A simple telescope consisting of an objective of focal length 60 cm and a single eye lens of focal length 5 cm is focused on a distant object in such a way that parallel rays emerge from eye lens. If the object subtends an angle of `2^(@)` at the objective, the angular width of the image is (Let `tan theta square theta` assuming small).

Options

A. `62^(@)`
B. `48^(@)`
C. `24^(@)`
D. `12^(@) C`

Answer

C

Explanation

Given : Focal length of the objective `f_(o) = 60 cm`
Focal length of the eye piece, `f_(e) = 5 cm`
angle subtended `theta = 2^(@)`
From the formula of magnification
`m = (f_(o))/(f_(e)) = (60)/(5) = 12`
Hence, angular width of the image is given by
`beta = a.m = 2 xx 12 = 24^(@)`