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Q. The magnifying power of a microscope with an objective of $5 \,mm$ focal length is $400$ . The length of its tube is $20\, cm$. Then the focal length of the eye-piece is

Ray Optics and Optical Instruments

Solution:

If nothing is said then it is considered that final image
is formed at infinite and $m _\infty \frac{\left(L_{\infty}-f_{o}-f_{e}\right) \cdot D}{f_{o} f_{e}} \simeq \frac{L D}{f_{0} f_{e}}$
$\Rightarrow 400=\frac{20 \times 25}{0.5 \times f_{e}} $
$\Rightarrow f_{e}=2.5 \,cm$