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

With the lens separation at \( L = 150 \, cm} \) and a magnifying power of \( M = 24 \), the following equations apply:\[L = f_o + f_e\]and\[M = \frac{f_o}{f_e}\]From the magnification formula, we can express \( f_e \) as:\[f_e = \frac{f_o}{M} = \frac{f_o}{24}\]Substituting this into the lens separation equation yields:\[150 = f_o + \frac{f_o}{24}\]This simplifies to:\[150 = \frac{25f_o}{24}\]Solving for \( f_o \):\[f_o = \frac{150 \times 24}{25} = 144 \, cm}\]The objective lens has a focal length of 144 cm.