Question

A planet is observed by an astronomical refracting telescope having and objective of focal length 16 m and an eye piece of focal length 2 cm then :

• A. The distance between the objective and the eye piece is 15 m
• B. The angular magnification of the planet is 600
• C. The image of planet in inverted
• D. The objective is smaller than the eye piece

Answer 1

Answer: (C)

The image of planet in inverted

Answer 2

Distance between objective and eye piece for normal adjustment is

$L = F_{0} + F_{e} = 16m + 2cm = 16.02m$

Angular magnification or magnifying power,

$\frac{\beta}{\alpha} = M = \frac{F_{O}}{F_{e}} = \frac{16}{0.02} = 800$