### Generation of wave function for a free particle

Simple harmonic motion:

If an object repeats the process or path at a fixed interval of time is known as periodic motion or Uniform circular motion. It is also called the Simple harmonic motion.

We know that the wave is the study of infinite S.H.M.

Let us consider a particle is moving with uniform velocity in a circular path with radius $A$ i.e. particle is doing simple harmonic motion. Let at any instant $t$ particle move from position $P_{1}$ to $P_{2}$. So vector resolution of $P_{2}$ position is:

From below figure-

Horizontal Component i.e. $x$ component of position vector $P_{2}$:

$x=A sin(\omega t - \phi) \qquad (1)$

Vertical Component i.e. $y$ component of position vector $P_{2}$:

$y=A cos (\omega t - \phi) \qquad (2)$
 Simple harmonic motion of a particle
According to Max Born hypothesis-

“Wave is a complex quantity which is represented by a wave function $\varphi$ ”

i.e. wave function mathematically can be represented as

$\varphi= x+iy \qquad (3)$

Substitute the value of $x$ component and $y$ component in above equation-

$\varphi= A sin (\omega t - \phi)+ i A cos (\omega t - \phi)$

$\varphi=A [sin (\omega t - \phi)+ i cos (\omega t - \phi) ]$

$\varphi=A e^ {i(\omega t - \phi)} \qquad (4)$

Where $\phi$is the phase of the wave. The value of $\phi$ can be found by the relation between phase difference and path difference of wave plane progressive wave.

$\phi =\frac {2\pi}{\lambda}\cdot x$

$\phi = k\cdot x$

Now substitute the value of $\phi$ from above in equation $(4)$. So wave function equation can be written as-

$\varphi=A e^ {i(\omega t - k \cdot x)}$
 Propagation of a wave along the x-axis
Let a particle of mass m be in motion along the positive x-direction with accurately known momentum $p$ and total energy $E$.

So from the equation of plane progressive wave-

$\psi(x,t)=Ae^{-i\omega (t-\frac{x}{v})} \qquad(5)$

$\psi(x,t )=Ae^{-i(\omega t-kx)}$

$\psi(x,t )=Ae^{i(kx-\omega t)} \qquad(6)$

According to Planck’s hypothesis-

$E=h\nu \qquad(7)$

$E=\frac{h}{2 \pi }\cdot 2 \pi \nu$

$E=\hbar.\omega$

$\omega=\frac{E}{\hbar } \qquad(8)$

According to de Broglie hypothesis-

$\lambda= \frac{h}{p}$

$p =\frac{h}{\lambda }$

$p =\frac{h}{2 \pi }\cdot \frac{2 \pi}{\lambda}$

$p=\hbar\cdot k$

$k=\frac{p}{\hbar}\qquad(9)$

Now put the value of $\omega$ and $k$ from equation $(8)$  and equation $(9)$in equation$(6)$

$\psi (x,t)=Ae^{ i(\frac{p}{\hbar }x-\frac{E}{\hbar }t)}$

$\psi (x,t )=Ae^{\frac{i}{\hbar}(px-Et)}$

The three-dimensional equation of wave function of a free particle:

$\psi (\overrightarrow{r},t)=Ae^{\frac{i}{\hbar}(\overrightarrow{p}x-Et)}$

### 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