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Q. The focal lengths of the objective and eye-lens of a microscope are $1\,cm$ and $5\,cm$ respectively. If the magnifying power for the relaxed eye is $45$, then the length of the tube is

Ray Optics and Optical Instruments

Solution:

By using $m_{\infty}=\frac{\left(L_{\infty}-f_{o}-f_{e}\right) \cdot D}{f_{o} \times f_{e}}$
$\Rightarrow 45=\frac{\left(L_{\infty}-1-5\right) \times 25}{1 \times 5} $
$\Rightarrow L_{\infty}=15 \,cm$