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 tub is
Correct option is C. $$15\, cm$$
by using $$m_{\infty}=\dfrac{(L_{\infty}-f_o-f_e).D}{f_of_e}$$
$$\Rightarrow 45=\dfrac{( L_{\infty}-1-5) \times 25}{1\times 5}\Rightarrow L_{\infty}=15\ cm$$.