Question

The angle subtended by the moon, at the objective lens of an astronomical telescope, is $0.5\text{rad}$. The focal length of the objective lens and the eyepiece of the telescope are $30\text{cm}$ and $10\text{cm}$ respectively. Find the angle subtended by the final image, at the eyepiece, in case of the least strain on the eye.

Answer 1

Given:
$\alpha =0.5\text{rad}$
$f_o=30\text{cm}$
$f_e=10\text{cm}$

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,

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

$\Rightarrow \frac{1}{(-\infty )}-\frac{1}{u_e}=\frac{1}{10}$

$\Rightarrow u_e=-10\text{cm}$
Also, from geometry
$\tan \alpha =\frac{h^{\prime }}{f_o}$,
for small angle $\alpha ,\tan \alpha \to \alpha$
$\Rightarrow h^{\prime }=\alpha f_o$
So,

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

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

$\Rightarrow \beta =\frac{0.5\times 30}{10}=1.5\text{rad}$

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