Question

Four lenses having the focal length of $+15 \,cm , 20\, cm$, $+150\, cm$, and $+250\, cm$, respectively, are provided to make an astronomical telescope. The focal length of the eyepiece to produce the largest magnification, should be

• A. +250 cm
• B. +150 cm
• C. +20 cm
• D. +15 cm

Answer 1

Answer: (D)

Magnifying power of an astronomical telescope is given by
$|m|=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)$
where $f_0$ is focal length of objective, $f_e$ is focal length of eyepiece, $D$ is least distance of distinct vision.
From this formula we observe that
Magnifying power $\propto \frac{1}{\text { focal length of eye-piece }}$
Hence, to produce the largest magnification, the focal length of eye-piece must be smallest.
Hence, $f_e=+15 \,cm$