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(C) The angular magnification $M$ of a telescope is given by the ratio of the focal length of the objective lens $(f_o)$ to the focal length of the ey

Vedclass · Accepted Answer

$24$

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(C) The angular magnification $M$ of a telescope is given by the ratio of the focal length of the objective lens $(f_o)$ to the focal length of the eyepiece $(f_e)$: $M = \frac{\beta}{\alpha} = \frac{f_o}{f_e}$ Where $\alpha$ is the angle subtended by the object at the objective and $\beta$ is the angle subtended by the image at the eyepiece. Given: $f_o = 60 \ cm$,$f_e = 5 \ cm$,and $\alpha = 2^\circ$. Substituting the values: $\frac{\beta}{2^\circ} = \frac{60}{5}$ $\frac{\beta}{2^\circ} = 12$ $\beta = 12 	imes 2^\circ = 24^\circ$. Therefore,the angular size of the image is $24^\circ$.

In a simple telescope,the focal length of the objective lens is $f_o = 60 \ cm$ and the focal length of the eyepiece is $f_e = 5 \ cm$. If the rays coming from the object make an angle of $2^\circ$ at the objective,then the angular magnification (angular size of the image) is ........... $^\circ$.

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