Question

The focal lengths of the objective and eye-piece of a compound microscope are $1$ cm and $5$ cm respectively. An object is placed at a distance of $1.1$ cm from the objective has its final image formed at least distance of distinct vision, then the magnifying power is

• A. $20$
• B. $30$
• C. $50$
• D. $60$

Answer 1

Answer: (D)

$60$

$f_0=1cm$

$f_e=5cm$

$v_0=1.1cm$

$v_e=25cm$

$M=?$

$M=\frac{v_0}{u_0}(1+\frac{D}{f_e})$

$\frac{1}{v_0}-\frac{1}{(-1.1)}=\frac{1}{1}$

$v_0=11cm$

$M=\frac{11}{1.1}(1+\frac{\text{/}{25}^5}{\text{/}5})=6\times 10=60$