Question

(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 3.48×10⁶m, and the radius of the lunar orbit is 3.8×10⁸m.

Answer 1

The focal length of the objective lens,fo=15m=15×10²cm
Focal length of the eyepiece,fe=1.0cm
(a)The magnification of a telescope is given as \(\alpha = \)\(\frac{f_o}{f_e}\)=15×10²=1500
Hence, the angular magnification of the given refracting telescope is 1500.
(b)Diameter of the moon,d=3.48×10⁶m
The radius of the lunar orbit,r0=3.8×10⁸m
Let d' be the diameter of the image of the moon formed by the objective lens. The angle subtended by the diameter of the moon is equal to the angle subtended by the image.
\(\frac{d}{r_o}\)=\(\frac{d'}{f_o}\)=\(\frac{3.48\times 10^6}{3.8\times 10^8}\)=\(\frac{d'}{15}\)

∴ d'=\(\frac{3.48}{3.8}\)×10^-2×15=13.74×10^-2m=13.74cm
Hence, the diameter of the moon's image formed by the objective lens is 13.74cm.