Question

The angle made by the moon on an astronomical telescope is $0.1\text{rad}$. Focal length of the objective lens and the eyepiece of the telescope are $10\text{cm}$ and $5\text{cm}$ respectively. Find the angle subtended by the final image at the eyepiece, in case of the least strain on the eye.

Answer 1

Given:
$\alpha =0.1\text{rad}$
$f_o=10\text{cm}$
$f_e=5\text{cm}$


In case of the least strain on the eye, the final image is formed at infinity.

So, $v_e=-\infty$

On using lens formula for eyepiece,

$\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$

$\Rightarrow \frac{1}{(-\infty )}-\frac{1}{u_e}=\frac{1}{5}$

$\Rightarrow u_e=-5\text{cm}$

So,

$\beta =\frac{|h^{\prime }|}{|u_e|}$

$\Rightarrow \beta =\frac{\alpha f_o}{|u_e|}$

$\Rightarrow \beta =\frac{0.1\times 10}{5}=0.2\text{rad}$

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