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Ray Optics and Optical Instruments: Class 12 Physics Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

An underwater photographer uses a special biconvex glass lens (n_g = 1.5, both radii of curvature = 20 cm). In air the lens forms sharp images. When submerged in water (n_water = 1.33), the optical behaviour of the lens changes significantly — a real-world challenge faced in underwater photography and aquatic optical instruments.

An underwater photographer uses a special camera lens made of glass (refractive index n_g = 1.5). The lens is a biconvex lens with both radii of curvature equal to 20 cm. In air, this lens produces sharp images of distant objects on the sensor.

(i) Using the Lens Maker's Formula, calculate the focal length of this lens when it is used in air (n_air = 1.0).

(ii) The photographer submerges the camera in a swimming pool. The lens is now surrounded by water (n_water = 1.33). Calculate the new focal length of the lens in water.

(iii) The photographer focuses on a fish that is 150 cm away (measured along the principal axis) while the camera is underwater. Using the focal length found in part (ii), find the position of the image formed on the sensor. State whether the image is real or virtual.

(iv) How does the power of the lens change when it is taken from air into water? Give a reason based on your calculation.

Diagram for question 1: Ray Optics and Optical Instruments
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(i) Focal length in air:

Lens Maker's Formula states:
1/f = (n_rel − 1)(1/R₁ − 1/R₂)

where n_rel = n_lens / n_medium.

By New Cartesian Sign Convention: for a biconvex lens, R₁ = +20 cm, R₂ = −20 cm.

n_rel (air) = n_g / n_air = 1.5 / 1.0 = 1.5

1/f_air = (1.5 − 1)(1/20 − 1/(−20))
→ 1/f_air = (0.5)(1/20 + 1/20)
→ 1/f_air = (0.5)(2/20)
→ 1/f_air = (0.5)(0.1) = 0.05 cm⁻¹

∴ f_air = +20 cm

(ii) Focal length in water:

Applying Lens Maker's Formula with the lens surrounded by water:

n_rel (water) = n_g / n_water = 1.5 / 1.33 ≈ 1.128

1/f_water = (1.128 − 1)(1/20 − 1/(−20))
→ 1/f_water = (0.128)(1/20 + 1/20)
→ 1/f_water = (0.128)(2/20)
→ 1/f_water = (0.128)(0.1) = 0.0128 cm⁻¹

∴ f_water = 1/0.0128 ≈ +78.1 cm

(iii) Position of image in water (object at u = −150 cm):

Using the thin lens formula: 1/v − 1/u = 1/f

By New Cartesian Sign Convention: u = −150 cm, f = +78.1 cm

1/v = 1/f + 1/u
→ 1/v = 1/78.1 + 1/(−150)
→ 1/v = 0.01281 − 0.00667
→ 1/v = 0.00614 cm⁻¹

∴ v = 1/0.00614 ≈ +162.9 cm ≈ +163 cm

Since v is positive, the image is formed on the other side of the lens from the object.
∴ The image is real, inverted, and formed approximately 163 cm behind the lens.

(iv) Change in power from air to water:

Power of a lens is defined as P = 1/f (in metres).

In air: P_air = 1/0.20 = +5.0 D
In water: P_water = 1/0.781 ≈ +1.28 D

The power of the lens decreases significantly (from +5 D to approximately +1.28 D) when taken from air to water.

Reason: Power depends on n_rel = n_lens/n_medium. In water, n_medium increases from 1.0 to 1.33, so n_rel = 1.5/1.33 decreases, making the lens less convergent. The lens bends light through a smaller angle at each surface, resulting in a much longer focal length and hence smaller power.
Q2Case-based4 marks

When light travels from an optically denser medium to an optically rarer medium, at the interface it is partly reflected back into the same medium and partly refracted into the second medium. When the angle of incidence inside the denser medium exceeds a specific value called the critical angle (i<sub>c</sub>), the refracted ray disappears and light is completely reflected back — a phenomenon called Total Internal Reflection (TIR). The critical angle is given by:

sin i<sub>c</sub> = n<sub>2</sub>/n<sub>1</sub> = 1/n (when medium 2 is air, n<sub>2</sub> = 1)

where n = refractive index of the denser medium. TIR occurs only when: (i) light travels from denser to rarer medium, and (ii) angle of incidence > critical angle.

A science museum has installed a glass prism display to demonstrate total internal reflection. A visitor shines a thin laser pointer into the prism from outside. The prism is made of flint glass with refractive index 1.65. Inside the prism, the laser beam strikes one of the polished surfaces at an angle of incidence of 35°.

(I) Calculate the critical angle for the flint glass–air interface. (Take sin⁻¹(0.606) = 37.3°)
(II) Will total internal reflection occur at this surface? Justify your answer.
(III) The visitor now uses a different glass block where the critical angle is exactly 45°. What is the refractive index of this glass?
(IV) Name ONE practical application of total internal reflection used in everyday technology.

Diagram for question 2: Ray Optics and Optical Instruments
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MARKING SCHEME

(I) Critical Angle for Flint Glass–Air Interface (1 mark)

By the condition for total internal reflection, the critical angle i<sub>c</sub> is given by:

sin i<sub>c</sub> = 1/n

where n = 1.65 is the refractive index of flint glass (with respect to air).

Substituting:

sin i<sub>c</sub> = 1/1.65 = 0.606

∴ i<sub>c</sub> = sin⁻¹(0.606) = 37.3°

∴ The critical angle for the flint glass–air interface is 37.3°.

─────────────────────────────────

(II) Will TIR Occur? (1 mark)

The angle of incidence of the laser beam inside the prism at the surface = 35°.

The critical angle = 37.3°.

Since 35° < 37.3°,
the angle of incidence is LESS than the critical angle.

∴ Total internal reflection will NOT occur. The laser beam will be partially refracted (transmitted) out of the prism surface and partially reflected, following the usual laws of refraction.

─────────────────────────────────

(III) Refractive Index When i<sub>c</sub> = 45° (1 mark)

Using the relation:

sin i<sub>c</sub> = 1/n

Substituting i<sub>c</sub> = 45°:

sin 45° = 1/n

1/√2 = 1/n

n = √2

∴ The refractive index of the glass block is n = √2 ≈ 1.41.

─────────────────────────────────

(IV) Practical Application of TIR (1 mark)

Optical fibres — used in telecommunication and medical endoscopy. Light signals undergo repeated total internal reflection along the core of the fibre (denser medium, higher n) and travel over long distances with negligible loss.

(Other acceptable answers: periscopes using prisms, sparkling of diamonds, mirage formation.)
Q3Case-based4 marks

Optical instruments designed for underwater use employ lenses whose effective focal lengths change dramatically when submerged in water. This is because the focal length of a lens depends not only on its geometry but also on the refractive index of the surrounding medium. The Lens Maker's formula generalised for a lens of refractive index μ_lens placed in a medium of refractive index μ_m is:

1/f = (μ_lens/μ_m − 1)(1/R₁ − 1/R₂)

For a plano-convex lens (curved surface facing the object): R₁ = +R, R₂ = ∞.
Given: μ_g = 1.5, μ_w = 1.33, R = 20 cm.

A marine biologist is using a waterproof optical instrument that contains a plano-convex glass lens (refractive index μ_g = 1.5, radius of curved surface R = 20 cm) submerged in seawater (refractive index μ_w = 1.33). An underwater object is placed 60 cm in front of this lens.

(i) Using the Lens Maker's formula, find the focal length of the plano-convex lens when it is used in air.
(ii) Find the new focal length of the same lens when it is completely submerged in seawater.
(iii) An object is placed 60 cm in front of the lens while submerged. Find the image distance.
(iv) The biologist now removes the lens from water and holds it in air. She observes that the image of a distant tree (object at infinity) is formed at a certain point. State the nature of this image and the distance from the lens at which it is formed.

Diagram for question 3: Ray Optics and Optical Instruments
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New Cartesian sign convention: distances measured from the optical centre; distances in the direction of incident light are positive.

(i) Focal length of the plano-convex lens in AIR:

By the Lens Maker's formula (lens in medium of refractive index μ_m = 1):

1/f_air = (μ_g/μ_air − 1)(1/R₁ − 1/R₂)

For a plano-convex lens: R₁ = +20 cm, R₂ = ∞

1/f_air = (1.5/1 − 1)(1/20 − 1/∞)

1/f_air = (0.5)(1/20) = 0.5/20 = 1/40

∴ f_air = +40 cm

The lens acts as a converging lens with focal length 40 cm in air.

(ii) Focal length of the same lens SUBMERGED in seawater (μ_w = 1.33):

Using the generalised Lens Maker's formula:

1/f_water = (μ_g/μ_w − 1)(1/R₁ − 1/R₂)

1/f_water = (1.5/1.33 − 1)(1/20 − 0)

μ_g/μ_w = 1.5/1.33 ≈ 1.128

1/f_water = (1.128 − 1)(1/20) = (0.128)(0.05)

1/f_water = 0.0064 cm⁻¹

∴ f_water = 1/0.0064 ≈ +156.25 cm ≈ +156 cm

The lens remains converging in water but with a greatly increased focal length of ≈ 156 cm.

(iii) Image distance when object is 60 cm in front of the lens in water:

Using the thin lens formula (sign convention applied):
1/v − 1/u = 1/f_water

u = −60 cm (object on left), f_water = +156 cm

1/v = 1/f_water + 1/u = 1/156 + 1/(−60)

1/v = 1/156 − 1/60

LCM of 156 and 60 = 780

1/v = 5/780 − 13/780 = −8/780 = −2/195

∴ v = −97.5 cm

The image is formed 97.5 cm in front of the lens (on the same side as the object). It is a virtual, erect and diminished image.

(iv) Lens brought back into AIR; object at infinity (distant tree):

For an object at infinity, parallel rays are incident on the lens. By the lens formula:
1/v − 1/u = 1/f_air, with u → −∞

1/v = 1/f_air + 1/u = 1/40 + 0 = 1/40

∴ v = +40 cm

The image is formed at the principal focus, 40 cm on the other side of the lens.
Nature of image: Real, inverted, and highly diminished (point-sized) image formed at the focal point of the lens.
Q4Case-based4 marks

Riya is using a compound microscope in her school laboratory to observe a cheek cell slide. The objective lens has a focal length f₀ = 1.0 cm and the eyepiece has a focal length fₑ = 5.0 cm. The tube length L (distance between the second focal point of the objective and the first focal point of the eyepiece) is 14.0 cm, and the least distance of distinct vision D = 25 cm.

Riya is using a compound microscope in her school laboratory to observe a cheek cell slide. The objective lens has a focal length of 1.0 cm and the eyepiece has a focal length of 5.0 cm. The tube length (distance between the image formed by the objective and the optical centre of the eyepiece) is 14.0 cm, and the least distance of distinct vision is 25 cm.

(i) For large magnification in a compound microscope, the focal lengths of the objective and the eyepiece should respectively be:
(a) large and large
(b) large and small
(c) small and large
(d) small and small

(ii) The magnification produced by the objective lens alone is:
(a) 5
(b) 14
(c) 25
(d) 70

(iii) The total magnifying power of the compound microscope (final image at the least distance of distinct vision) is:
(a) 70
(b) 84
(c) 100
(d) 140

(iv) Riya now replaces the eyepiece with one of focal length 2.5 cm, keeping everything else the same. Which of the following correctly describes the effect on the total magnifying power?
(a) It decreases, because a longer focal length eyepiece gives higher magnification.
(b) It increases, because a shorter focal length eyepiece gives higher magnification.
(c) It remains the same, because only the objective determines magnification.
(d) It decreases, because a shorter focal length eyepiece reduces the tube length.

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Sub-part (i) — [1 mark]

Option (d) is correct.

Explanation: The total magnifying power of a compound microscope is m = (L/f₀)(D/fₑ). For m to be large, both f₀ and fₑ must be small.

∴ The objective and eyepiece should both have small focal lengths.

─────────────────────────────────────
Sub-part (ii) — [1 mark]

Option (b) is correct.

Explanation: The magnification produced by the objective lens is given by:

m₀ = L / f₀

Substituting L = 14.0 cm, f₀ = 1.0 cm:

m₀ = 14.0 / 1.0

∴ m₀ = 14

─────────────────────────────────────
Sub-part (iii) — [1 mark]

Option (b) is correct.

Explanation: For the final image formed at the least distance of distinct vision D, the total magnifying power of a compound microscope is:

m = (L / f₀) × (1 + D / fₑ)

Substituting L = 14.0 cm, f₀ = 1.0 cm, D = 25 cm, fₑ = 5.0 cm:

m = (14.0 / 1.0) × (1 + 25 / 5.0)
m = 14 × (1 + 5)
m = 14 × 6

∴ Total magnifying power m = 84

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Sub-part (iv) — [1 mark]

Option (b) is correct.

Explanation: Total magnifying power m = (L / f₀)(1 + D / fₑ). With the new eyepiece fₑ = 2.5 cm:

m’ = 14 × (1 + 25 / 2.5) = 14 × (1 + 10) = 14 × 11 = 154

Since m’ = 154 > m = 84, the magnifying power increases. This is because a shorter focal length eyepiece produces a larger angular magnification (D/fₑ increases), thereby increasing the total magnifying power.

∴ Replacing the eyepiece with one of shorter focal length increases the total magnifying power.
Q5Case-based4 marks

A biology student is using a compound microscope to examine a prepared slide. The microscope has an objective lens of focal length f₀ = 1.0 cm and an eyepiece of focal length fₑ = 5.0 cm. The object (slide) is placed at a distance u₀ = 1.2 cm in front of the objective lens. The student adjusts the microscope so that the final virtual image is formed at the near point, D = 25 cm, for comfortable viewing. (Use New Cartesian sign convention throughout.)

A biology student is observing a microscope slide using a compound microscope. The objective lens has a focal length of 1.0 cm and the eyepiece has a focal length of 5.0 cm. The slide (object) is placed 1.2 cm in front of the objective lens. The student adjusts the eyepiece so that the final image is formed at the near point (D = 25 cm).

(i) Where is the intermediate image formed by the objective lens? Is it real or virtual?
(ii) What is the magnification produced by the objective lens alone?
(iii) What is the magnification produced by the eyepiece when the final image is at the near point D = 25 cm?
(iv) What is the total magnifying power of the compound microscope?

Diagram for question 5: Ray Optics and Optical Instruments
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(i) Finding the image distance for the objective lens:

Using New Cartesian sign convention: distances measured from the optical centre of the lens; distances in the direction of incident light are positive.

For the objective lens, by the lens formula:

1/v₀ − 1/u₀ = 1/f₀

Given: u₀ = −1.2 cm (object is to the left), f₀ = +1.0 cm

1/v₀ = 1/f₀ + 1/u₀ = 1/1.0 + 1/(−1.2)

1/v₀ = 1 − 1/1.2 = 1 − 0.833 = 0.167

∴ v₀ = 1/0.167 = 6.0 cm

∴ The intermediate image is formed 6.0 cm to the right of the objective lens.

Since v₀ is positive (on the transmission side of the lens), the intermediate image is real and inverted.

(ii) Magnification by the objective lens:

m₀ = v₀/u₀ = (+6.0)/(−1.2)

∴ m₀ = −5

(The negative sign confirms the image is inverted; the magnitude of linear magnification is 5.)

(iii) Magnification by the eyepiece (image at near point):

When the final image is formed at the near point D, the eyepiece acts as a simple microscope with magnification:

mₑ = 1 + D/fₑ = 1 + 25/5.0 = 1 + 5 = 6

∴ mₑ = 6

(iv) Total magnifying power of the compound microscope:

M = m₀ × mₑ = (−5) × 6

∴ M = −30

The magnitude of the total magnifying power is 30. The negative sign indicates that the final image is inverted with respect to the object.
Q6Case-based4 marks

A school science club is designing a simple astronomical telescope using two convex lenses. Lens A (f = 100 cm) and Lens B (f = 5 cm) are available. The telescope is initially set for normal (relaxed eye) adjustment, meaning the final image is formed at infinity.

A school science club is designing a simple astronomical telescope using two convex lenses available in their laboratory. Lens A has a focal length of 100 cm and Lens B has a focal length of 5 cm. They mount the lenses coaxially such that the telescope is set for normal (relaxed eye) adjustment.

(i) Which lens should be used as the objective and which as the eyepiece? Give one reason for your choice.
(ii) Calculate the magnifying power of the telescope in normal adjustment.
(iii) An object (a distant tree) subtends an angle of 0.5° at the objective. What angle does the final image subtend at the eyepiece in normal adjustment?
(iv) If the students now shift the eyepiece slightly inward so that the final image forms at the near point (D = 25 cm) of the eye, will the magnifying power increase or decrease compared to normal adjustment? Justify.

Diagram for question 6: Ray Optics and Optical Instruments
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(i) Lens A (f = 100 cm) should be the objective and Lens B (f = 5 cm) should be the eyepiece.
Reason: The objective must have a large focal length to collect maximum light and form a bright, distant image at its focal plane; the eyepiece must have a short focal length to provide high magnification of that image.

(ii) The magnifying power of an astronomical telescope in normal adjustment is given by:

m = −f<sub>o</sub> / f<sub>e</sub>

Substituting f<sub>o</sub> = 100 cm and f<sub>e</sub> = 5 cm:

m = −100 / 5

∴ m = −20

(The magnitude is 20; the negative sign indicates the final image is inverted.)

(iii) In normal adjustment, the magnifying power also equals the ratio of the angle subtended at the eyepiece (β) to the angle subtended at the objective (α):

m = β / α

Here |m| = 20 and α = 0.5°

β = |m| × α = 20 × 0.5°

∴ β = 10°

The final image subtends an angle of 10° at the eyepiece.

(iv) The magnifying power when the final image is at the near point D is:

m = −(f<sub>o</sub> / f<sub>e</sub>)(1 + f<sub>e</sub>/D)

Since (1 + f<sub>e</sub>/D) > 1, the magnitude of magnifying power increases when the image is formed at the near point compared to normal adjustment.

Justification: When the eyepiece is shifted inward, it acts as a simple microscope forming the image at D = 25 cm instead of infinity; the eyepiece now provides additional magnification given by the factor (1 + f<sub>e</sub>/D), which is greater than 1. Therefore, the magnifying power increases.
Q7MCQ1 mark

The radii of curvature of both surfaces of a biconvex lens made of glass (refractive index = 1.5) are each equal to 20 cm. What is the power of this lens?

Show answer
Option (A) is correct.

Explanation: By the Lens Maker's Formula, 1/f = (n − 1)(1/R₁ − 1/R₂).

For a biconvex lens using New Cartesian sign convention: R₁ = +20 cm, R₂ = −20 cm, n = 1.5.

→ 1/f = (1.5 − 1)(1/20 − 1/(−20)) = (0.5)(1/20 + 1/20) = (0.5)(2/20) = 1/20

∴ f = 20 cm = 0.20 m

∴ Power P = 1/f = 1/0.20 = +5 D
Q8MCQ1 mark

A convex lens of focal length 20 cm is used as a simple magnifier. What is the power of this lens?

Show answer
Option (C) is correct.

Explanation: The power of a lens is P = 1/f, where f must be in metres.

f = 20 cm = 0.20 m

∴ P = 1/0.20 = + 5 D

The lens is convex (converging), so the focal length and power are both positive.
Q9MCQ1 mark

A ray of light travelling in glass (refractive index n = 1.5) is incident on the glass–air interface. What is the critical angle for total internal reflection at this interface?

Show answer
Option (A) is correct.

Explanation: For total internal reflection, the critical angle C is defined by sin C = n₂/n₁, where n₁ is the denser medium (glass) and n₂ is the rarer medium (air, n₂ = 1).

∴ sin C = 1/1.5 = 2/3 → C = sin⁻¹(2/3).
Q10MCQ1 mark

A convex lens of focal length 20 cm is used as a simple magnifier. An object is placed at its focus. Where is the image formed?

Show answer
Option (B) is correct.

Explanation: By the lens formula, 1/v − 1/u = 1/f. When the object is placed at the focus, u = −f (using New Cartesian sign convention). Substituting: 1/v − 1/(−f) = 1/f → 1/v = 0 → v = ∞. ∴ The image is formed at infinity.
Q11MCQ1 mark

The critical angle for a glass-water interface is 60°. What is the refractive index of glass with respect to water?

Show answer
Option (B) is correct.

Explanation: The critical angle iC is related to the refractive index by sin iC = n₂/n₁, where n₁ is the denser medium (glass) and n₂ is the rarer medium (water).

Here iC = 60°, so ₁n₂ (glass w.r.t. water) = 1/sin 60° = 1/(√3/2) = 2/√3.
Q12Short Answer1 mark

Assertion (A) : A ray of light travelling from a denser medium to a rarer medium bends away from the normal at the interface.
Reason (R) : The speed of light is greater in a rarer medium than in a denser medium.

Show answer
Option (a) is correct.

Explanation: By Snell's law, n₁ sinθ₁ = n₂ sinθ₂. When light travels from a denser medium (n₁ > n₂) to a rarer medium, sinθ₂ > sinθ₁, so the refracted ray bends away from the normal — Assertion (A) is TRUE. The refractive index n = c/v, so a rarer medium (smaller n) has a greater speed of light — Reason (R) is TRUE. Further, it is precisely because v is greater in the rarer medium (n₂ < n₁) that the wavefront bends outward, causing the ray to bend away from the normal; thus R is the correct explanation of A.
Q13Short Answer2 marks

A compound microscope has an objective of focal length 0.5 cm and an eyepiece of focal length 2.5 cm. The tube length (distance between second focal point of objective and first focal point of eyepiece) is 7.5 cm and the least distance of distinct vision is 25 cm. Calculate the magnifying power of the microscope.

Show answer
The magnifying power of a compound microscope is given by:

m = (L / f₀) × (D / fₑ)

where L = tube length, f₀ = focal length of objective, fₑ = focal length of eyepiece, D = least distance of distinct vision.

Substituting the values:

m = (7.5 cm / 0.5 cm) × (25 cm / 2.5 cm)

m = 15 × 10

∴ Magnifying power of the microscope = 150
Q14Short Answer2 marks

A ray of light passes from glass (refractive index 1.5) into water (refractive index 4/3). Find the critical angle for total internal reflection at the glass-water interface.

Show answer
Using New Cartesian sign convention and the relation for critical angle:

When light travels from a denser medium (glass) to a rarer medium (water), total internal reflection occurs when the angle of incidence exceeds the critical angle C, given by:

sin C = n₂/n₁

where n₁ = refractive index of glass = 1.5 and n₂ = refractive index of water = 4/3.

Substituting:

sin C = (4/3) / (3/2) = (4/3) × (2/3) = 8/9

∴ C = sin⁻¹(8/9) ≈ 62.7° ≈ 63°
Q15Short Answer2 marks

A biconvex lens is made of glass of refractive index 1.5. Both surfaces have equal radii of curvature of 20 cm. Calculate the focal length of the lens.

Show answer
Using the Lens Maker's Formula:

1/f = (n − 1)(1/R₁ − 1/R₂)

Using New Cartesian sign convention: for a biconvex lens, R₁ = +20 cm (first surface, centre of curvature on transmission side) and R₂ = −20 cm (second surface, centre of curvature on incidence side).

Substituting values:

1/f = (1.5 − 1)(1/20 − 1/(−20))

1/f = (0.5)(1/20 + 1/20)

1/f = (0.5)(2/20) = (0.5)(1/10) = 1/20

∴ f = 20 cm
Q16Short Answer2 marks

A double-convex lens is made of glass of refractive index 1.5. Both surfaces have equal radii of curvature of 20 cm. Using the lens maker's formula, calculate the focal length of the lens.

Show answer
Using the New Cartesian sign convention, for a double-convex lens: R₁ = +20 cm, R₂ = −20 cm, n = 1.5.

By the Lens Maker's Formula:

1/f = (n − 1)(1/R₁ − 1/R₂)

Substituting the values:

1/f = (1.5 − 1)(1/20 − 1/(−20))

1/f = (0.5)(1/20 + 1/20)

1/f = (0.5)(2/20) = (0.5)(1/10) = 1/20

∴ f = 20 cm
Q17Short Answer3 marks

A glass slab of thickness 3.0 cm and refractive index 1.5 is placed on a printed page lying on a table. A student looks vertically down through the slab and observes the print. (i) By how much does the print appear to shift upward? (ii) The student now replaces the slab with a glass prism (apex angle A = 60°) made of the same glass (n = 1.5). A ray of light is incident on one face of the prism at the angle of minimum deviation. Calculate the angle of minimum deviation (δ_m). (iii) Using the result from (ii), find the speed of light inside this glass. (Given: c = 3 × 10⁸ m/s, sin 30° = 0.5, sin 45° ≈ 0.707, sin 60° ≈ 0.866)

Diagram for question 17: Ray Optics and Optical Instruments
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New Cartesian sign convention: distances measured from the refracting surface/optical centre; upward shift is a real-depth vs apparent-depth concept.

(i) Apparent Shift Due to Glass Slab [1 mark]

When a slab of thickness t and refractive index n is placed over an object, the apparent depth formula gives:

Apparent depth = t / n

Shift (upward) = Real thickness − Apparent depth
= t − t/n = t(1 − 1/n)

Substituting t = 3.0 cm, n = 1.5:

Shift = 3.0 × (1 − 1/1.5)
= 3.0 × (1 − 0.667)
= 3.0 × 0.333

∴ Apparent shift = 1.0 cm (the print appears raised by 1.0 cm)

(ii) Angle of Minimum Deviation [2 marks]

By the prism formula, the refractive index at minimum deviation is:

n = sin((A + δ_m)/2) / sin(A/2)

Here A = 60°, so sin(A/2) = sin 30° = 0.5.

Substituting n = 1.5:

1.5 = sin((60° + δ_m)/2) / 0.5

→ sin((60° + δ_m)/2) = 1.5 × 0.5 = 0.75

Now sin 45° ≈ 0.707 and sin 60° ≈ 0.866; since 0.75 lies between these,

(60° + δ_m)/2 = sin⁻¹(0.75) ≈ 48.6°

However, for an exact CBSE-standard calculation: using the standard result for n = 1.5 and A = 60°,

sin((A + δ_m)/2) = n × sin(A/2) = 1.5 × sin 30° = 1.5 × 0.5 = 0.75

(A + δ_m)/2 = 48.59° ≈ 48.6°

∴ A + δ_m = 97.2°
∴ δ_m = 97.2° − 60°

∴ δ_m ≈ 37.2° (approximately 37°)

[Alternate accepted approach: candidates may note n = √2 ≈ 1.414 gives δ_m = 30° exactly; for n = 1.5 the answer δ_m ≈ 37° is accepted with correct working.]

(iii) Speed of Light Inside Glass [1 mark]

The refractive index is defined as:

n = c / v → v = c / n

Substituting c = 3 × 10⁸ m/s, n = 1.5:

v = (3 × 10⁸) / 1.5

∴ v = 2 × 10⁸ m/s

The speed of light inside the glass prism is 2 × 10⁸ m/s.
Q18Short Answer3 marks

A wildlife photographer uses a camera fitted with a convex lens of focal length 200 mm. To photograph a distant bird, she attaches a 'teleconverter' — a concave lens of focal length 500 mm — just behind the main lens (i.e., in contact with it).

(i) Calculate the equivalent focal length of the lens combination.

(ii) The bird is perched on a branch 10 m away from the camera. Using the equivalent focal length found in (i), calculate the image distance (position of the film/sensor) at which the camera must be focused.

(iii) The photographer then removes the teleconverter and uses only the original convex lens (f = 200 mm) to image the same bird at 10 m. Compare the size of the image formed on the film in the two cases and state which arrangement gives a larger image. Justify your answer.

Diagram for question 18: Ray Optics and Optical Instruments
Show answer
New Cartesian Sign Convention: All distances are measured from the optical centre (pole) of the lens; distances in the direction of incident light are positive, and opposite to it are negative.

─────────────────────────────
Part (i): Equivalent focal length of the combination
─────────────────────────────

For two thin lenses in contact, the equivalent focal length f is given by:

1/f = 1/f₁ + 1/f₂

Here,
f₁ = +200 mm (convex lens)
f₂ = −500 mm (concave lens, negative sign by sign convention)

→ 1/f = 1/(+200) + 1/(−500)

→ 1/f = 5/1000 − 2/1000 = 3/1000

∴ f = 1000/3 mm ≈ 333 mm

─────────────────────────────
Part (ii): Image distance for the bird at 10 m
─────────────────────────────

Using the thin lens formula: 1/v − 1/u = 1/f

Given:
Object distance u = −10 m = −10000 mm (object is to the left; negative)
f = +1000/3 mm

→ 1/v = 1/f + 1/u

→ 1/v = 3/1000 + 1/(−10000)

→ 1/v = 30/10000 − 1/10000 = 29/10000

∴ v = 10000/29 mm ≈ 344.8 mm ≈ 345 mm

The film/sensor must be placed approximately 345 mm behind the combination.

─────────────────────────────
Part (iii): Comparison of image sizes
─────────────────────────────

For the convex lens alone (f₁ = 200 mm, u = −10000 mm):

→ 1/v = 1/200 + 1/(−10000) = 50/10000 − 1/10000 = 49/10000

∴ v = 10000/49 mm ≈ 204.1 mm

Linear magnification is given by m = v/u (in magnitude, for a real image):

|m|_combination = (10000/29) / 10000 = 1/29 ≈ 0.0345

|m|_convex alone = (10000/49) / 10000 = 1/49 ≈ 0.0204

Since |m|_combination > |m|_convex alone, the image formed by the combination (convex + teleconverter concave lens) is larger.

∴ The combination with the teleconverter (equivalent f ≈ 333 mm) produces a larger image of the bird on the film.

Physical reason: Attaching a concave teleconverter increases the effective focal length of the system. A longer focal length gives a greater linear magnification for a distant object, which is exactly why teleconverters are used in wildlife photography to 'zoom in' on distant subjects.
Q19Short Answer3 marks

A biology student is using a compound microscope to observe a prepared slide. The objective lens has a focal length of 0.4 cm and the eyepiece has a focal length of 5 cm. The length of the microscope tube (distance between the second focal point of the objective and the first focal point of the eyepiece) is 14 cm, and the student adjusts the eyepiece so that the final image forms at the near point (D = 25 cm) of her eye.

(i) Calculate the magnifying power of the microscope under these conditions.
(ii) The student's teacher points out that if the eyepiece is readjusted so the final image forms at infinity (image at infinity / relaxed eye), the magnifying power changes. Without detailed calculation, state whether the magnifying power increases or decreases, and give a physical reason for this change.
(iii) The student then replaces the objective lens with another convex lens of the same material but having both radii of curvature doubled. How does the focal length of the new objective change? Use the Lens Maker's equation to justify your answer.
(iv) With the new objective (from part iii), state — with justification — whether the overall magnifying power of the microscope increases or decreases compared to the original setup (with final image at near point as in part i).

Diagram for question 19: Ray Optics and Optical Instruments
Show answer
(i) Magnifying power of compound microscope:

The magnifying power when the final image is at the near point D is given by:

m = (L / f₀) × (1 + D / fe)

where L = tube length = 14 cm, f₀ = 0.4 cm, fe = 5 cm, D = 25 cm.

Substituting:

m = (14 / 0.4) × (1 + 25 / 5)
= 35 × (1 + 5)
= 35 × 6

∴ Magnifying power m = 210

(ii) When the final image is formed at infinity (relaxed eye), the magnifying power formula becomes:

m∞ = (L / f₀) × (D / fe)

Comparing with part (i): m = (L / f₀)(1 + D/fe) = (L/f₀)(D/fe) + (L/f₀)

Since (L/f₀) is a positive quantity, m > m∞.

∴ The magnifying power decreases when the final image is at infinity.

Physical reason: When the final image is at the near point, the eyepiece acts as a magnifier with the object inside its focal length, providing extra angular magnification (the "+1" term). When the image is at infinity, the eyepiece provides only D/fe magnification (no "+1" contribution), so the total magnification is smaller.

(iii) Lens Maker's equation:

1/f = (n − 1)(1/R₁ − 1/R₂)

For the original objective with radii R₁ and R₂:
1/f₀ = (n − 1)(1/R₁ − 1/R₂)

For the new objective, both radii are doubled: R₁' = 2R₁, R₂' = 2R₂:

1/f₀' = (n − 1)(1/2R₁ − 1/2R₂)
= (n − 1) × (1/2)(1/R₁ − 1/R₂)
= (1/2) × (1/f₀)

∴ f₀' = 2f₀ = 2 × 0.4 = 0.8 cm

The focal length of the new objective doubles.

(iv) With the new objective f₀' = 0.8 cm, and keeping L = 14 cm, D = 25 cm, fe = 5 cm:

m' = (L / f₀') × (1 + D / fe)
= (14 / 0.8) × 6
= 17.5 × 6
= 105

Since m' = 105 < m = 210,

∴ The overall magnifying power decreases.

Justification: The magnifying power of a compound microscope is inversely proportional to the focal length of the objective lens (m ∝ 1/f₀). Doubling f₀ halves the contribution of the objective, thereby halving the total magnifying power.
Q20Short Answer3 marks

A ray of light passes from glass (refractive index 1.5) into water (refractive index 4/3).
(a) State the two conditions necessary for total internal reflection to occur.
(b) Calculate the critical angle for the glass–water interface.
(c) If the angle of incidence at the glass–water boundary is 62°, state with reason whether the ray will undergo total internal reflection or refraction.

Show answer
(a) Conditions for Total Internal Reflection (TIR):
(i) The ray of light must travel from a denser medium to a rarer medium (i.e., from higher refractive index to lower refractive index).
(ii) The angle of incidence in the denser medium must be greater than the critical angle for that pair of media.

(b) By Snell's law at the critical angle C, the refracted ray grazes the interface (angle of refraction = 90°):

n<sub>glass</sub> sin C = n<sub>water</sub> sin 90°

∴ sin C = n<sub>water</sub> / n<sub>glass</sub> = (4/3) / (3/2) = (4/3) × (2/3) = 8/9

∴ C = sin<sup>−1</sup>(8/9) = sin<sup>−1</sup>(0.889) ≈ 62.7°

(c) The angle of incidence given is 62°.
Since 62° < 62.7° (critical angle), the condition i > C is NOT satisfied.
∴ The ray will undergo refraction (it will pass from glass into water), not total internal reflection.
Q21Short Answer3 marks

An underwater photographer is working inside a large aquarium tank. She uses a thin plano-convex glass lens (refractive index n<sub>g</sub> = 1.5) with its curved surface facing the object. The radius of curvature of the curved surface is 20 cm. In air, this lens has a well-known focal length. However, when the lens is completely submerged in water (refractive index n<sub>w</sub> = 4/3), both the object and the image space are in water.

(i) Using the Lens Maker's formula, calculate the focal length of this plano-convex lens when it is used in air.
(ii) Calculate the focal length of the same lens when it is completely submerged in water.
(iii) An object is placed 90 cm in front of the submerged lens. Using New Cartesian sign convention and the focal length found in part (ii), find the position of the image formed by the submerged lens and state its nature.
(iv) The photographer observes that the effective power of her underwater camera system changes drastically compared to in air. Give a physical reason why the focal length of a convex lens increases when it is submerged in water, even though the lens geometry remains unchanged.

Diagram for question 21: Ray Optics and Optical Instruments
Show answer
New Cartesian Sign Convention: All distances measured from the optical centre (pole); distances in the direction of incident light are positive, opposite are negative.

(i) Focal length of plano-convex lens in AIR

By the Lens Maker's Formula:
1/f = (n<sub>rel</sub> − 1)(1/R<sub>1</sub> − 1/R<sub>2</sub>)

For a plano-convex lens with curved surface facing the object:
R<sub>1</sub> = +20 cm (curved surface), R<sub>2</sub> = ∞ (flat surface)
n<sub>rel</sub> = n<sub>g</sub>/n<sub>air</sub> = 1.5/1 = 1.5

1/f<sub>air</sub> = (1.5 − 1)(1/20 − 1/∞)
→ 1/f<sub>air</sub> = (0.5)(1/20 − 0)
→ 1/f<sub>air</sub> = 0.5/20 = 1/40

∴ f<sub>air</sub> = +40 cm

(ii) Focal length of the same lens SUBMERGED in water

When the lens is in water, n<sub>rel</sub> = n<sub>g</sub>/n<sub>w</sub> = 1.5/(4/3) = 1.5 × (3/4) = 9/8

Applying the Lens Maker's Formula again (same geometry: R<sub>1</sub> = +20 cm, R<sub>2</sub> = ∞):
1/f<sub>water</sub> = (n<sub>rel</sub> − 1)(1/R<sub>1</sub> − 1/R<sub>2</sub>)
→ 1/f<sub>water</sub> = (9/8 − 1)(1/20 − 0)
→ 1/f<sub>water</sub> = (1/8)(1/20)
→ 1/f<sub>water</sub> = 1/160

∴ f<sub>water</sub> = +160 cm

(iii) Position and nature of image for submerged lens (object at u = −90 cm)

Using the Thin Lens Formula:
1/v − 1/u = 1/f

Given: u = −90 cm, f = +160 cm

1/v = 1/f + 1/u
→ 1/v = 1/160 + 1/(−90)
→ 1/v = 1/160 − 1/90
→ 1/v = (90 − 160)/(160 × 90)
→ 1/v = −70/14400
→ 1/v = −7/1440

∴ v = −1440/7 ≈ −205.7 cm ≈ −205.7 cm

Nature of image:
Since v is negative, the image is formed on the same side as the object (i.e., on the side of the incident light). The image is virtual and erect.

Magnification: m = v/u = (−205.7)/(−90) ≈ +2.3 (magnified)

∴ The image is formed approximately 205.7 cm in front of the lens (on the same side as the object); the image is virtual, erect, and magnified.

(iv) Physical reason for increased focal length in water

The ability of a lens to converge or diverge light depends on the RELATIVE refractive index between the lens material and the surrounding medium, not on the absolute refractive index of the lens alone.

When the lens is in air, the relative refractive index n<sub>rel</sub> = 1.5/1 = 1.5, giving a large bending of light at each surface (large refraction).

When the lens is submerged in water, n<sub>rel</sub> = 1.5/(4/3) = 1.125, which is much closer to 1. The contrast between the lens and the surrounding medium is greatly reduced, so light rays are bent through a much smaller angle at each surface.

Since the refracting power at each surface decreases, the lens converges parallel rays at a much farther point — hence the focal length increases dramatically (from 40 cm to 160 cm). In general, if n<sub>rel</sub> → 1, f → ∞ and the lens loses all refracting power.
Q22Short Answer3 marks

A student sets up a simple camera using a convex lens of focal length 10 cm to photograph a tree that is 50 m away from the lens.

(i) At what distance behind the lens should the photographic film be placed to get a sharp image of the tree?
(ii) If the tree is 20 m tall, what is the height of the image formed on the film?
(iii) State whether the image formed is real or virtual, and erect or inverted.
(iv) The student then moves closer and photographs a flower placed 15 cm in front of the same lens. Calculate the new image distance.

Diagram for question 22: Ray Optics and Optical Instruments
Show answer
Using New Cartesian Sign Convention: distances measured from the optical centre of the lens; distances in the direction of incident light are positive, and opposite are negative.

Given: f = +10 cm (convex lens)

─────────────────────────────────
(i) Position of the image of the tree:
─────────────────────────────────
By the Lens Formula: 1/v − 1/u = 1/f

u = −5000 cm (tree is 50 m = 5000 cm in front of the lens)

Substituting:
1/v − 1/(−5000) = 1/10
1/v + 1/5000 = 1/10
1/v = 1/10 − 1/5000 = 500/5000 − 1/5000 = 499/5000

∴ v = 5000/499 ≈ 10.02 cm

The film should be placed ≈ 10.02 cm behind the lens (on the other side of the object).

─────────────────────────────────
(ii) Height of the image on the film:
─────────────────────────────────
Magnification m = v/u

m = (+10.02)/(−5000) = −0.002004 ≈ −1/500

Height of image h' = m × h
h' = (−1/500) × 20 m = −0.04 m = −4 cm

∴ The height of the image on the film is 4 cm (the negative sign indicates it is inverted).

─────────────────────────────────
(iii) Nature of the image:
─────────────────────────────────
Since v is positive (image forms on the opposite side of the lens from the object), the image is REAL and INVERTED.

─────────────────────────────────
(iv) Image distance for the flower at u = −15 cm:
─────────────────────────────────
By the Lens Formula: 1/v − 1/u = 1/f

1/v − 1/(−15) = 1/10
1/v + 1/15 = 1/10
1/v = 1/10 − 1/15 = 3/30 − 2/30 = 1/30

∴ v = +30 cm

The image of the flower is formed 30 cm behind the lens.
Q23Short Answer3 marks

A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 30 cm. An object is placed 60 cm from this lens combination.
(i) Find the equivalent focal length of the combination.
(ii) Find the position of the final image formed by the combination.
(iii) State whether the image is real or virtual.

Diagram for question 23: Ray Optics and Optical Instruments
Show answer
(i) Finding equivalent focal length:

When two thin lenses are placed in contact, the equivalent power is:
P = P₁ + P₂

∴ 1/F = 1/f₁ + 1/f₂

Here, f₁ = +20 cm (convex lens) and f₂ = −30 cm (concave lens).

→ 1/F = 1/20 + 1/(−30)

→ 1/F = 3/60 − 2/60 = 1/60

∴ F = +60 cm

The equivalent focal length of the combination is +60 cm. [1 mark]

(ii) Finding the position of the final image:

Using New Cartesian sign convention: all distances are measured from the optical centre of the lens combination; distances in the direction of incident light are positive.

For the combination (treated as a single lens), the lens formula is:
1/v − 1/u = 1/F

Given: u = −60 cm (object on the left), F = +60 cm.

→ 1/v − 1/(−60) = 1/60

→ 1/v + 1/60 = 1/60

→ 1/v = 1/60 − 1/60 = 0

∴ v → ∞

∴ The final image is formed at infinity. [1 mark]

(iii) Nature of the image:

Since the image is formed at infinity (rays emerge as a parallel beam), it is a real image formed on the same side as the outgoing light.

∴ The image is real and formed at infinity (the object is placed at the focus of the equivalent combination). [1 mark]
Q24Short Answer3 marks

A wildlife photographer uses a concave mirror of focal length 20 cm as a reflector behind a flash lamp to concentrate light onto a distant subject. While testing the setup indoors, she notices that when a small bulb (treated as a point source) is placed at 30 cm in front of the mirror, a bright patch of light forms on a wall behind the bulb.

(i) Using the New Cartesian sign convention and the mirror formula, find the position and nature of the image formed by the mirror. (2 marks)

(ii) The photographer now wants the reflected beam to be perfectly parallel (i.e., the image should form at infinity). Where should she place the bulb? Justify using the mirror formula. (1 mark)

(iii) In the original position (object at 30 cm), the magnification produced by the mirror is calculated. State whether the image is magnified or diminished and whether it is real or virtual. (1 mark)

Diagram for question 24: Ray Optics and Optical Instruments
Show answer
New Cartesian Sign Convention: All distances are measured from the pole P of the mirror. Distances measured in the direction of incident light are positive; opposite to incident light are negative.

Given: Concave mirror, focal length f = −20 cm (concave mirror, negative sign)
Object distance u = −30 cm (object in front of mirror, negative sign)

─────────────────────────────────────
Part (i): Position and nature of image (2 marks)
─────────────────────────────────────

By the mirror formula:

1/v + 1/u = 1/f

Substituting:

1/v + 1/(−30) = 1/(−20)

1/v = 1/(−20) − 1/(−30)

1/v = −1/20 + 1/30

1/v = (−3 + 2)/60

1/v = −1/60

∴ v = −60 cm

The image is formed 60 cm in front of the mirror (on the same side as the object).

Magnification:
m = −v/u = −(−60)/(−30) = −2

∴ The image is real (v is negative, formed in front of the mirror), inverted, and magnified (|m| = 2 > 1). This is consistent with the bright patch observed on the wall behind the bulb — the reflected rays converge and then diverge to form the patch on the wall beyond the bulb.

─────────────────────────────────────
Part (ii): Object position for parallel reflected beam (1 mark)
─────────────────────────────────────

For the reflected beam to be perfectly parallel, the image must form at infinity, i.e., v → ∞.

By the mirror formula:
1/v + 1/u = 1/f

As v → ∞, 1/v → 0:
0 + 1/u = 1/f
u = f = −20 cm

∴ The bulb must be placed at the focus, i.e., 20 cm in front of the concave mirror.

Justification: When the object is placed at the focus of a concave mirror, incident rays after reflection emerge as a parallel beam (image at infinity). This is the principle used in reflector torches and flash concentrators.

─────────────────────────────────────
Part (iii): Nature of image at u = −30 cm (1 mark)
─────────────────────────────────────

From Part (i): m = −2

Since |m| = 2 > 1 → the image is magnified.
Since m is negative → the image is inverted.
Since v = −60 cm (negative, in front of mirror) → the image is real.

∴ The image is real, inverted, and magnified (twice the size of the object).
Q25Short Answer3 marks

A school science exhibition features a display where a beam of white light passes through a glass prism (refractive index for red light n_r = 1.514 and for violet light n_v = 1.523, prism angle A = 60°) and produces a beautiful spectrum on a screen. A student observes that the violet light bends more than the red light.

(i) State the phenomenon responsible for the splitting of white light into its component colours by the prism.
(ii) Using the formula for angle of deviation at minimum deviation condition, calculate the angle of minimum deviation (δ_m) for RED light in this prism. [Given: sin 30° = 0.5, sin⁻¹(0.757) = 49.2°]
(iii) Explain in one sentence WHY violet light deviates more than red light in a glass prism.
(iv) If the same prism is immersed in a liquid of refractive index 1.50, predict with reason whether the dispersion of white light will INCREASE, DECREASE, or remain the SAME compared to when the prism is in air.

Diagram for question 25: Ray Optics and Optical Instruments
Show answer
ANSWER (CBQ — Ray Optics: Dispersion by Prism) [4 marks]

(i) The phenomenon is DISPERSION of light.
Dispersion is the splitting of white light into its constituent colours (wavelengths) when it passes through a prism, due to different colours having different speeds (and hence different refractive indices) in the glass medium. [½ mark]

(ii) Calculation of angle of minimum deviation for RED light:

Using the prism formula at minimum deviation:

n = sin((A + δ_m) / 2) / sin(A / 2)

For red light: n_r = 1.514, A = 60°

→ sin(A/2) = sin 30° = 0.500

→ 1.514 = sin((60° + δ_m) / 2) / 0.500

→ sin((60° + δ_m) / 2) = 1.514 × 0.500 = 0.757

→ (60° + δ_m) / 2 = sin⁻¹(0.757) = 49.2°

→ 60° + δ_m = 98.4°

∴ δ_m (red) = 98.4° − 60° = 38.4° [1½ marks]

(iii) Reason for violet deviating more than red:

The refractive index of glass is greater for violet light (shorter wavelength) than for red light (longer wavelength) — n_v > n_r — because violet light travels more slowly in glass and is bent (refracted) more at each surface of the prism.
[1 mark]

(iv) Effect of immersing the prism in liquid (n_liquid = 1.50):

The dispersion will DECREASE (reduce significantly).

Reason: The angle of deviation of any ray through a prism depends on the RELATIVE refractive index n_relative = n_glass / n_medium. When the prism is surrounded by a liquid of refractive index 1.50 (close to both n_r = 1.514 and n_v = 1.523), the relative refractive indices for red and violet become very close to 1:

n_relative(red) = 1.514 / 1.50 ≈ 1.009
n_relative(violet) = 1.523 / 1.50 ≈ 1.015

The difference (n_v − n_r) effectively reduces from 0.009 (in air) to 0.006 (in liquid), so the angular spread between red and violet — i.e., the angular dispersion — decreases markedly. The prism loses most of its dispersive power when immersed in the liquid.
[1 mark]

∴ Dispersion DECREASES when the prism is immersed in the given liquid.
Q26Short Answer3 marks

A wildlife photographer uses a compound microscope to examine a thin biological sample. The objective lens has focal length f₀ = 0.8 cm and the eyepiece has focal length fₑ = 2.5 cm. The tube length (distance between second focal point of objective and first focal point of eyepiece) is L = 12 cm. The final image is formed at the near point (D = 25 cm) of the observer's eye.

(i) Write the expression for the total magnification of a compound microscope when the final image is formed at the near point, identifying each symbol.

(ii) Calculate the magnification produced by the objective lens alone.

(iii) Calculate the total magnification of the microscope.

(iv) The photographer now switches to a higher-power objective of focal length f₀ = 0.4 cm, keeping all other parameters unchanged. Without full calculation, state and justify whether the total magnification will increase, decrease, or remain the same.

Show answer
(i) The total magnification of a compound microscope when the final image is formed at the near point is:

m = m₀ × mₑ = (L / f₀) × (1 + D / fₑ)

where:
• L = tube length (distance between second focal point of objective and first focal point of eyepiece)
• f₀ = focal length of objective lens
• fₑ = focal length of eyepiece
• D = least distance of distinct vision (near point) = 25 cm
• m₀ = magnification by objective; mₑ = magnification by eyepiece

[1 mark]

(ii) Magnification produced by objective lens alone:

m₀ = L / f₀

Substituting: m₀ = 12 cm / 0.8 cm

∴ m₀ = 15

[1 mark]

(iii) Total magnification:

First, magnification by eyepiece:
mₑ = 1 + D / fₑ = 1 + 25 / 2.5 = 1 + 10 = 11

Total magnification:
m = m₀ × mₑ = 15 × 11

∴ m = 165

[1 mark]

(iv) The total magnification will INCREASE.

Justification: From the expression m = (L / f₀)(1 + D / fₑ), the total magnification is inversely proportional to f₀. When f₀ is reduced from 0.8 cm to 0.4 cm (i.e., halved), the objective magnification m₀ doubles from 15 to 30, since m₀ = L/f₀. All other factors (L, fₑ, D) remain unchanged, so the total magnification doubles to 330.

Physical reason: A shorter focal length objective brings its rear focal point closer to the lens, so the real image of the object forms further inside the tube — i.e., the intermediate image is more magnified — resulting in a larger final magnification.

∴ Total magnification increases (doubles) when f₀ is halved.

[1 mark]
Q27Short Answer3 marks

A nature photographer uses a compound microscope to observe a tiny insect. The objective lens has a focal length of 0.8 cm and the eyepiece has a focal length of 5 cm. The insect is placed 1.0 cm in front of the objective lens. The final image is formed at the near point of the eye (D = 25 cm).

(i) Where is the intermediate image formed by the objective lens? (Show working.)
(ii) What is the magnification produced by the objective lens alone?
(iii) What is the magnification produced by the eyepiece alone (image at near point)?
(iv) The photographer then replaces the eyepiece with one of focal length 2.5 cm. State and explain how the total magnifying power of the microscope changes.

Diagram for question 27: Ray Optics and Optical Instruments
Show answer
Using New Cartesian Sign Convention: distances measured from the optical centre of each lens; distances in the direction of incident light are positive.

(i) Image formed by the objective lens:

By the lens formula: 1/v − 1/u = 1/f

For the objective lens: u = −1.0 cm, f<sub>o</sub> = +0.8 cm

1/v = 1/f<sub>o</sub> + 1/u = 1/0.8 + 1/(−1.0)

1/v = 1.25 − 1.00 = 0.25

∴ v = +4.0 cm

The intermediate image is formed 4.0 cm on the other side of the objective lens (real and inverted).

(ii) Magnification by the objective lens:

m<sub>o</sub> = v/u = 4.0/(−1.0)

∴ m<sub>o</sub> = −4 (magnitude 4; negative sign indicates inverted image)

(iii) Magnification by the eyepiece (image at near point D = 25 cm):

For a simple lens used as a magnifier with the final image at the near point:

m<sub>e</sub> = 1 + D/f<sub>e</sub> = 1 + 25/5

∴ m<sub>e</sub> = +6

(iv) Effect of replacing the eyepiece (f<sub>e</sub> = 2.5 cm):

New magnification of eyepiece: m<sub>e</sub>' = 1 + D/f<sub>e</sub>' = 1 + 25/2.5 = 1 + 10 = 11

Original total magnifying power: |M| = |m<sub>o</sub>| × m<sub>e</sub> = 4 × 6 = 24

New total magnifying power: |M'| = |m<sub>o</sub>| × m<sub>e</sub>' = 4 × 11 = 44

The total magnifying power increases (from 24 to 44). This is because the eyepiece magnification m<sub>e</sub> = 1 + D/f<sub>e</sub> is inversely related to f<sub>e</sub>; a shorter focal length eyepiece produces greater angular magnification, thereby increasing the overall magnifying power of the microscope.
Q28Short Answer3 marks

(a) State the two conditions necessary for total internal reflection.

(b) A ray of light travels from glass (refractive index n = 1.5) into water (refractive index n = 1.33). Calculate the critical angle for this glass–water interface.

Show answer
(a) The two conditions necessary for total internal reflection are:

(i) The ray of light must travel from a denser medium into a rarer medium (i.e., from a medium of higher refractive index to one of lower refractive index).

(ii) The angle of incidence in the denser medium must be greater than the critical angle (i > C) for that pair of media.

[1 mark — both conditions stated]

(b) By Snell's law applied at the critical angle condition:

At the critical angle C, the refracted ray grazes the interface (angle of refraction = 90°).

Snell's law: n₁ sin C = n₂ sin 90°

∴ sin C = n₂ / n₁

Here, the ray travels from glass (n₁ = 1.5) into water (n₂ = 1.33).

Substituting:

sin C = 1.33 / 1.5 = 0.8867

∴ C = sin⁻¹(0.8867)

∴ C ≈ 62.5°

[½ mark — correct formula; ½ mark — correct substitution; ½ mark — ∴ C ≈ 62.5°; ½ mark — identifying glass as denser medium with correct ratio]

Note: Since n_glass > n_water, glass is the denser medium, so total internal reflection is possible at this interface when i > 62.5°.
Q29Short Answer3 marks

A monochromatic ray of light undergoes minimum deviation through an equilateral glass prism. The angle of minimum deviation is found to be equal to the angle of the prism.

(a) Calculate the refractive index of the material of the prism.
(b) What is the angle of refraction inside the prism at minimum deviation?
(c) If the same prism is immersed in a liquid of refractive index 1.4, will total internal reflection be possible at the prism face for this ray? Justify your answer.

Diagram for question 29: Ray Optics and Optical Instruments
Show answer
Given: Equilateral prism → angle of prism A = 60°.
Angle of minimum deviation δ<sub>m</sub> = A = 60°.

(a) Refractive index of the prism material:

By the prism formula at minimum deviation:
n = sin((A + δ<sub>m</sub>)/2) / sin(A/2)

Substituting A = 60°, δ<sub>m</sub> = 60°:
n = sin((60° + 60°)/2) / sin(60°/2)
n = sin 60° / sin 30°
n = (√3/2) / (1/2)

∴ n = √3 ≈ 1.732

(b) Angle of refraction inside the prism at minimum deviation:

At minimum deviation, the refracted ray inside the prism is parallel to the base, and both refracting angles are equal:
r₁ = r₂ = r = A/2

∴ r = 60°/2 = 30°

(c) Whether TIR is possible when the prism is immersed in liquid of refractive index n<sub>L</sub> = 1.4:

For TIR at the prism–liquid interface, the angle of incidence at that face must exceed the critical angle C.

At minimum deviation r₁ = r₂ = 30°, so the ray strikes each face at an angle of incidence of 30° (measured from the normal inside the prism).

The critical angle C for the prism–liquid interface is given by:
sin C = n<sub>L</sub> / n = 1.4 / √3 = 1.4 / 1.732 ≈ 0.808
C ≈ 53.9° ≈ 54°

Since the angle of incidence at the face (30°) is less than the critical angle (≈ 54°), the condition for TIR (i > C) is NOT satisfied.

∴ Total internal reflection will NOT occur at the prism face when it is immersed in the liquid of refractive index 1.4.
Q30Short Answer3 marks

A biconvex lens made of glass (refractive index n = 1.5) has both radii of curvature equal to 30 cm. A student places this lens in contact with a concave lens of unknown focal length. She observes that when a luminous object is placed 40 cm in front of the combination, the final image is formed at 120 cm on the same side as the object.

(i) Using the Lens Maker's formula, find the focal length of the biconvex lens.
(ii) Find the focal length of the concave lens in the combination.
(iii) Find the power of the combination and state whether the combination behaves as a converging or a diverging lens.
(iv) If the concave lens is now removed and only the biconvex lens is kept, determine the position of the image of the same object (placed 40 cm away) and state the nature of the image formed.

Diagram for question 30: Ray Optics and Optical Instruments
Show answer
(i) Focal length of the biconvex lens:

By the Lens Maker's Formula:
1/f = (n − 1)(1/R₁ − 1/R₂)

Using New Cartesian Sign Convention:
For a biconvex lens, R₁ = +30 cm, R₂ = −30 cm, n = 1.5

1/f₁ = (1.5 − 1)(1/30 − 1/(−30))
= (0.5)(1/30 + 1/30)
= (0.5)(2/30)
= 1/30

∴ f₁ = +30 cm

(ii) Focal length of the concave lens:

For the combination, u = −40 cm (object on left), v = −120 cm (image on same side as object → virtual, so v = −120 cm)

Using lens formula for the combination:
1/f_comb = 1/v − 1/u
= 1/(−120) − 1/(−40)
= −1/120 + 1/40
= −1/120 + 3/120
= 2/120
= 1/60

∴ f_comb = +60 cm

For lenses in contact:
1/f_comb = 1/f₁ + 1/f₂
1/60 = 1/30 + 1/f₂
1/f₂ = 1/60 − 1/30 = 1/60 − 2/60 = −1/60

∴ f₂ = −60 cm (concave lens, as expected)

(iii) Power of the combination:

P = 1/f_comb (in metres)
= 1/(0.60 m)

∴ P = +1.67 D

Since P > 0 (and f_comb = +60 cm > 0), the combination behaves as a converging lens.

(iv) Image position using only the biconvex lens (f = +30 cm, u = −40 cm):

1/v − 1/u = 1/f
1/v − 1/(−40) = 1/30
1/v + 1/40 = 1/30
1/v = 1/30 − 1/40 = 4/120 − 3/120 = 1/120

∴ v = +120 cm

The image is formed 120 cm on the other side of the lens.

Nature of image: Since v is positive (real side) and u is negative, the image is real and inverted.
Magnification m = v/u = 120/(−40) = −3
|m| = 3 > 1, so the image is magnified (enlarged).

∴ The image is real, inverted, and magnified, formed at 120 cm from the biconvex lens on the side opposite to the object.

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