PhysicsCBSE Class 12 · Sample paper, session 2025-26 · 3 marks · Short answer

CBSE Class 12 Physics sample paper 2025-26: question 25

From the official CBSE sample question paper for 2025-26, section C, question 25, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q25 · 3 marks · Short answerRay Optics and Optical Instruments

A compound microscope consists of an objective lens of focal length 0.82 cm and an eyepiece lens of focal length 2.9 cm. An object is placed 0.91 cm from the objective lens. The image is formed at the near point (25 cm) from the eye.
(I) Calculate that the angular magnification of the microscope.
(II) Draw the ray diagram of compound microscope in normal adjustment.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

First locate the image formed by the objective using the lens formula.

Hint 2

For final image at near point, eyepiece magnification is $1 + D / f_{e}$ .

Hint 3

In normal adjustment the intermediate image lies at the focus of the eyepiece.

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) Objective: $\frac{1}{v _{o}} - \frac{1}{u _{o}} = \frac{1}{f _{o}}$ with $u_{o} = - 0.91$ cm, $f_{o} = 0.82$ cm: $\frac{1}{v _{o}} = \frac{1}{0.82} - \frac{1}{0.91} = 1.2195 - 1.0989 = 0.1206 \Rightarrow v_{o} \approx 8.3$ cm.
1 mark
Image at near point: $M = m_{o} m_{e} = \frac{v _{o}}{u _{o}} (1 + \frac{D}{f _{e}}) = \frac{8.3}{- 0.91} (1 + \frac{25}{2.9}) = (- 9.12) (9.62) \approx - 87.7$ (magnitude ≈ 88, image inverted).
1 mark
(II) Ray diagram (normal adjustment): object AB placed just beyond $F_{o}$ of the objective; objective forms a real, inverted, magnified image A′B′ exactly at the focus $F_{e}$ of the eyepiece; rays from A′B′ emerge parallel from the eyepiece, so the final, magnified image is formed at infinity (inverted with respect to the object). Arrows on rays, both lenses and foci labelled.
1 mark

Answer: (I) $v_{o} \approx 8.3$ cm, M ≈ −87.7; (II) ray diagram with final image at infinity

Where the marks are

Finds image distance of objective ≈ 8.3 cm (1 mark)
Magnification $M = (v_{o} / u_{o}) (1 + D / f_{e}) \approx - 87.7$ (1 mark)
Correct labelled ray diagram in normal adjustment (1 mark)

Revise the chapter

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

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 paperSample paper 2025-26: every question, by section.

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CBSE Class 12 Physics sample paper 2025-26: question 25

Hints

Step-by-step solution

Revise the chapter

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