A microscope with a tube length of $180mm$achieves a total magnification of $800X$with a $40X$objectives and a $20X$eye piece. The microscope is focused for viewing with a related eye. how far is the sample from the objective lens?

We have given that:

Tube length of microscope$L=180mm$,

Magnification of objectlocalid="1650116668170" $m_{obj}=-40$

and focal length of eye$f_{eye}=20$.

We need to find the distance of the sample from the objective lens.

By using this formula, we will get the value for $f_{obj}$

$m_{obj}=-\frac{L}{f_{obj}}$

multiplying both the sides by $f_{obj}$,

localid="1648889301173" $f_{obj}\times m_{obj}=-\frac{L}{\bcancel{f_{obj}}}\bcancel{\times f_{obj}}$

$f_{obj}=-\frac{L}{m_{obj}}$

$f_{obj=-\frac{L}{m_{obj}}}$

By substituting the value in equation,

$f_{obj}=-\frac{180}{-40}$

$f_{obj}=4.5mm$.

Here,$f_{obj}$is the focal length of eye.

Now Let us find value for $f_{eye}$,

$M_{eye}=\frac{25}{f_{eye}}$

$f_{eye}=\frac{25}{20}$

$f_{eye}=1.25cm\to 12.5mm.$

Let us find value for $f_{eye}$,

$M_{eye}=\frac{25}{f_{eye}}$

$f_{eye}=\frac{25}{20}$

$f_{eye}=1.25cm\to 12.5mm$

As localid="1650116930783" $S^{\text{'}}=L-f_{eye}$

Substituting the given values,

localid="1650116940212" $S^{\text{'}}=180mm-12.5mm$

localid="1650116917597" $S^{\text{'}}=167.5mm$

Here,$S^{\text{'}}$is the distance of image from lens.

Finally,

$S={\left(\frac{1}{4.5}-\frac{1}{167.5}\right)}^{-1}$

$S=4.624mm$.

Here,$S$is the distance of the sample from the objective lens.