In a compound microscope, the magnified virtual image is formed at a distance of 25 cm from the eye-piece. The focal length of its objective lens is 1 cm. If the magnification is 100 and the tube length of the microscope is 20 cm, then the focal length of the eye-piece lens (in cm) is __________.
Answer (integer)
6
Solution
L = 20, f<sub>0</sub> = 1cm, M = 100<br><br>$M = {{{v_0}} \over {{u_0}}}\left( {1 + {D \over {{f_e}}}} \right)$<br><br>$\therefore$ $M = {L \over {{f_0}}}\left( {1 + {D \over {{f_e}}}} \right)$ [v<sub>0</sub> $\approx$ L, u<sub>0</sub> $\approx$ f<sub>0</sub>]
<br><br>$\Rightarrow$ ${{20} \over 1}\left( {1 + {{25} \over {{f_e}}}} \right)$ = 100
<br><br>on solving we get<br><br>f<sub>e</sub> = 6.25 cm
About this question
Subject: Physics · Chapter: Optics · Topic: Lenses and Optical Instruments
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