The electromagnetic wave equations in non-conducting media :
For electric field vector:
$\nabla^{2} \overrightarrow{E}=\frac{1}{v^{2}} \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} \qquad(1)$
For magnetic field vector:
$\nabla^{2} \overrightarrow{B}=\frac{1}{v^{2}} \frac{\partial^{2} \overrightarrow{B}}{\partial t^{2}} \qquad(2)$
The wave equation of electric field vector:
$\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$
The wave equation of magnetic field vector:
$\overrightarrow{B}(\overrightarrow{r},t)=B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$
Now the solution of electromagnetic wave for electric field vector.
Differentiate with respect to $t$ of equation $(3)$
$\frac{\partial \overrightarrow{E}}{\partial t}=i \omega E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$
Again differentiate with respect to $t$ of above equation:
$\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}=i^{2} \omega^{2} E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$
$\frac{\partial^{2} \overrightarrow{E}}{\partial^{2} t}=- \omega^{2} \overrightarrow{E}(\overrightarrow{r},t)$
Now substitute the value of the above equation in equation$(1)$
$\nabla^{2} \overrightarrow{E}=\frac{-\omega^{2}}{v^{2}} \overrightarrow{E}(\overrightarrow{r},t)$
$\nabla^{2} \overrightarrow{E}=-(\frac{\omega}{v})^{2} \overrightarrow{E}(\overrightarrow{r},t)$
$\nabla^{2} \overrightarrow{E}=-\alpha^{2} \overrightarrow{E}(\overrightarrow{r},t) \qquad (\because \alpha=\frac{\omega}{v} )$
Where $\alpha$ - Wave propagation Constant
$\nabla^{2} \overrightarrow{E} + \alpha^{2} \overrightarrow{E}(\overrightarrow{r},t)=0 $
This is the solution of the electromagnetic wave equation in non-conducting media for the electric field vector.
Now the component form of the above equation:
$(\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} +\frac{\partial^{2}}{\partial z^{2}})(\hat{i}E_{x}+\hat{j}E_{y}+\hat{k}E_{z}) \\ =- \alpha^{2}(\hat{i}E_{x}+\hat{j}E_{y}+\hat{k}E_{z}) \qquad(5)$
If the wave is propagating along $z$ direction. Then for uniform-plane electromagnetic waves-
$\frac{\partial}{\partial x}=\frac{\partial}{\partial y}=0$
$\frac{\partial^{2}}{\partial x^{2}}=\frac{\partial^{2}}{\partial y^{2}}=0$
$E_{z}=0$
Now the equation $(5)$ can be written as:
$\frac{\partial^{2}}{\partial x^{2}} (\hat{i}E_{x}+\hat{j}E_{y})=- \alpha^{2}(\hat{i}E_{x}+\hat{j}E_{y})$
Now separate the above equation in $x$ and $y$ components so
$\frac{\partial^{2} E_{x}}{\partial z^{2}}=- \alpha^{2}E_{x}$
$\frac{\partial^{2}E_{y}}{\partial z^{2}} =- \alpha^{2}E_{y}$
The solution of electromagnetic wave for magnetic field vector can find out by following the above method.
Therefore $x$ and $y$ components of the solution of the electromagnetic wave equation for magnetic field vector can be written as. i.e.
$\frac{\partial^{2} B_{x}}{\partial z^{2}}=- \beta^{2}B_{x} \qquad \left( \because \beta=\frac{\omega}{v} \right)$
$\frac{\partial^{2}B_{y}}{\partial z^{2}} =- \beta^{2}B_{y}$
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
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