Question

Magnifying power of a telescope is $40$ in case when the final image is formed at infinity. Find its maximum magnifying power, if focal length of eyepiece is $6.25\text{cm}$.

• A. $45$
• B. $50$
• C. $55$
• D. $60$

Answer 1

The correct option is B $50$

Given:
$f_e=6.25\text{cm}$

We know that magnifying power when the final image is formed at infinity is given by,

$m=\frac{f_o}{f_e}$

$\Rightarrow 40=\frac{f_o}{f_e}$

Also, maximum magnifying power of the telescope is achieved when the final image is formed at the least distance of distinct vision.

So, maximum magnifying power is given by,

$m^{\prime }=\frac{f_o}{f_e}(1+\frac{f_e}{D})$

$\Rightarrow m^{\prime }=40(1+\frac{6.25}{25})=50$

Hence, option $\left(B\right)$ is the correct answer.