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

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

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

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

A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at
a) the least distance of distinct vision (25 cm) and
b) infinity? What is the magnifying power of the microscope in each case?

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

Work backwards: start with the eyepiece, where you know where the final image must be.

Hint 2

The objective's image distance equals the lens separation minus the eyepiece object distance.

Hint 3

For infinity, the intermediate image lies at the eyepiece focus; use $M = (v_{o} /∣ u_{o} ∣) (D / f_{e})$ .

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) Final image at D = 25 cm
For the eyepiece: $v_{e} = - 25$ cm, $f_{e} = 6.25$ cm. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ :
$\frac{1}{u _{e}} = \frac{1}{v _{e}} - \frac{1}{f _{e}} = - \frac{1}{25} - \frac{1}{6.25} = - \frac{1}{5}$
½ mark
$u_{e} = - 5$ cm. So the image formed by the objective lies 5 cm from the eyepiece.
Its distance from the objective is $v_{o} = 15 - 5 = 10$ cm.
½ mark
For the objective: $v_{o} = + 10$ cm, $f_{o} = 2$ cm.
$\frac{1}{u _{o}} = \frac{1}{v _{o}} - \frac{1}{f _{o}} = \frac{1}{10} - \frac{1}{2} = - \frac{4}{10}$
½ mark
$u_{o} = - 2.5$ cm. The object must be placed 2.5 cm in front of the objective.
½ mark
Magnifying power:
$M = \frac{v _{o}}{∣ u _{o} ∣} (1 + \frac{D}{f _{e}}) = \frac{10}{2.5} (1 + \frac{25}{6.25}) = 4 \times 5 = 20$
½ mark
(b) Final image at infinity
The objective's image must lie at the focus of the eyepiece.
$v_{o} = 15 - 6.25 = 8.75$ cm
½ mark
$\frac{1}{u _{o}} = \frac{1}{8.75} - \frac{1}{2} = \frac{2 - 8.75}{17.5} = - \frac{6.75}{17.5}$
½ mark
$u_{o} = - \frac{17.5}{6.75} \approx - 2.59$ cm. The object must be placed about 2.6 cm from the objective.
½ mark
Magnifying power:
$M = \frac{v _{o}}{∣ u _{o} ∣} \cdot \frac{D}{f _{e}} = \frac{8.75}{2.59} \times \frac{25}{6.25} \approx 3.375 \times 4 = 13.5$
1 mark

Answer: (a) Object 2.5 cm from objective, M = 20; (b) object ≈ 2.59 cm (≈2.6 cm) from objective, M ≈ 13.5

Where the marks are

Eyepiece lens formula with $v_{e} = - 25$ cm (½ mark)
$u_{e} = - 5$ cm, hence $v_{o} = 10$ cm (½ mark)
Objective lens formula substitution (½ mark)
$u_{o} = - 2.5$ cm (½ mark)
$M = 20$ (½ mark)
For infinity, $v_{o} = 15 - 6.25 = 8.75$ cm (½ mark)
Objective lens formula substitution (½ mark)
$u_{o} \approx - 2.6$ cm (½ mark)
$M \approx 13.5$ (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 2024-25: every question, by section.

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

Hints

Step-by-step solution

Revise the chapter

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