What is the energy density in the electromagnetic wave in free space?
The total energy stored in electromagnetic waves per unit volume due to the electric field and the magnetic field is called energy density in the electromagnetic wave in free space.
$U=\epsilon_{0} E^{2}=\frac{B^{2}}{\mu_{0}}$
Derivation of Energy density in electromagnetic waves in free space:
The energy per unit volume due to the electric field is
$U_{E}= \frac{1}{2} \overrightarrow{E}.\overrightarrow{D} \qquad(1)$
The energy per unit volume due to the magnetic field is
$U_{B}= \frac{1}{2} \overrightarrow{B}.\overrightarrow{H} \qquad(2)$
The total energy density of electromagnetic waves is
$U=U_{E}+U_{B} \qquad(3)$
Now substitute the value of $U_{E}$ and $U_{B}$ in equation$(3)$ then we get
$U=\frac{1}{2} \left( \overrightarrow{E}.\overrightarrow{D}+\overrightarrow{B}.\overrightarrow{H} \right)$
$U=\frac{1}{2} \left( \overrightarrow{E}.\epsilon_{0}\overrightarrow{E}+\overrightarrow{B}.\frac{1}{\mu_{0}}\overrightarrow{B} \right) \qquad ( \because \overrightarrow{B}= \mu_{0} \overrightarrow{H} \:OR \: \overrightarrow{D}= \epsilon_{0}\overrightarrow{E} )$
$U=\frac{1}{2} \left( \epsilon_{0} E^{2}+\frac{B^{2}}{\mu_{0}} \right) \qquad ( \because \overrightarrow{E}\overrightarrow{E}= E^{2} \:OR \: \overrightarrow{B}.\overrightarrow{B}=B^{2})$
$U=\frac{1}{2} \left( \epsilon_{0} E^{2}+\frac{E^{2}}{c^{2} \mu_{0}} \right) \qquad ( \because B=\frac{E}{c})$
$U=\frac{1}{2} \left( \epsilon_{0} E^{2}+\frac{E^{2}}{c^{2} \mu_{0}} \right) $
$U=\frac{1}{2} \left( \epsilon_{0} E^{2}+\epsilon_{0} E^{2} \right) \qquad ( \because c=\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}})$
$U=\frac{1}{2} \left(2 \epsilon_{0} E^{2} \right) $
$U= \epsilon_{0} E^{2} $
Similarly, the energy density of electromagnetic waves in free space in terms of the magnetic field $B$ can be written as:
$U= \frac{B^{2}}{\mu_{0}} $
The average value of energy density in the electromagnetic waves in free space:
Now we will find the average value of energy density in the electromagnetic wave in free space from the above equation $U= \epsilon_{0} E^{2} $. So we get
$\left< U \right> = \epsilon_{0} \left< E^{2} \right>$
$\left< U \right> = \epsilon_{0} \frac{E_{0}^{2}}{2} \qquad \left (\because \left< E^{2} \right>=\frac{E_{0}^{2}}{2} \right)$
$\left< U \right> = \epsilon_{0} E_{rms}^{2} \qquad \left (\because E_{rms}^{2}=\frac{E_{0}^{2}}{2} \right) \qquad (4)$
We know that
$\left< \overrightarrow{S} \right> = \frac{E_{rms}^{2}}{Z_{0}} .\hat{n} \qquad (5)$
Now divide the equation $(5)$ by equation$(4)$
$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\frac{\frac{E_{rms}^{2}}{Z_{0}} .\hat{n}}{\epsilon_{0} E_{rms}^{2}}$
$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\frac{\hat{n}}{\epsilon_{0} Z_{0}}$
$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\frac{\hat{n}}{\sqrt{\epsilon_{0} \mu_{0}}} \qquad(\because z_{0}= \sqrt{\frac{\mu_{0}}{\epsilon_{0}}})$
$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\hat{n} c \qquad(\because c= \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}})$
$ \left< \overrightarrow{S} \right>=\hat{n} c \left< U \right> $
The energy flow per unit area per unit time in an electromagnetic wave is the product of energy density, speed of light, and the direction of propagation.
The ratio of the energy densities of the electric field and magnetic field:
So from above equation $U_{E}=\epsilon_{0} E^{2}$ and equation $U_{B}=\frac{B^{2}}{\mu_{0}}$, we can find the ratio between them i.e.
$\frac{U_{E}}{U_{B}}=\frac{\epsilon_{0} E^{2}}{\frac{B^{2}}{\mu_{0}}}$
$\frac{U_{E}}{U_{B}}=\frac{\epsilon_{0} \mu_{0} E^{2}}{B^{2}}$
$\frac{U_{E}}{U_{B}}=\frac{c^{2}}{c^{2}}$
$\frac{U_{E}}{U_{B}}=1$
$U_{E}=U_{B}$
So the energy density of the electric field is the same as the energy density of the magnetic field.
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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