Question:medium

The magnifying power of a telescope is \( 9 \). When adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The ratio of the focal length of the objective lens to the focal length of the eyepiece is:

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The magnification of an astronomical telescope is given by \( M = \frac{f_o}{f_e} \). The larger the objective lens focal length, the higher the magnification.
Updated On: Nov 26, 2025
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