### Frame of References (Inertial Frame and Non Inertial Frame)

It is assumed that space is continuous and the motion of particles in space can be described by their position at different instants of time. The position of a particle is known as a point in space. These points are described by the coordinate system in space.

When point (position of particle in space) and the time are taken together then it is called an Event.
The coordinate system of a particle which describe the position of any particle relative to it, then such coordinate system is known as Frame of Reference or System of Reference.

Absolute Space:

The absolute space is those frame of reference relative to which every motion and position should be measured.

Types of Frame of Reference:

According to the motion of particles frame of reference is divided into two categories

1. Inertial frame of reference
2. Non-inertial frame of reference
Inertial frame of reference:

Those unaccelerated frame of reference, in which Newton's first and second laws hold, are called inertial frames.
In an inertial frame, body does not experience any force so according to second law of Newton-

$F=ma \qquad (1)$

$F=0$

so from equation $(1)$

$ma=0$

$a=0$

$\frac{d^2r}{dt^2}=0\qquad (2)$

In component form, above equation, can be written as:

$\frac{d^2x}{dt^2}=0;\:\:\:\frac{d^2y}{dt^2}=0;\:\:\:\frac{d^2z}{dt^2}=0\qquad (3)$

Discussion of Inertial Frame in terms of relative frame of reference:

Let us consider an inertial frame $S$ and another frame $S'$ which is moving with constant velocity $v$ relative to frame $S$. If the position of the origins of the two frames coincide, then in the two frames the position vector of any particle $P$ at any instant $t$ can be related by the following expression-

 Inertial Frame of Reference

$\overrightarrow{r}=\overrightarrow{oo'}+\overrightarrow{r'}$

$\overrightarrow{r}=\overrightarrow{v}t+\overrightarrow{r'} \qquad \left \{ {\overrightarrow{oo}'=\overrightarrow{v}t} \right \}$

Differentiate the above equation

$\frac{d\overrightarrow{r}}{dt}=\overrightarrow{v}+\frac{d\overrightarrow{r'}}{dt}$

Again differentiate the above equation

$\frac{d^2\overrightarrow{r}}{dt^2}=\frac{d^2\overrightarrow{r'}}{dt^2}$

$a=a'$

Now we conclude that "If a frame is an inertial frame then all those frames which are moving with constant velocity relative to the first frame are also inertial"

Non-inertial Frame:
Those frames of reference in which Newton's law of inertia does not hold are called non-inertial frames.
All the accelerated and rotating frames are the non-inertial frames of reference.

According to Newton's Second Law, the force $F$ applied on a body of mass $m$ is given by

$F_{i}=ma_{i}\qquad (1)$

Newton's Second Law is not valid when a body of mass $m$ self accelerated and accelerated body will observe the acceleration $a_{N}$. Hence

$F_{i}\neq ma_{N}\qquad (2)$

If no external force is acting on a particle. Even then in the accelerated frame, It will appear that a force is acting on it. This force is called pseudo force or fictitious force. The direction of force is opposite to acceleration.

$F_{o}\neq ma_{o}\qquad (3)$

Discussion of Non-inertial Frame in term of relative frame of reference:

Let us consider, an inertial frame $S$ and another frame $S'$ is moving with an acceleration $a_{0}$ relative to frame $S$.
 Non-inertial Frame of Reference
$\overrightarrow{r_{i}}=\overrightarrow{r_{N}}+\frac{1}{2}a_{0}t^{2}$

Differentiate the above equation with respect to $'t'$

$\frac{d\overrightarrow{r_{i}}}{dt}=\frac{d\overrightarrow{r_{N}}}{dt}+a_{0}t$

Again differentiate the above equation with respect to $'t'$

$\frac{d^{2}\overrightarrow{r_{i}}}{dt^{2}}=\frac{d^{2}\overrightarrow{r_{N}}}{dt^{2}}+a_{0}$

$a_{i}=a_{N}+a_{0}$

$a_{i}-a_{0}=a_{N}$

$ma_{i}-ma_{0}=ma_{N}$

$F_{i}+F_{0}=F_{N}$

This formula gives the observed force $F_{N}$ in the accelerated system.

### 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