An object is 10.0 mm from the objective of a certain compound microscope. The lenses are 300 mm apart, and the intermediate image is 50.0 mm from the eyepiece. What overall magnification is produced by the instrument?

1. The focal length of the eyepiece,$f_{ev}=50\text{mm}$.$f_{ev}=50\text{mm}$
2. The object's distance from the microscope,$p=10\text{mm}$.
3. The separation between the lens system,$\left(f_{ob}+s+f_{ev}\right)=300\text{mm}$.

The ratio of the height of the image to the height of the object is given as the lateral magnification of the lens or the mirror. Now, the magnification of the mirror or the lens can also be given as the negative value of the ratio of the image distance to the object distance from the mirror. In a compound microscope, the total magnification is the product of the objective and ocular lenses.

Formula:

The mirror equation,

$\frac{1}{f}=\frac{1}{i}+\frac{1}{p}$ (i)

Where f = focal length, p = object distance from the mirror, i = image distance.

The overall magnification of a compound microscope,

$m_{\theta }=-\frac{s}{f_{ob}}\frac{25\mathrm{cm}}{f_{ey}}$ (ii)

Where s is the tube length of the microscope, fob is the focal length of the objective, and fey is the focal length of the eyepiece.

Calculate the intermediate image distance as,

$i=\left(f_{ob}+s+f_{ey}\right)-f_{ey}$

By substituting the given value, the value of the image distance can be given as follows:

role="math" localid="1663026969553" $\begin{array}{rcl}\frac{1}{f_{ob}}& =& \left(\frac{1}{10}+\frac{1}{250}\right)\\ f_{ob}& =& \frac{250\times 10}{250+10}\\ & =& \frac{250}{26}\\ & =& 9.6153\text{cm}\approx 9.62\text{mm}\\ & & \end{array}$

Now, the tube length of the compound microscope can be given as follows:

$\begin{array}{rcl}s& =& \left(f_{ob}+s+f_{ev}\right)-f_{ob}-f_{ev}\\ & =& 300-9.62-50\\ & =& 240\text{mm}\\ & & \end{array}$

Now, the overall magnification of the compound microscope can be given using the above data in equation (ii) as follows:

$\begin{array}{rcl}M& =& -\frac{240}{9.62}\times \frac{25\times 10}{50}\\ & =& -\frac{240}{9.62}\times 5\\ & =& -124.74\\ & \approx & -125\\ & & \end{array}$

Therefore, the overall magnification of the instrument is -125.