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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 eyepiece lens (in cm ) is

0 2 Views | Posted 2 months ago
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    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago

    M = L/f? [1 + D/fe]
    ⇒ 100 = 20/Fe [1 + 25/Fe]
    ⇒ 1 + 25/Fe = 5
    ⇒ Fe = 25/4 = 6.25 cm

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