Question

A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0cm. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon fromed by the objective? The diameter of the moon is $3.5 \times 10^{6}$m and the radius of the lunar orbit round the earth is $3.8 \times 10^{8,}m$:

• A. 10 cm
• B. 12.5 cm
• C. 15 cm
• D. 17.5 cm

Answer 1

Answer: (D)

17.5 cm

Answer 2

: As $u > > f_{0,}v = f_{0} = 19m$

Now, $u = - 3.8 \times 10^{8}m.$ therefore magnification produced by the objective is

$m_{0} = \frac{v}{u} = - \frac{19}{3.8 \times 10^{8}} = - 0.5 \times 10^{- 7}$

$\therefore$ Diameter of the image of moon is

$3.5 \times 10^{6} \times 0.5 \times 10^{- 7} = 0.175m = 17.5cm$