An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eye piece 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 eye piece are

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$f_{o} = 30 cm and f_{e} = 6 cm$

$f_{o} = 45 cm and f_{e} = - 9 cm$

$f_{o} = 7 .2 cm and f_{e} = 5 cm$

$f_{o} = 50 cm and f_{e} = 10 cm$

answer is D.

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Detailed Solution

$In this case | m | = \frac{f_{o}}{f_{e}} = 5 - - - - (i) and length of telescope = f_{0} + f_{e} = 36 - - - (ii) Solving (i) and (ii), we get f_{e} = 6 cm, f_{0} = 30 cm$

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