### Parallel Plate Air Capacitor and Its Capacitance

Parallel Plate Air Capacitor→

A parallel-plate capacitor consists of two long, plane, metallic plates mounted on two insulating stands and placed at a small distance apart in a vacuum (or air). The plates are exactly parallel to each other.
 Parallel Plate Air Capacitor
Derivation of the capacitance of parallel plate capacitor in the air→

Let us consider, Two plates $X$ and $Y$ are separated at a small distance $d$ in the vacuum (or air). If the area of plates is $A$ and the plates $X$ and $Y$ have charge $+q$ and $-q$ respectively. If the surface charge density on each plate is $\sigma$ then electric field intensity at a point between two parallel plates is

$E=\frac{\sigma}{\epsilon_{\circ}}$

$E=\frac{q}{\epsilon_{\circ}A}\qquad(1) \qquad (\because \sigma=\frac{q}{A})$

The potential difference between the parallel plates capacitor in air (or vacuum) is

$V=E.d$

Now substitute the value of $E$ from equation$(1)$ in the above equation

$V=\frac{q}{\epsilon_{\circ}A}.d \qquad(2)$

The capacitance of the parallel plate capacitor in air (or vacuum) is

$C_{\circ}=\frac{q}{V} \qquad(3)$

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

$C_{\circ}=\frac{q}{\frac{q}{\epsilon_{\circ}A}.d}$

$C_{\circ}=\frac{\epsilon_{\circ}A}{d}$

If the space between the plates is filled with some dielectric medium of dielectric constant $K$, then the electric field between the plates is

$E=\frac{\sigma}{K \: \epsilon_{\circ}}$

The Potential difference between the parallel plates when dielectric medium $(K)$ is present between them

$V=\frac{q}{K \: \epsilon_{\circ} \: A}.d$

Now, The capacitance of the parallel-plate capacitor when the dielectric medium $(K)$ is present between them

$C=\frac{K\: \epsilon_{\circ} \: A}{d}$

### 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 then 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 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 then 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 light w

### 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

### Particle in one dimensional box (Infinite Potential Well)

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below. Particle in One-Dimensional Box(Infinite Potential Well) The potential energy inside the one -dimensional box can be represented as $\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$ $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$ If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as: $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$ \$\frac{d^{2} \psi(x)}{d x