Derivation of energy flow in the electromagnetic wave in free space:
The Poynting vector is given by
$\overrightarrow{S}=\overrightarrow{E} \times \overrightarrow{H} \qquad(1)$
$\overrightarrow{S}=\frac{1}{\mu_{0}} ( \overrightarrow{E} \times \overrightarrow{B} ) \qquad(2) \qquad (\because \overrightarrow{B}= \mu_{0} \overrightarrow{H})$
We know that the characteristic impedance equation i.e.
$\overrightarrow{B}=\frac{1}{\mu_{0}c}(\hat{n} \times \overrightarrow{E}) \qquad(3)$
Now substitute the value of $\overrightarrow{B}$ in equation$(2)$
$\overrightarrow{S}=\frac{1}{\mu_{0}c} [\overrightarrow{E} \times (\hat{n} \times \overrightarrow{E})]$
$\overrightarrow{S}=\frac{1}{\mu_{0}c} [(\overrightarrow{E}.\overrightarrow{E}) \hat{n}- (\overrightarrow{E}.\hat{n}) \overrightarrow{E})] \qquad(4)$
As $\overrightarrow{E}$ is perpendicular to $\hat{n}$ so $\overrightarrow{E} . \hat{n}=0$ then we get for above equation$(4)$
$\overrightarrow{S}=\frac{1}{\mu_{0}c} E^{2} \hat{n}$
From the above equation, we can conclude $\overrightarrow{S}$ has the same direction as $\hat{n}$ which is the direction of wave propagation.
Energy flow in an electromagnetic wave takes place in the direction of the propagation of the wave.
Here $\mu_{0} c=Z_{0}$ (The Characteristic impedance of free space)
$\overrightarrow{S}=\frac{1}{Z_{0}} E^{2} \hat{n}$
This is the equation of energy flow in the electromagnetic wave in free space.
The average energy flow over one period of the electromagnetic wave in free space:
Now the average energy flow of the above equation
$ \left< \overrightarrow{S} \right> =\frac{1}{Z_{0}} \left< E^{2} \right> \hat{n} \qquad(5)$
We know the electric field vector wave equation i.e. $\overrightarrow{E}=E_{0} e^{i(\overrightarrow{k}. \overrightarrow{r} -\omega t)} $
So the value of $\left< E^{2} \right>$ from above equation:
$\left< E^{2} \right>= \left< Re[E_{0} e^{i(\overrightarrow{k}. \overrightarrow{r} -\omega t)}]^{2} \right>$
$\left< E^{2} \right>= \left< E_{0}^{2}\: cos^{2}(\overrightarrow{k}. \overrightarrow{r} -\omega t) \right>$
For one period or cycle of electromagnetic wave the value of $cos^{2}(\overrightarrow{k}. \overrightarrow{r} -\omega t)=\frac{1}{2}$ then we get
$\left< E^{2} \right>= \frac{E_{0}^{2}}{2}$
$\left< E^{2} \right>= (\frac{E_{0}}{2})^{2}$
$\left< E^{2} \right>= E_{rms}^{2} \qquad \left (\because E_{rms} = \frac{E_{0}}{2}\right )$
Now substitute the value of $ \left< E^{2} \right>$ in equation $(5)$ then we get
$ \left< \overrightarrow{S} \right> =\frac{E_{rms}^{2}}{Z_{0}} \hat{n} $
This average energy flow equation over one period of the electromagnetic wave in free space.
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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