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Question

(D) For an astronomical telescope with the final image at infinity (normal adjustment),the angular magnification is given by $m = \frac{f_o}{f_e} = 5$

Vedclass · Accepted Answer

$f_o = 30 \, cm$ and $f_e = 6 \, cm$

Vedclass · Answer

(D) For an astronomical telescope with the final image at infinity (normal adjustment),the angular magnification is given by $m = \frac{f_o}{f_e} = 5$.This implies $f_o = 5f_e$.The length of the telescope tube is the distance between the objective and the eyepiece,which is $L = f_o + f_e = 36 \, cm$.Substituting $f_o = 5f_e$ into the length equation: $5f_e + f_e = 36 \, cm$.$6f_e = 36 \, cm$,which gives $f_e = 6 \, cm$.Now,calculating $f_o$: $f_o = 5 	imes 6 \, cm = 30 \, cm$.Therefore,the focal lengths are $f_o = 30 \, cm$ and $f_e = 6 \, cm$.

For a distant object,the angular magnification of an astronomical telescope is $5$. The distance between the objective and the eyepiece is $36 \, cm$,and the final image is formed at infinity. What are the focal lengths of the objective $(f_o)$ and the eyepiece $(f_e)$?

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