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. `+155cm`
C. `+15 cm`
D. `25 cm`

Answer

C

Explanation

Magniifying power of an astronomical telescope is given by
`|m|(f_(o))/(f_(e))(1+(f_(e))/(D))`
where`f_(o)` 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, to produce the largest magnification the focal length of eye- piece must be smallest.