Four lenses of focal length `+15 cm , +20 cm, +150 cm` and `+250cm` are available for making an astronomical telescope. To produce the largest magnification, the focal length of the eye-piece should be

Options

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

Answer

D

Explanation

Magnifying power of an astronmical telesope is given by
`|m| = (f_(0))/(f_(e)) (1+(f_(e))/(D))`
wher `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 `prop (1)/("focal length of eye-piece")`
Hence, tp produce the larger magnification the focal length of eye-piece must be smallest.
Hence, `f_(e) =+15 cm`