Question

Magnifying power of a terrestrial telescope is $40$ in the case of normal adjustment. Find its maximum magnifying power, if focal length of eyepiece is $6.25\text{cm}$.

• A. $46.25$
• B. $50$
• C. $44$
• D. $48.25$

Answer 1

The correct option is B $50$

Given,

$m=40;f_e=6.25\text{cm}$

Magnifying power of the terrestrial telescope in normal adjustment is,

$m=\frac{f_0}{f_e}$

$\Rightarrow 40=\frac{f_0}{6.25}$

$\therefore f_0=250\text{cm}$

In the case of maximum magnifying power,

$m=\frac{f_0}{f_e}(1+\frac{f_e}{D})$

Where, $D=\text{Least distance of distinct vision}$

$m=40(1+\frac{6.25}{25})=40\times \frac{5}{4}[\because D=25\text{cm}]$

$\therefore m=50$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.