Question

Find the magnifying power of a telescope, when the final image is formed at least distance of distinct vision. Focal length of used lenses are $12.5\text{cm}$ and $100\text{cm}$.

• A. $10$
• B. $12$
• C. $18$
• D. $22$

Answer 1

The correct option is B $12$

We know the focal length of the objective is larger than the focal length of the eyepiece.

$\therefore f_0=100\text{cm}$ and $f_e=12.5\text{cm}$

As, magnifying power,

$m=\frac{f_0}{f_e}(1+\frac{f_e}{D})$

$\Rightarrow m=\frac{100}{12.5}(1+\frac{12.5}{25})$

$\Rightarrow m=8\times \frac{3}{2}=12$

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

Key concept : In a telescope, the focal length of the objective lens is larger than the focal length of the eyepiece.