Question

In a compound microscope, the objective lens and eyepiece have a focal length of $0.95cm$ and $5cm$ respectively and are kept at a distance of $20cm$ . If the last image is formed at a distance of $25cm$ from the eyepiece, calculate magnifying power.

• A. $94$
• B. $84$
• C. $75$
• D. $88$

Answer 1

Answer: (A)

$f_0=0.95m=\frac{V_0}{u_0}\left(1+\frac{D}{f_e}\right)$
$f_e=5cm{u}_e=\frac{25}{6}$
For eyepiece
$m_e=1+\frac{D}{f_e}=\frac{v_e}{u_e},u_e=\frac{v_e}{6}$
$v_0=20-\frac{25}{6}=\frac{95}{6}$
$m_0=\frac{v_o}{u_o}=\frac{v_o}{f_o}=\frac{\frac{95}{6}}{0.95}\approx \frac{94}{6}$
$m_{net}=\frac{94}{6}\times 6=94$