Question

The magnifying power of an astronomical telescope is 24. In normal adjustment, the distance between its two lenses is 150 cm. Find the focal length of the objective lens.

Hint:

When solving problems involving optical instruments like telescopes, always check if the system is in normal adjustment as it simplifies the use of formulae.

Answer 1

Given that the distance between the lenses, \( L = 150 \, cm} \), and the magnifying power, \( M = 24 \), we have: \[ L = f_o + f_e \] and \[ M = \frac{f_o}{f_e} \] By rearranging and substituting from the magnification formula: \[ f_e = \frac{f_o}{M} = \frac{f_o}{24} \] Substituting into the lens distance equation gives: \[ 150 = f_o + \frac{f_o}{24} \] \[ 150 = \frac{25f_o}{24} \] \[ f_o = \frac{150 \times 24}{25} = 144 \, cm} \] Therefore, the focal length of the objective lens is 144 cm.