The angular magnification of an astronomical | vedclass

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Question

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

Vedclass · Accepted Answer

$40$

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(B) For an astronomical telescope with the final image at infinity (normal adjustment),the angular magnification $m$ is given by:$m = \frac{f_o}{f_e} = 10 \implies f_o = 10 f_e$The length of the telescope $L$ in normal adjustment is the sum of the focal lengths of the objective and the eyepiece:$L = f_o + f_e = 44 \, cm$Substituting $f_o = 10 f_e$ into the length equation:$10 f_e + f_e = 44$$11 f_e = 44$$f_e = 4 \, cm$Now,calculate the focal length of the objective lens $f_o$:$f_o = 10 	imes 4 = 40 \, cm$

The angular magnification of an astronomical telescope is $10$. The final image is formed at infinity. The length of the telescope is $44 \, cm$. The focal length of the objective lens is .......... $cm$.

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