(i) A giant refracting telescope at an observatory has an objective lens of focal length 15 m . If an eyepiece of focal length 1.0 cm is used, what is angular magnification of the telescope ?
(ii) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? the diameter of the moon is `3.48xx10^(6)m`, and the radius of lunar orbit is `3.8xx10^(8) m`.

Explanation

Focal length of the objective lens, fo=15m = `15 xx 102cm`
Focal length of the eyepiece, fe=1.0cm
a) The angular magnificationo of a telescope is given as:
`a=f_(0)/f_(c)`
`=(15 xx 10^(2))/1.0 = 1500`
Hence, the angular magnification of the given refracting telescope is 1500.
b) Diameter of the moon, d`=3.48 xx 106 m`
Radius of the lunar orbit, r0=3.8 xx 108m
Let `d^(')` be the diameter of the image of its moon formed by the objective lens.
The angle subtended by the diameter of the moon is equals to the angle subtended by the image.
`d/r_(0) = d^(')/f_(0)`
`(3.48 xx 10^(6))/(3.8 xx 10^(8)) = d^(')/15`
`therefore d^(')=3.48/3.8 xx 10^(-2) xx 15`
`=13.74 xx 10^(-2) m = 13.74 cm`
Hence, the diameter of the moon's image formed by the objective lens is 13.74 cm