The focal lengths of the objective and eye-pi | vedclass

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(C) The magnification $m$ of a telescope is given by the ratio of the angle subtended by the image at the eye $(\beta)$ to the angle subtended by the

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$25$

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(C) The magnification $m$ of a telescope is given by the ratio of the angle subtended by the image at the eye $(\beta)$ to the angle subtended by the object at the objective $(\alpha)$.$m = \frac{\beta}{\alpha} = \frac{f_o}{f_e}$Given:Focal length of objective, $f_o = 100 \ cm$Focal length of eye-piece, $f_e = 2 \ cm$Angle subtended by the moon at the eye (object angle), $\alpha = 0.5^o$Substituting the values into the magnification formula:$\frac{\beta}{0.5^o} = \frac{100}{2}$$\frac{\beta}{0.5^o} = 50$$\beta = 50 	imes 0.5^o = 25^o$Therefore, the angle subtended by the moon's image is $25^o$.

The focal lengths of the objective and eye-piece of a telescope are $100 \ cm$ and $2 \ cm$ respectively. The moon subtends an angle of $0.5^o$ at the eye. If it is viewed through the telescope, the angle subtended by the moon's image will be......$^o$.

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