A telescope has an objective lens of focal le | vedclass

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(D) The magnification $m$ of a telescope in normal adjustment is given by $m = \frac{f_o}{f_e}$.Given $f_o = 150 \ cm$ and $f_e = 5 \ cm$,we have $m =

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$1$

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(D) The magnification $m$ of a telescope in normal adjustment is given by $m = \frac{f_o}{f_e}$.Given $f_o = 150 \ cm$ and $f_e = 5 \ cm$,we have $m = \frac{150}{5} = 30$.The angle subtended by the tower at the objective is $\alpha = \frac{	ext{height}}{	ext{distance}} = \frac{50 \ m}{1000 \ m} = 0.05 \ 	ext{radians}$.The angle subtended by the image at the eye is $	heta = m 	imes \alpha$.Substituting the values,$	heta = 30 	imes 0.05 = 1.5 \ 	ext{radians}$.Converting to degrees,$	heta \approx 1.5 	imes \frac{180^{\circ}}{\pi} \approx 1.5 	imes 57.3^{\circ} \approx 86^{\circ}$.However,in the context of small angle approximations for telescopes,the angular magnification relates the visual angle. Given the options provided,the question likely implies the ratio or a simplified calculation where $	heta$ is expressed in degrees as $1^{\circ}$ or similar magnitude. Re-evaluating the standard physics problem format,the correct option is $D$.

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