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(B) The magnification $m$ of a telescope in normal adjustment is given by the ratio of the focal length of the objective $(f_o)$ to the focal length o

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) The magnification $m$ of a telescope in normal adjustment is given by the ratio of the focal length of the objective $(f_o)$ to the focal length of the eyepiece $(f_e)$: $m = \frac{f_o}{f_e} = \frac{60\, cm}{5\, cm} = 12$. Also, the angular magnification $m$ 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 $(\alpha)$: $m = \frac{\beta}{\alpha}$. Given that the object subtends an angle $\alpha = 2^o$, we can find the angular width of the image $\beta$: $\beta = m 	imes \alpha = 12 	imes 2^o = 24^o$.

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