Question

The focal length of objective and eye lens of a microscope are $4cm$ and $8cm$ respectively. If the least distance of distinct vision is $24cm$ and object distance is $4.5cm$ from the objective lens, then the magnifying power of the microscope will be

• A. $18$
• B. $32$
• C. $64$
• D. $20$

Answer 1

The correct option is B $32$

For objective lens $\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}$
$\Rightarrow \frac{1}{(+4)}=\frac{1}{v_0}-\frac{1}{(-4.5)}\Rightarrow v_0=36cm$
$\therefore |m_D|=\frac{v_0}{u_0}(1+\frac{D}{f_e})=\frac{36}{4.5}(1+\frac{24}{8})=32$