Q142 / 160 · AP EAMCET 2023 (15 May Shift 2) · Physics
The magnifying power of a telescope with tube length 60 $\mathrm{cm}$ is 5 . Then the focal length of its eye piece is
- A$20 \mathrm{~cm}$
- B$40 \mathrm{~cm}$
- C$30 \mathrm{~cm}$
- Correct answerD$10 \mathrm{~cm}$
Solution
The magnifying power of a telescope is given by
$\begin{aligned}
& m=\frac{f_0}{f_e}=5 \\
& f_0=5 f_e \\
& f_0+f_e=60 \mathrm{~cm} \Rightarrow 5 f_e+f_e=60 \\
& 6 f_e=60 \Rightarrow f_e=10 \mathrm{~cm}
\end{aligned}$
The focal length of its eye piece, $f_e=10 \mathrm{~cm}$
Hence, the correct answer is option: 'D'.