A small telescope has an objective lens of focal length 100 cm and an eye-piece of focal length 5 cm. The magnifying power of the telescope when the final image is formed at the least distance of distinct vision will be

24

The correct answer is Option (1) → 24

$\text{Given: } f_o = 100~\text{cm},~ f_e = 5~\text{cm},~ D = 25~\text{cm (least distance of distinct vision)}$

$\text{Magnifying power of telescope for final image at least distance:}$

$M = - \frac{f_o}{f_e} \left(1 + \frac{f_e}{D} \right)$

Substitute values:

$M = - \frac{100}{5} \left(1 + \frac{5}{25} \right)$

$M = - 20 \left(1 + 0.2 \right)$

$M = - 20 \cdot 1.2 = -24$

$\text{Answer: } M = 24$