Question

A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 cm tall tower at a distance of 1 Km is observed through this telescope in normal setting then the image of the tower is $\theta$ then $\theta$ is close to;

• A. ${60}^{\circ }$
• B. ${86}^{\circ }$
• C. ${90}^{\circ }$
• D. $2^{\circ }$

Answer 1

The correct option is A ${60}^{\circ }$

Magnification from mirror

$m=\frac{h_i}{h_o}=\frac{-v}{u}$

Whereas

$h_i=$ Height of image from principal axis

$h_o=$ Height of object from principal axis

$u=$ Object distance

$v=$ Image distance

Angular magnification $m=\frac{f_o}{f_e}=\frac{150}{5}=30$

$\therefore \frac{\tan \beta }{\tan \alpha }=30\Rightarrow \tan \beta =\tan \alpha \times 30$

$\tan \beta =(\frac{50}{1000}\times 30)=\frac{15}{10}=\frac{3}{2}$

$\tan \beta =\frac{3}{2}\Rightarrow \tan \beta ={\tan }^{-1}\left(\frac{3}{2}\right)$

$\theta =\beta \simeq {60}^{\circ }$