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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. - Physics

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Question

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.

  1. Calculate the angular magnification of the microscope.  (2)
  2. Draw the ray diagram of a compound microscope in normal adjustment. (1)
Diagram
Numerical
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Solution

I. Given, fo ​= 0.82 cm, uo ​= −0.91 cm, fe​ = 2.9 cm, D = 25 cm

Using the lens formula:

`1/v_o = 1/f_o - 1/u_o`

`1/v_o = 1/0.82 - 1/(-0.91)`

`1/v_o` = 1.2195 + 1.0989

`1/v_o` = 2.3184

vo ​= `1/2.3184`

vo ​= 0.4313 cm

Objective magnification:

mo ​= `v_o/u_o`

= `0.4313/(-0.91)`

= −0.474

Eyepiece magnification:

`m_e = 1 + D/f_e`

= `1 + 25/2.9`

= 9.62

Total angular magnification:

M = mo ​× me

= −0.474 × 9.62

= −4.56

II. 

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Notes

The Answer to Part I in the board paper solution is incorrect. As per the calculation, the correct answer is −4.56, not -87.7.

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2025-2026 (March) Board Sample Paper
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