Principle: The principle of the compound microscope is based on the magnification of an image by using two lenses.

Construction: A compound microscope consists of two convergent lenses (i.e. objective lens $O$ and eye-piece lens $e$) placed coaxially in a double tube system. The objective lens is an achromatic convergent lens system of short focal length and short aperture. The other eye-piece lens $e$ is also an achromatic convergent lens system of large focal length and large aperture. The observation is taken through the eye-piece lens by the observer. The eye-piece lens is fitted outer side of a movable tube and the inner side connects with a non-movable tube in which the objective lens is fitted on another side of the non-movable tube. The separation between the objective or eye-piece lens can be changed by an arrangement, this is known as rack and pinion arrangement.

Working: Suppose a small object $ab$ is placed slightly away from the first focus $f_{\circ}$ of the objective lens which forms a real, inverted, and magnified image $a_{1}b_{1}$. Now adjust the eye-piece lens by moving like this, that the image $a_{1}b_{1}$ lies in between the optical center and the second focal length $f_{e}$ of the eye-piece lens. This image $a_{1}b_{1}$ works as an object for the eye-piece lens which forms a magnified, virtual, and final image $a_{2}b_{2}$. The final image $a_{2}b_{2}$ is generally formed at the least distance $D$ of distinct vision, although it can be formed anywhere between this position and infinity.

Angular Magnification Or Magnifying Power($M$):

The angular magnification or magnifying power can be defined as the ratio of the angle subtended by the image at the eye ($\beta$) to the angle subtended by the object at the eye when placed at least distance of distinct vision ($\alpha$)

$M= \frac{Angle \: subtended \: by \: the \: image \: at \: the \: eye \: (\beta)}{Angle \: subtended \: by \: the \: object \: at \: the \: eye \: when \\ placed \: at \: least \: distance \: of \: distinct \: vision \: (\alpha)}$

$M=\frac{\beta}{\alpha} \approx \frac{tan \beta}{tan \alpha} \quad (1)$

From figure

$tan \beta = \frac{a_{2}b_{2}}{a_{2} e} $

$tan \alpha = \frac{a_{2}a_{3}}{a_{2}e}$

Now subtitute these values in equation $(1)$, then

$M=\frac{\frac{a_{2}b_{2}}{a_{2} e}}{\frac{a_{2}a_{3}}{a_{2}e}}$

$M=\frac{a_{2}b_{2}}{a_{2}a_{3}}$

Here $a_{2}a_{3} = ab$

So the above equation can be written as

$M=\frac{a_{2}b_{2}}{ab}$

$M=\frac{a_{2}b_{2}}{ab} \frac{a_{1}b_{1}}{a_{1}b_{1}}$

$M=\frac{a_{2}b_{2}}{a_{1}b_{1}} \frac{a_{1}b_{1}}{ab}$

Here $m_{e}=\frac{a_{2}b_{2}}{a_{1}b_{1}}$ and $m_{\circ}= \frac{a_{1}b_{1}}{ab}$

Now substitute the values of $m_{e}$ and $m_{\circ}$ in the above equation

$M=m_{e} \times m_{\circ} \qquad(1)$

Where

$m_{e} \rightarrow$ The linear magnification produced by eye-piece lens system

$m_{\circ} \rightarrow$ The linear magnification produced by the object lens system

So now again from the figure

The linear magnification produced by object lens system $m_{\circ} = -\frac{v_{\circ}}{u_{\circ}}$

The linear magnification produced by eye-piece lens system $m_{e} = \frac{D}{u_{e}}$

Substitute the value of $m_{\circ}$ and $m_{e}$ in equation $(1)$

$M= -\frac{v_{\circ}}{u_{\circ}} \left( \frac{D}{u_{e}} \right) \qquad(2)$

Adjustment of a Compound Microscope:

1.) Adjustment for Clear Vision: In this final image an object is formed at least a distance of distinct vision $D$. For this configuration,

On substitution $u=-u_{e}$, $v=-D$ and $f=f_{e}$ in the lens formula for eyepiece lens

$\frac{1}{-D}+\frac{1}{u_{e}}=\frac{1}{f_{e}}$

$\frac{1}{u_{e}}=\frac{1}{f_{e}} + \frac{1}{D}$

$\frac{D}{u_{e}}= \left( \frac{D}{f_{e}} + 1 \right)$

Now substitute the value of $\frac{D}{u_{e}}$ in equation $(2)$

$M= -\frac{v_{\circ}}{u_{\circ}} \left(1+ \frac{D}{f_{e}} \right) $

The length of the microscope tube in this setup

$L=$ Distance between the object and the eye-piece lenses

$L= v_{\circ} + |u_{e}|$

2.) Adjustment for Relaxed Eye: In this configuration, the final image of an object is formed in a relaxed eye position i.e. at $\infty$. In this setup, the eye-piece lens system is moved back until the image of object $ab$, formed by object lens,i.e., $a_{1}b_{1}$ fall at (coincide with)second focus $f'_{e}$ of the eye-piece lens system. Mathematically this situation comes when $u_{e} = f_{e}$

Thus, the magnifying power in this position,

$M=-\frac{v_{\circ}}{u_{\circ}} \left( \frac{D}{f_{e}} \right)$

For this set the length of the microscope tube,

$L=v_{\circ}+f_{e}$

$m_{e} \rightarrow$ The linear magnification produced by eye-piece lens system

$m_{\circ} \rightarrow$ The linear magnification produced by the object lens system