Magnifying Power

In an astronomical telescope adjusted for parallel rays: - Magnifying power - Length Given: So:

The magnifying power of an astronomical telescope in normal adjustment is given by: where: - is the focal length of the objective, - is the focal length of the eyepiece. We are given that the magnifying power , so: Thus, we can express the focal length of the objective in terms of the focal length of the eyepiece:
Step 1: Use the total length of the telescope
In normal adjustment, the total length of the telescope is the sum of the focal lengths of the objective and the eyepiece: Substitute into this equation:
Step 2: Find
Now that we know , we can calculate using : Thus, the focal lengths of the objective and the eyepiece are: Therefore, the correct answer is option (D).