PhysicsCBSE Class 12 · Board exam 2023, set 55/1/1 · 5 marks · Long answer

CBSE Class 12 Physics 2023: question 33(b)

Set in the CBSE Class 12 board exam of 2023, paper 55/1/1, section D, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerRay Optics and Optical Instruments

(i) A spherical surface of radius of curvature R separates two media of refractive indices $n_{1}$ and $n_{2}$ . A point object is placed in front of the surface at distance u in medium of refractive index $n_{1}$ and its image is formed by the surface at distance v, in the medium of refractive index $n_{2}$ . Derive a relation between u and v.
(ii) A solid glass sphere of radius $6 \cdot 0$ cm has a small air bubble trapped at a distance $3 \cdot 0$ cm from its centre C as shown in the figure. The refractive index of the material of the sphere is $1 \cdot 5$ . Find the apparent position of this bubble when seen through the surface of the sphere from an outside point E in air.

Figure in the paper: Circle (glass sphere) with centre O; point C (air bubble) lies on the horizontal line at 3 cm from O on the right; point E is outside the sphere on the left on the same line through O and C. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 55/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Use exterior angles of the triangles formed by the object, the image, the centre and the point of incidence. Use the small-angle approximation.

Hint 2

For (ii), light travels from glass into air, so n₁ = 1.5 and n₂ = 1.

Hint 3

Measure u and R from the surface facing E. Both the bubble and the centre lie on the side opposite to the observer, so both are negative.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Figure and assumptions. A convex spherical surface has pole M and centre of curvature C, with medium $n_{1}$ on the object side and $n_{2}$ on the other side. The object O lies on the principal axis. A ray ON strikes the surface at N, where NC is the normal. It refracts and meets the axis at the image I. The aperture is small, so all angles are small and $tan θ \approx sin θ \approx θ$ . The distance of N from the axis is taken as MN.
1 mark
In $△ N O C$ , the angle of incidence $i$ is an exterior angle:
$i = ∠ N O M + ∠ N C M \approx \frac{M N}{O M} + \frac{M N}{M C}$ ...(1)
½ mark
In $△ N C I$ , $∠ N C M$ is an exterior angle, so $∠ N C M = r + ∠ N I M$ . Hence
$r = ∠ N C M - ∠ N I M \approx \frac{M N}{M C} - \frac{M N}{M I}$ ...(2)
½ mark
By Snell's law, $n_{1} sin i = n_{2} sin r$ . For small angles this becomes $n_{1} i = n_{2} r$ .
Substituting (1) and (2) and cancelling MN:
$\frac{n _{1}}{O M} + \frac{n _{2}}{M I} = \frac{n _{2} - n _{1}}{M C}$
½ mark
Apply the Cartesian sign convention: $O M = - u$ , $M I = + v$ , $M C = + R$ . This gives
$\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$
½ mark
(ii) Data. The bubble is seen through the nearer surface (the one facing E). Light goes from the bubble (glass) into air, so $n_{1} = 1.5$ and $n_{2} = 1$ .
The bubble is 3.0 cm inside that surface, and the centre lies on the same side as the bubble. Hence $u = - 3$ cm and $R = - 6$ cm.
½ mark
Substituting:
$\frac{1}{v} - \frac{1.5}{- 3} = \frac{1 - 1.5}{- 6}$
½ mark
$\frac{1}{v} = \frac{0.5}{6} - \frac{1.5}{3} = \frac{0.5 - 3}{6} = - \frac{2.5}{6}$
½ mark
$v = - 2.4$ cm.
The bubble appears 2.4 cm from the surface, inside the sphere, on the same side as the bubble. It therefore looks closer to the surface than it really is.
½ mark

Answer: (i) $\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$ ; (ii) v = −2.4 cm, i.e. the bubble appears 2.4 cm from the surface, inside the sphere

Where the marks are

Diagram with small-aperture assumption (1 mark)
Expression for i using the exterior-angle property (½ mark)
Expression for r (½ mark)
Snell's law for small angles and substitution (½ mark)
Sign convention and final relation n₂/v − n₁/u = (n₂ − n₁)/R (½ mark)
Correct values n₁ = 1.5, n₂ = 1, u = −3 cm, R = −6 cm (½ mark)
Correct substitution (½ mark)
Correct simplification (½ mark)
v = −2.4 cm, inside the sphere (½ mark)

Revise the chapter

Chapter: Ray Optics and Optical Instruments (notes, formulas, where marks are lost and every past question)
Topic: Refraction at Spherical Surfaces and Lenses

Practise this chapterBoard-style questions from Ray Optics and Optical Instruments, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 55/1/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Physics 2023: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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