Principle Construction, Working and Angular Magnification of Simple Microscope

Principle of Simple Microscope:

The principle of the simple microscope is based on the magnification of an image by using a simple convex lens.

Construction:

A simple microscope consists of one convergent lens only. The object is placed between the lens and its focal length, and the eye is placed just behind the lens. Then the eye sees a magnified, erect, and virtual image on the same side as the object at the least distance of distinct vision $(D)$ from the eye, and the image is then seen most distinctly.

Working:

If the small object $ab$ is placed between a lens $O$ and its first focus $f$ then Its magnified virtual image $a_{1}b_{1}$ is formed at a distance $D$ from the lens. Since the eye is just behind the lens, the distance of image $a_{1}b_{1}$ from the eye is also $D$.

Angular Magnification Or Magnifying Power($M$):

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 the least distance of distinct vision ($\alpha$) is called the angular magnification or magnifying power.

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

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

From figure

$tan \beta = \frac{ab}{oa} $

$tan \alpha = \frac{a_{1}b_{2}}{a_{1}o}$

Here $a_{1}b_{2} = ab$

$tan \alpha = \frac{ab}{a_{1}o}$

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

$M=\frac{\frac{ab}{ao}}{\frac{ab}{a_{1}o}}$

$M=\frac{a_{1}o}{ao}$

Here $ao = u$ (Distance between object and optical center of the lens) and $a_{1}o = D$ (Least Distance of distinct vision), then the above equation can be written as

$M=\frac{D}{u} \qquad(2)$

We know that the lens formula $\frac{1}{v}-\frac{1}{u} = \frac{1}{f}$

Now put
$v=-D$ (The image $a'b'$ is being formed at a distance $D$ from lens)
$u=-u$

$\frac{1}{-D}-\frac{1}{-u} = \frac{1}{f}$

Multiply $D$ in the above equation

$-\frac{D}{D}-\frac{D}{-u} = \frac{D}{f}$

$-1-\frac{D}{-u} = \frac{D}{f}$

$\frac{D}{u} =1 + \frac{D}{f} \qquad(3)$

From equation $(2)$ and equation $(3)$, then

$M=1 + \frac{D}{f} $

If eye is kept at distance $d$ from lens then $v=-(D-d)$, and the magnifying power will be

$M=1+\frac{D-d}{f}$

To see with a relaxed eye, the image $a'b'$ should be formed at infinity. In this case, the object $ab$ will be at the focus of the lens, i.e. $u=f$ then magnifying power

$M= \frac{D}{f} $