Question

The length of the compound microscope is $14cm$. The magnifying power for relaxed eye is $25$. If the focal length of eye lens is $5cm$, then the object distance for objective lens will be

• A. $1.8cm$
• B. $1.5cm$
• C. $2.1cm$
• D. $2.4cm$

Answer 1

The correct option is A $1.8cm$

$L_{\infty }=v_0+f_e\Rightarrow 14=v_0+5\Rightarrow v_0=9cm$
Magnifying power of microscope for relaxes eye
$m=\frac{v_0}{u_0}.\frac{D}{f_e}$ or $25=\frac{9}{u_0}.\frac{25}{5}$ or $u_0=\frac{9}{5}=1.8cm$