In electromagnetic waves, the electric field vector and magnetic field vector are mutually perpendicular to each other (Proof)
The general solution of the wave equation for the electric field vector and magnetic field vector are respectively given below
$\overrightarrow{E}= E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(1)$
$\overrightarrow{B}= B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(2)$
Here $E_{\circ}$ and $B_{\circ}$ are the complex amplitude of electric field vector $\overrightarrow{E}$ and magnetic field vector $\overrightarrow{B}$ respectively and $\overrightarrow{k}$ is the propagation constant.
Now
$\overrightarrow{\nabla} \times \overrightarrow{E}= \left( \hat{i} \frac{\partial}{\partial x} + \hat{i} \frac{\partial}{\partial x} +\hat{i} \frac{\partial}{\partial x} \right). \left( \hat{i}E_{x} + \hat{j}E_{y} + \hat{k}E_{z} \right) $
$\overrightarrow{\nabla} \times \overrightarrow{E} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
E_{x} & E_{y} & E_{z} \\
\end{vmatrix}$
$\overrightarrow{\nabla} \times \overrightarrow{E} = \hat{i} \left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] - \hat{j} \left[ \frac{\partial E_{z}}{\partial x} -\frac{\partial E_{x}}{\partial z} \right] + \hat{k} \left[ \frac{\partial E_{y}}{\partial x} -\frac{\partial E_{x}}{\partial y } \right] \qquad(3)$
Here $E_{x}$, $E_{y}$ and $E_{z}$ are the component of electric field vector $\overrightarrow{E}$ in $x$, $y$, and $z$ direction. So the component form of the equation $(1)$
$\overrightarrow{E_{x}}= E_{\circ x} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$
$\overrightarrow{E_{y}}= E_{\circ y} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$
$\overrightarrow{E_{z}}= E_{\circ z} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$
We know that:
$\overrightarrow{k}.\overrightarrow{r}=\left( \hat{i}k_{x} + \hat{j}k_{y} + \hat{k}k_{z} \right). \left( \hat{i}x + \hat{j}y +\hat{k}z \right)$
$\overrightarrow{k}.\overrightarrow{r}=x k_{x} + yk_{y} + zk_{z} $
So above equation can be written as:
$\overrightarrow{E_{x}}= E_{\circ x} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(4)$
$\overrightarrow{E_{y}}= E_{\circ y} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(5)$
$\overrightarrow{E_{z}}= E_{\circ z} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(6)$
Now find that derivative from the equation $(4)$, equation $(5)$, and equation $(6)$ then substitute these values in equation $(3)$, So we get
$\left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] = \frac{\partial E_{z}}{\partial y} - \frac{\partial E_{y}}{\partial z} $
$\left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] = \frac{\partial}{\partial y} \left( E_{\circ z} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \right) -\frac{\partial}{\partial z} \left( E_{\circ y} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \right) $
$\left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] = \left(i k_{y} E_{z} - i k_{z} E_{y} \right) $
$\left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] = i \left( k_{y} E_{z} - k_{z} E_{y} \right) \qquad(7)$
Similarly
$\left[ \frac{\partial E_{z}}{\partial x} -\frac{\partial E_{x}}{\partial z} \right] = i \left( k_{x} E_{z} - k_{z} E_{x} \right) \qquad(8)$
$\left[ \frac{\partial E_{y}}{\partial x} -\frac{\partial E_{x}}{\partial y} \right] = i \left( k_{x} E_{y} - k_{y} E_{x} \right) \qquad(9)$
Now substitute the value of equation $(7)$, equation $(8)$, and equation $(9)$ in equation $(3)$
$\overrightarrow{\nabla} \times \overrightarrow{E} = i \left[\hat{i} \left( k_{y} E_{z} - k_{z} E_{y} \right) - \hat{j} \left( k_{x} E_{z} - k_{z} E_{x} \right) _ \hat{k} \left( k_{x} E_{y} - k_{y} E_{x} \right) \right]$
$\overrightarrow{\nabla} \times \overrightarrow{E} = i \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
k_{x} & k_{y} & k_{z} \\
E_{x} & E_{y} & E_{z} \\
\end{vmatrix}$
$\overrightarrow{\nabla} \times \overrightarrow{E} = i \left( \overrightarrow{k} \times \overrightarrow{E} \right) \qquad(10)$
According to Maxwell's third equation
$\overrightarrow{\nabla} \times \overrightarrow{E}= - \frac{\partial \overrightarrow{B}}{\partial t}$
$\overrightarrow{\nabla} \times \overrightarrow{E}= - \frac{\partial}{\partial t} \left( B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \right) \quad \left\{From \: equation \: (2)\right\}$
$\overrightarrow{\nabla} \times \overrightarrow{E}= i \omega \overrightarrow{B} \qquad(11)$
From equation $(10)$ and equation $(11)$
$i \left( \overrightarrow{k} \times \overrightarrow{E} \right) = i \omega \overrightarrow{B}$
$ \left( \overrightarrow{k} \times \overrightarrow{E} \right) = \omega \overrightarrow{B} \qquad(12)$
$\therefore$ Magnetic field vector $(\overrightarrow{B})$ is perpendicular to both electric field vector $(\overrightarrow{E})$ and propagation of wave vector $(\overrightarrow{k})$.
Similarly, from $\overrightarrow{\nabla} \times \overrightarrow{B}$, we get
$ \left( \overrightarrow{k} \times \overrightarrow{B} \right) = -\frac{\omega}{c} \overrightarrow{E} \qquad(13)$
Thus, In an electromagnetic wave, the electric field and magnetic field vector are perpendicular to each other and also to the direction of propagation of the wave.
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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