In compound microscope the magnification is $95$, and the distance of object from objective lens $\frac{1}{3.8}\text{cm}$ and focal length of objective is $\frac{1}{4}\text{cm}.$ What is the magnification of eye piece when final image is formed at least distance of distinct vision ?

The correct option is A $5$
Given,

$\text{Magnification of compound microscope}M=95\text{Distance of object}\left(u_o\right)=\frac{1}{3.8}\text{cm}\text{Focal length}\left(f_o\right)=\frac{1}{4}\text{cm}$

Net Magnification,

$M=m_o\times m_e$ and

$m_o=\frac{f_o}{u_o+f_o}$

$\Rightarrow M=\left(\frac{f_{\circ }}{u+f_{\circ }}\right)\times m_e$

$\Rightarrow M=\frac{\left(\frac{1}{4}\right)}{(-\frac{1}{3.8}+\frac{1}{4})}\times m_e$

$\Rightarrow 95=\frac{\left(\frac{1}{4}\right)}{\left(\frac{-20+19}{76}\right)}\times m_e$

$\Rightarrow 95=\frac{76}{4\times (-1)}\times m_e$

$\Rightarrow m_e=\frac{95\times 4(-1)}{76}$

$\Rightarrow m_e=-5$

Magnitude of magnification, $\left|\begin{array}{c}{m}_e\end{array}\right|=5$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.