PhysicsCBSE Class 12 · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 12 Physics sample paper 2026-27: question 32(A)

From the official CBSE sample question paper for 2026-27, section E, question 32(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(A) · 5 marks · Long answerRay Optics and Optical Instruments

A compound microscope consists of an objective lens of focal length 2 cm and an eyepiece lens of focal length 5 cm. The object is placed 2.2 cm from the objective lens.

I. Calculate the position of the image formed by the objective lens.
II. The eyepiece is so placed that the final image is formed at infinity. Find the magnifying power of the microscope.
III. Explain briefly why the compound microscope provides much higher magnification compared to a simple magnifier.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

Treat the objective as a single convex lens and apply $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ with the object distance taken as negative.

Hint 2

For a final image at infinity, the eyepiece acts as a simple magnifier with magnification $D / f_{e}$ .

Hint 3

Multiply the objective's magnification $v_{o} /∣ u_{o} ∣$ by the eyepiece's magnification.

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. Image by the objective. Use the lens formula with the sign convention: $f_{o} = + 2$ cm and $u_{o} = - 2.2$ cm (the object is on the incident side).
$\frac{1}{f _{o}} = \frac{1}{v _{o}} - \frac{1}{u _{o}} \Rightarrow \frac{1}{2} = \frac{1}{v _{o}} + \frac{1}{2.2}$
1 mark
$\frac{1}{v _{o}} = \frac{1}{2} - \frac{1}{2.2} = \frac{2.2 - 2}{4.4} = \frac{0.2}{4.4} = \frac{1}{22}$
So $v_{o} = + 22$ cm. A real, inverted image forms 22 cm from the objective, on the side opposite to the object.
1 mark
II. Magnifying power, final image at infinity. For normal adjustment of a compound microscope:
$M = m_{o} \times m_{e} = \frac{v _{o}}{∣ u _{o} ∣} \cdot \frac{D}{f _{e}}$ , with $D = 25$ cm.
1 mark
$M = \frac{22}{2.2} \times \frac{25}{5} = 10 \times 5 = 50$
The magnifying power of the microscope is 50.
1 mark
III. Why the magnification is higher.
- In a compound microscope the objective first forms a large, real, magnified image of the tiny object.
- The eyepiece then magnifies this image again.
- The total magnification is the product $m_{o} \times m_{e}$ , so it is far larger than what a single lens (simple magnifier) can give.
1 mark

Answer: I. $v_{o} = 22$ cm from the objective (real image); II. $M = 50$ ; III. The total magnification is the product of the objective and eyepiece magnifications.

Where the marks are

Applies the lens formula with correct signs ( $u_{o} = - 2.2$ cm, $f_{o} = 2$ cm) (1 mark)
Obtains $v_{o} = 22$ cm (1 mark)
Writes $M = \frac{v _{o}}{∣ u _{o} ∣} \cdot \frac{D}{f _{e}}$ for the final image at infinity (1 mark)
Calculates $M = 50$ (1 mark)
Explains the two-stage magnification: total $M = m_{o} \times m_{e}$ (1 mark)

Revise the chapter

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 2026-27: question 32(A)

Hints

Step-by-step solution

Revise the chapter

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