Question

The angular magnification of a refracting telescope is 40. When the object and final image are both at infinity, the distance between the eyepiece and the objective is 143.5 cm. The telescope is used to view a distant radio tower. The real image of the tower, formed by the objective, is 6.0 mm in height. The focal point of the eyepiece is positioned at the real image. What is the angle subtended by the final image of the tower?
a: 0.15 rad
b: 0.17 rad
c: 0.21 rad
d: 0.19 rad
e: 0.23 rad

Answer 1

(a)

f_o = focal length of objective lens

f_e = focal length of eyepiece

d = distance between eyepiece and objective lens = 143.5 cm

m = angular magnification = 40

h = height = 6 mm = 0.6 cm

Using the formula :

f_o + f_e = d

f_o + f_e = 143.5                  Eq-1

Using the formula for angular magnification:

m = f_o/f_e

f_o = 40 f_e                  Eq-2

Using Eq-1 and Eq-2

40 f_e + f_e = 143.5

f_e = 3.5 cm

f_o = 140 cm

b)

angle = theta = h/f_e = 0.6 / 3.5 = 0.17 rad