PhysicsCBSE Class 12 · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 12 Physics sample paper 2024-25: question 33(I)

From the official CBSE sample question paper for 2024-25, section E, question 33(I), for 5 marks. Try it first; the hints and our step-by-step solution are below.

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

a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is angular magnification of the telescope in normal adjustment?
b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 1 0^{6}$ m, and the radius of lunar orbit is $3.8 \times 1 0^{8}$ m.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

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

Hint 1

Convert both focal lengths to the same unit before dividing.

Hint 2

The moon is very far away, so where does the objective form its image?

Hint 3

The image and the moon subtend the same angle at the objective; write that angle in two ways.

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.

(a) Given $f_{o} = 15$ m and $f_{e} = 1.0$ cm $= 0.01$ m.
In normal adjustment: $M = \frac{f _{o}}{f _{e}} = \frac{15}{0.01} = 1500$
1 mark
(b) Angle subtended by the moon at the objective:
$α = \frac{diameter of moon}{radius of lunar orbit} = \frac{3.48 \times 1 0 ^{6}}{3.8 \times 1 0 ^{8}}$ rad
1½ marks
The moon is effectively at infinity, so the objective forms its image in the focal plane. If $d$ is the image diameter, it subtends the same angle at the objective:
$α = \frac{d}{f _{o}} = \frac{d}{15}$
1½ marks
Equating the two expressions:
$d = \frac{3.48 \times 1 0 ^{6}}{3.8 \times 1 0 ^{8}} \times 15 \approx 0.137$ m $= 13.7$ cm
1 mark

Answer: (a) 1500; (b) diameter of moon's image ≈ 0.137 m = 13.7 cm

Where the marks are

Uses $M = f_{o} / f_{e}$ to get 1500 (1 mark)
Angle subtended by moon = diameter/orbit radius (1.5 marks)
Image angle = $d / f_{o}$ with image in focal plane (1.5 marks)
Correct result $d \approx 13.7$ cm (1 mark)

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Chapter: Ray Optics and Optical Instruments (notes, formulas, where marks are lost and every past question)

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CBSE Class 12 Physics sample paper 2024-25: question 33(I)

Hints

Step-by-step solution

Revise the chapter

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