### 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

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

### Numerical Aperture and Acceptance Angle of the Optical Fibre

Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig

### Fraunhofer diffraction due to a single slit

Let $S$ be a point monochromatic source of light of wavelength $\lambda$ placed at the focus of collimating lens $L_{1}$. The light beam is incident normally from $S$ on a narrow slit $AB$ of width $e$ and is diffracted from it. The diffracted beam is focused at the screen $XY$ by another converging lens $L_{2}$. The diffraction pattern having a central bright band followed by an alternative dark and bright band of decreasing intensity on both sides is obtained. Analytical Explanation: The light from the source $S$ is incident as a plane wavefront on the slit $AB$. According to Huygens's wave theory, every point in $AB$ sends out secondary waves in all directions. The undeviated ray from $AB$ is focused at $C$ on the screen by the lens $L_{2}$ while the rays diffracted through an angle $\theta$ are focussed at point $p$ on the screen. The rays from the ends $A$ and $B$ reach $C$ in the same phase and hence the intensity is maximum. Fraunhofer diffraction due to

### Electromagnetic wave equation in free space

Maxwell's Equations: Maxwell's equation of the electromagnetic wave is a collection of four equations i.e. Gauss's law of electrostatic, Gauss's law of magnetism, Faraday's law of electromotive force, and Ampere's Circuital law. Maxwell converted the integral form of these equations into the differential form of the equations. The differential form of these equations is known as Maxwell's equations. $\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\rho}{\epsilon_{0}}$ $\overrightarrow{\nabla}. \overrightarrow{B}=0$ $\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$ $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}$ Modified Form: $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \left(\overrightarrow{J}+ \epsilon \frac{ \partial \overrightarrow{E}}{\partial t} \right)$ For free space