JEE Main 2021 · Physics · Ray and Wave Optics · hard

In a compound microscope the objective and eyepiece have focal lengths 1 cm and 5 cm, and their separation is 15 cm. The final image is formed at the…

Numerical answer

In a compound microscope the objective and eyepiece have focal lengths

1\ \mathrm{cm}

and

5\ \mathrm{cm}

, and their separation is

15\ \mathrm{cm}

. The final image is formed at the least distance of distinct vision,

25\ \mathrm{cm}

from the eye (eye close to the eyepiece). The magnitude of the total magnification is

Show answer and solution

Answer: 59

Eyepiece:

v_e=-25

f_e=5

\dfrac1{u_e}=\dfrac1{v_e}-\dfrac1{f_e}=-\dfrac{6}{25}\Rightarrow |u_e|=\dfrac{25}{6}\ \mathrm{cm}

, and

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

. The objective's image is therefore at

v_0=15-\dfrac{25}{6}=\dfrac{65}{6}\ \mathrm{cm}

. Objective:

\dfrac1{u_0}=\dfrac1{v_0}-\dfrac1{f_0}=\dfrac{6}{65}-1=-\dfrac{59}{65}

, so

m_0=\dfrac{v_0}{|u_0|}=\dfrac{65}{6}\cdot\dfrac{59}{65}=\dfrac{59}{6}

. Total

M=\dfrac{59}{6}\times6=59

Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake