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(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} =

Vedclass · Accepted Answer

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

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(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 $f_o = 5f_e$ ... $(i)$. The length of the telescope tube is the sum of the focal lengths of the objective and the eyepiece: $L = f_o + f_e = 36 \, cm$ ... $(ii)$. Substituting $(i)$ into $(ii)$,we get $5f_e + f_e = 36 \, cm$,which simplifies to $6f_e = 36 \, cm$. Thus,$f_e = 6 \, cm$. Substituting $f_e$ back into $(i)$,we get $f_o = 5 	imes 6 \, cm = 30 \, cm$. Therefore,the focal lengths are $f_o = 30 \, cm$ and $f_e = 6 \, cm$.

An astronomical telescope has an angular magnification of magnitude $5$ for distant objects. The separation between the objective and the eyepiece is $36 \, cm$ and the final image is formed at infinity. The focal length $f_o$ of the objective and the focal length $f_e$ of the eyepiece are:

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