Question

The magnifying power of a telescope is 9. When adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal lengths of lenses are:

• A. 10 cm, 10 cm
• B. 15 cm, 5 cm
• C. 18 cm, 2 cm
• D. 11 cm, 9 cm

Hint:

Use formulas: \( M = \frac{f_o}{f_e} \), and \( L = f_o + f_e \) for telescope adjusted for parallel rays.

Answer 1

The Correct Option is C

In an astronomical telescope adjusted for parallel rays: - Magnifying power \( M = \frac{f_o}{f_e} \) - Length \( L = f_o + f_e \) Given: \[ M = 9, \quad L = 20 \Rightarrow f_o + f_e = 20, \quad \frac{f_o}{f_e} = 9 \] So: \[ f_o = 9 f_e \Rightarrow 9f_e + f_e = 20 \Rightarrow 10f_e = 20 \Rightarrow f_e = 2 \Rightarrow f_o = 18 \]