If we need a magnification of $$375$$ from a compound microscope of tube length $$150mm$$ and an objective of focal length $$5\,mm$$, the focal length of the eye-piece, should be close to :
Correct option is D. $$22$$ mm
Given $$M=375,L=150mm,f_0=5mm,fe=?,d=25cm$$.
$$M$$ for a compound microscope is given by
$$M=\dfrac{v_0}{+u_0}\left(1+\dfrac{d}{fe}\right)$$
focal length of objective lens is small $$u_0\simeq f_0$$. Alos as focal length of eye is small $$v_0\simeq L$$
$$\therefore M=\dfrac{L}{+f_0}\left(1+\dfrac{d}{f_e}\right)$$
$$\Rightarrow 375=\dfrac{150}{5}\left(1+\dfrac{250}{f_e}\right)$$
$$\Rightarrow \dfrac{375}{30}=\left(1+\dfrac{250}{f_e}\right)$$
$$\Rightarrow \dfrac{375}{30}-1=\dfrac{250}{f_e}$$
$$\Rightarrow \dfrac{345}{30}=\dfrac{250}{f_e}$$
$$\Rightarrow f_e=2.2cm$$
$$\boxed{F_e=22mm}$$