JEE Main 2018 · Physics · Ray and Wave Optics · medium

A compound microscope has an objective of focal length 1 cm and an eyepiece of focal length 5 cm. A small object is placed 1.2 cm from the objective…

Single correct option

A compound microscope has an objective of focal length

1\ \mathrm{cm}

and an eyepiece of focal length

5\ \mathrm{cm}

. A small object is placed

1.2\ \mathrm{cm}

from the objective, and the final virtual image is formed at the least distance of distinct vision,

25\ \mathrm{cm}

, from the eye. The magnifying power of the microscope is

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Answer: (C) $30$

Objective:

\dfrac1{v_o}=\dfrac1{f_o}+\dfrac1{u_o}=1-\dfrac1{1.2}=\dfrac16

(in

\mathrm{cm^{-1}}

), so

v_o=6\ \mathrm{cm}

and the magnification is

m_o=\dfrac{v_o}{|u_o|}=\dfrac{6}{1.2}=5

. Eyepiece with the final image at the near point:

m_e=1+\dfrac{D}{f_e}=1+\dfrac{25}{5}=6

. Total magnifying power

M=m_om_e=5\times6=30

. For a final image at infinity the eyepiece factor would be

D/f_e=5

and the product would be

25

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake