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(D) For an astronomical telescope with the final image at infinity,the angular magnification is given by $m = -\frac{f_o}{f_e}$.Given the magnitude of

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$30 \, cm$ and $6 \, cm$

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(D) For an astronomical telescope with the final image at infinity,the angular magnification is given by $m = -\frac{f_o}{f_e}$.Given the magnitude of magnification is $5$,we have $|m| = \frac{f_o}{f_e} = 5$,which implies $f_o = 5f_e$.The separation between the objective and the eyepiece (tube length) is $L = f_o + f_e = 36 \, cm$.Substituting $f_o = 5f_e$ into the equation: $5f_e + f_e = 36 \, cm$.$6f_e = 36 \, cm \Rightarrow f_e = 6 \, cm$.Now,calculating $f_o$: $f_o = 5 	imes 6 \, cm = 30 \, cm$.Thus,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 $36 \, cm$. The final image is formed at infinity. The focal length $f_o$ of the objective and $f_e$ of the eyepiece are

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