(a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eyepiece of focal length 1.0cm is used, what is the angular magnification of the telescope?
(b) 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 `3.48 xx 10^(6)m`, and the radius of lunar orbit is `3.8xx10^(8)m`.

Explanation

`rArr` Here, focal length of objective `f_0 = 15 m `
Focal length of eye-piece `f_e` = 1 cm = 0.01 m
(a) Angular magnification,
`m = (f_0)/(f_e) = (15)/(0.01) = 1500`
Suppose, diamter of image is d hence angle made by moon is `alpha` then
`alpha =("Diameter of moon")/("Orbital radius of moon") = (3.48 xx 10^6)/(3.8 xx 10^8)`
Angle subtended by image of objective is alos `alpha` Hence,
`therefore alpha = ("Diamter of image of moon")/(f_0) = d/f_0`
`therefore` From euqation (a) and (2),
`d/f_0 = (3.48 xx10^6)/(3.8 xx 10^8)`
`therefore d/15 = (3.48 xx 10^5)/(3.5 xx 10^8)`
`therefore d= (3.48 xx 10^6 xx 15)/(3.8 xx 10^8)`
`therefore d = 13.74 xx 10^(-2) m `
`therefore d = 13.74 cm `