A telescope has an eyepiece of focal length 5 | vedclass

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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

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 $(f_e)$: $M = \frac{f_o}{f_e}$ Also,the 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 $(\alpha)$: $M = \frac{\beta}{\alpha}$ Equating the two expressions: $\frac{\beta}{\alpha} = \frac{f_o}{f_e}$ $\beta = \alpha 	imes \frac{f_o}{f_e}$ Given: $\alpha = 2^o$,$f_o = 60\, cm$,$f_e = 5\, cm$. Substituting the values: $\beta = 2^o 	imes \frac{60}{5} = 2^o 	imes 12 = 24^o$ Therefore,the angular width of the image is $24^o$.

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