Question

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

• A. $0.1\text{rad}$
• B. $0.2\text{rad}$
• C. $0.4\text{rad}$
• D. $0.5\text{rad}$

Answer 1

The correct option is B $0.2\text{rad}$

Given,
$f_0=10\text{cm},f_e=5\text{cm}$

Angle subtended on objective, $\theta =0.1\text{rad}$.


Height of the image $\left(h\right)$ can be calculated using $\Delta ABO$,


$\tan \theta =\frac{h}{f_0}$

When $\theta$ is small, $\tan \theta \simeq \theta$

$\therefore \theta =\frac{h}{f_0}$

$\Rightarrow h=\theta f_0=0.1\times 10=1\text{cm}$

The angle formed by the final image on eyepiece is equal to the angle made by $AB$ on eyepiece.

$\therefore \tan \beta =\frac{AB}{f_e}=\frac{1}{5}$

$\Rightarrow \tan \beta \approx \beta =0.2\text{rad}[\because \beta \text{is small angle}]$

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

Key Note: Angle made by final image on the eyepiece is equal to angle made by first image on the eyepiece.