Question

The magnifying power of a telescope at normal adjustment with tube length $60\text{cm}$ is $5$. What is the focal length of its eye piece?

• A. $20\text{cm}$
• B. $\text{40 cm}$
• C. $\text{30 cm}$
• D. $\text{10 cm}$

Answer 1

The correct option is D $\text{10 cm}$

For telescope, at normal adjustment,

$\text{Tube length,}L=f_o+f_e=60\text{cm}$
$\text{magnification,}m=\frac{f_o}{f_e}=5$
$\Rightarrow f_o=5f_e$
$\therefore L=5f_e+f_e=60\text{cm}$
$\Rightarrow f_e=10\text{cm}$
$\therefore$ focal length of eye-piece, $f_e=10\text{cm}$

Hence, option $\left(D\right)$ is correct.