Question

A compound microscope has a magnifying power $48$. The focal length of its eye-piece is $5cm$. If final image is formed at least distance of distinct vision $\left(25cm\right)$ then magnification of objective lens is

• A. $-8$
• B. $8$
• C. $-200$
• D. $200$

Answer 1

The correct option is A $-8$

$M=-48$

$t_c=5cm,D=25cm$

Now, $M=M_0\times M_e=M_0\times [1+\frac{D}{f_e}]$

$\Rightarrow -48=M_0\times [1+\frac{25}{5}]$

$\Rightarrow -48=M_0\left[6\right]$

$\Rightarrow M_0=-8$