### Principle and Proof of Law of Conservation of Energy

Law of Conservation of Energy: According to this principle
The energy is neither created nor destroyed. The energy can be changed from one form to another form. i.e. when there is not any external influence is applied on the particle then the total energy of the particle is always conserved.

Proof of Conservation's Law of Energy: Let us consider, A particle is freely falling from height $h$ under the gravitational acceleration $g$. So the total energy of the particle at different points :
Calculation of Total Energy at Point $A$:

The initial velocity of the particle at point $A$ is $(v_{A})=0$

The kinetic energy of the particle at point $A$ is $(K_{A})=\frac{1}{2}mv_{A}^{2}$

$K_{A}=0\qquad \left( \because v_{A}=0 \right)$

The potential energy of the particle at point $A$ is $(U_{A})=mgh$

The total energy of a particle at point $A$ is $(E_{A})=K_{A}+U_{A}$

Now substitute the value of $K_{A}$ and $U_{A}$ in the above equation. Now the above equation can be written as:

$E_{A}=0+mgh$

$E_{A}=mgh \qquad(1)$

Calculation of Total Energy at Point $B$: If a particle travels the distance $x$ and reaches point $B$ then velocity at point $B$ can be calculated by the equation of motion. i.e.

$v_{B}^{2}=v_{A}^{2}+2gx$

But the initial velocity of the particle is zero. i.e $v_{A}=0$ then the above equation can be written as

$v_{B}^{2}=0+2gx$

$v_{B}^{2}=2gx$

The kinetic energy of the particle at point $B$ is $(K_{B})=\frac{1}{2}mv_{B}^{2}$

Now substitute the value of $v_{B}^{2}$ in the above equation. Now the above equation can be written as:

$K_{B}=\frac{1}{2}m \left( 2g x \right)$

$K_{B}=mgx$

The height of the particle at point $B$ is $(h-x)$ then the potential energy at point $B$ is $(U_{B})=mg(h-x)$

The total energy of a particle at point $B$ is $(E_{B})=K_{B}+U_{B}$

Now substitute the value of $K_{B}$ and $U_{B}$ in the above equation. Now the above equation can be written as:

$E_{B}=mgx+mg(h-x)$

$E_{B}=mgh \qquad(2)$

Calculation of Total Energy at Point $C$: Now the particle reaches point $C$ by traveling distance $h$ from the initial point $A$. This point $C$ is just before the collision from the surface. So the velocity of the particle at point $C$.

$v_{C}^{2}= v_{A}^{2}+2gh$

$v_{C}^{2}= 0+2gh \qquad \left( v_{A}=0 \right)$

$v_{C}^{2}= 2gh$

The kinetic energy of the particle at point $C$ is $(K_{C})=\frac{1}{2}mv_{C}^{2}$

Now substitute the value of $v_{C}^{2}$ in the above equation. Now the above equation can be written as:

$K_{C}=\frac{1}{2}(2mgh)$

$K_{C}=mgh$

At point $C$ the particle is just above the surface so the height $h$ of the object will be zero. i.e. $h=0$

The potential energy at point $C$ is $(U_{C})=0$

The total energy of the particle at point $C$ is $(E_{C})=K_{C}+U_{C}$

Now substitute the value of $K_{C}$ and $U_{C}$ in the above equation. Now the above equation can be written as:

$E_{C}=mgh+0$

$E_{C}=mgh \qquad(3)$

From the above equation $(1)$, equation $(2)$ and equation $(3)$, we get

$E_{A}=E_{B}=E_{C}$

Now we can conclude that the total energy of the particle is always conserved when there is no external influence applied. This is proof of the law of conservation of Energy.

### 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