The magnifying power of an astronomical teles | vedclass

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(A) For an astronomical telescope in normal adjustment, the final image is formed at infinity.In this state, the focal point of the objective lens coi

Vedclass · Accepted Answer

-$f_0 / f_e$

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(A) For an astronomical telescope in normal adjustment, the final image is formed at infinity.In this state, the focal point of the objective lens coincides with the focal point of the eyepiece.The magnifying power $(m)$ is defined as the ratio of the angle subtended by the image at the eye $(\beta)$ to the angle subtended by the object at the objective lens $(\alpha)$.Mathematically, $m = \beta / \alpha = -f_0 / f_e$, where $f_0$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.The negative sign indicates that the final image is inverted with respect to the object.

The magnifying power of an astronomical telescope for normal adjustment is given by:

Similar Questions

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