Question

If focal length of an objective and eyepiece of a telescope are $3cm$ and $6cm$ respectively. The object (tiny) is placed at $2cm$ from objective and final image is formed at infinity, then the magnifying power of microscope is

• A. $4$
• B. $8$
• C. $6$
• D. $12$

Answer 1

The correct option is B $8$

Magnification power $\left(m\right)=\frac{v_o}{u_o}\left(\frac{D}{f_e}\right)$
Where, $v_0$=image distance,
$u_0$= object distance
$f_e$= focal lenght of eyepiece lens
$D$= $25cm$
Given, $u_0=2cm,f_e=6cm$
On putting lens formula for objective ($v_0$)
$u_o=-2$
$\frac{1}{v_{)}}-\frac{1}{u_0}=\frac{1}{f}$
$\frac{1}{v_0}+\frac{1}{2}=\frac{1}{3}$
$\frac{1}{v_0}=\frac{1}{3}-\frac{1}{2}\Rightarrow v_0=-6$

So, magnification of lens
$=\frac{v_o}{u_o}\left(\frac{24}{6}\right)$
$=\frac{6}{3}\times \left(\frac{24}{6}\right)=8$