### Intensity of a wave

Definition of Intensity of a wave:
In a medium, the energy per unit area per unit time delivered perpendicuar to the direction of the wave propagation s caled the intensity of the wave. It is denoted by $I$.

If the energy $E$ is delivered in the time $t$ rom area $A$ perpendicular to the wave propagation, then

$I=\frac{E}{At} \qquad{1}$

Unit: $Joule/m^{2}-sec$ or $watt/m^{2}$

Dimensional formula: $[MT^{-3}]$

We know that the total mechanical energy of a vibrating particle is

$E=\frac{1}{2}m \omega^{2} a^{2}$

Where $\omega$ is the angular frequency and $a$ is the amplitude of the wave.

$E=\frac{1}{2}m (2\pi n)^{2} a^{2} \qquad \left( \omega=2\pi n \right)$

$E=2 \pi^{2} m n^{2} a^{2} \qquad(2)$

Where $m$ is the mass of the vibrating particle.

Now substitute the value of $E$ from equation $(2)$ to equation $(1)$. So the intensity of the wave

$I=\frac{2 \pi^{2} m n^{2} a^{2}}{At} \qquad(3)$

If the wave travels the distance $x$ in time $t$ with velocity $v$, Then

$t=\frac{x}{v}$

Now substitute the value of the above equation in equation $(3)$

$I=\frac{2 \pi^{2} m v n^{2} a^{2}}{Ax}$

$I=\frac{2 \pi^{2} m v n^{2} a^{2}}{V}$

Where $V$ is the volume of the corresponding medium during the wave propagation in time $t$.

$I=2 \pi^{2} \rho v n^{2} a^{2} \qquad \left(\because \rho=\frac{m}{V} \right)$

It is clear that for wave propagation in a medium with a constant velocity, i.e. wave's intensity is directly proportional to the square of amplitude and frequency both.

$I\propto a^{2}$ and $I \propto n^{2}$

### 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

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