The diameter of the moon is `3.5xx10^(3)km` and its distance from the earth is `3.8xx10^(5) km`. It is seen by a telescope having the focal length of the objective and the eye-piece as `4m` and `10cm` respectively. The diameter of the image of the moon will be approximately

Options

A. `15^(@)`
B. `20^(@0`
C. `30^(@)`
D. `35^(@)`

Answer

B

Explanation

Magnification of telescope (m) = `(f_(o))/(f_(e)) = (400)/(10)` = 40
Angle subtended by moon on objective
`alpha = (3.5 xx 10^(3) )/(3.8 xx 10^(5)) = (3.5)/(3.8) xx 10^(-2) ` rad.
Now, m = `("Angle subtended by image")/("Angle subtended by object")`
`m = (beta)/(alpha) rArr beta = 40 xx (3.5)/(3.8) xx 10^(-2) = 0.36` rad
`beta = 0.36 xx (180)/(pi) = 20.64=20^(@). " "` (approx.)