An astronomical telescope has an angular magnification of ma... | Physics | SolveeIt | Solveeit

Today, August 18

Question

An astronomical telescope has an angular magnification of magnitude 5 for far objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length $f_o$ of the objective and the focal length $f_e$ of the eyepiece are

A

$f_o$ = 45 cm and $f_e$ = - 9 cm

B

$f_o$ = 50 cm and $f_e$ = 10 cm

C

$f_o$ = 7.2 cm and $f_e$ = 5 cm

D

$f_o$ = 30 cm and $f_e$ = 6 cm

Answer

$f_o$ = 30 cm and $f_e$ = 6 cm

Explanation

Solution

Image formed by objective (I_1) is at second focus of it
because objective is focussed for distant objects. Therefore,
$\hspace25mm P_1 I_1=f_o$
Further $I_1$ should lie at first focus of eyepiece because final
image is formed at infinity.
$\therefore \hspace25mm P_2 I_1=f_e$
Given $\hspace20mm P_1 P_2=$ 36 cm
$\therefore \hspace20mm f_o+f_e=36 \hspace25mm ...(i)$
Further angular magnification is given as 5. Therefore,
$\hspace25mm \frac{f_o}{f_e}=5 \hspace20mm ...(ii)$
Solving Eqs. (i) and (ii), we get
$\hspace15mm f_o=30$ cm and $f_e=$ 6 cm
$\therefore$ Correct option is (d).