A simple telescope consisting of an objective | vedclass

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(B) The angular magnification $M$ of a telescope in normal adjustment (where parallel rays emerge from the eye-lens) is given by the ratio of the foca

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) The angular magnification $M$ of a telescope in normal adjustment (where parallel rays emerge from the eye-lens) is given by the ratio of the focal length of the objective $(f_o)$ to the focal length of the eye-lens $(f_e)$: $M = \frac{f_o}{f_e}$ Also,the angular magnification 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: $f_o = 60\,cm$ $f_e = 5\,cm$ $\alpha = 2^o$ Equating the two expressions for magnification: $\frac{f_o}{f_e} = \frac{\beta}{\alpha}$ $\frac{60}{5} = \frac{\beta}{2}$ $12 = \frac{\beta}{2}$ $\beta = 12 	imes 2 = 24^o$ Therefore,the angular width of the image is $24^o$.

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