: (a) What increased? (b) What did not? (c) Give two limiting causes.

Answers: Magnification increased fivefold; resolving power did not. Diffraction, atmospheric seeing, aberrations, focus or brightness may limit detail.

14. Compact Revision Notes

Which optical element?

Simple microscope: one convex lens.
Compound microscope: short-focus objective + eyepiece.
Astronomical refractor: long-focus objective + short-focus eyepiece.
Reflector: concave primary mirror + secondary + eyepiece.
Terrestrial telescope: objective + erecting system + eyepiece.

Image character

Simple microscope final: virtual, erect, enlarged.
Compound intermediate: real, inverted, enlarged.
Compound final: virtual, inverted.
Astronomical telescope final: virtual and inverted.
Terrestrial telescope final: virtual and erect.

Formula choice

Relaxed eye → infinity formula.
Maximum accommodation → near-point formula.
Microscope resolution → 0.61λ/NA.
Telescope resolution → 1.22λ/Dₐ.
Reflector power uses effective mirror focal length.

Sign convention

Use Cartesian signs in lens equations.
A virtual final image on the incident side has v<0.
Negative telescope magnification denotes inversion.
Power is often quoted as a positive magnitude with orientation stated separately.

Common mistakes and shortcuts

Do not mix D values.
D=25 cm is the eye's near point; Dₐ is telescope aperture.

Remember the +1.
Near-point simple microscope and eyepiece power contain 1+D/f.

Power is not resolution.
focal-length ratio sets power; aperture and wavelength set diffraction resolution.

Quick telescope solve.
If L and M are known: fₑ=L/(M+1), f₀=ML/(M+1).

Quick comparison.
At fixed λ, microscope RP∝NA and telescope RP∝aperture.

Orientation check.
Track each real-image inversion; an eyepiece does not automatically erect the image.