Question

The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the eye lens and objective is 20 cm. Determine the focal lengths of both the lenses.

Hint:

For telescopes, the magnifying power is the ratio of the focal lengths of the objective lens to the eye lens. The total distance between the lenses is the sum of their focal lengths.

Answer 1

The magnifying power \( M \) of a telescope is given by the formula: \[ M = \frac{f_o}{f_e} \] where \( f_o \) is the focal length of the objective lens and \( f_e \) is the focal length of the eye lens. We are given that the magnifying power \( M = 9 \) and the distance between the eye lens and the objective lens \( d = 20 \, \text{cm} \). This distance is the sum of the focal lengths of the two lenses: \[ d = f_o + f_e \] Thus, we have the system of equations: 1. \( M = 9 = \frac{f_o}{f_e} \) 2. \( d = f_o + f_e = 20 \, \text{cm} \) From the first equation, we can express \( f_o \) in terms of \( f_e \): \[ f_o = 9 f_e \] Substitute this into the second equation: \[ 9 f_e + f_e = 20 \] \[ 10 f_e = 20 \] \[ f_e = 2 \, \text{cm} \] Now substitute \( f_e = 2 \, \text{cm} \) into \( f_o = 9 f_e \): \[ f_o = 9 \times 2 = 18 \, \text{cm} \] Thus, the focal lengths of the lenses are: \[ \boxed{f_o = 18 \, \text{cm}} \quad \text{and} \quad \boxed{f_e = 2 \, \text{cm}} \]