Derivation of momentum of electromagnetic wave:
Maxwell's had also predicted that electromagnetic waves transport linear momentum in the direction of propagation. Let a particle which has mass $m$ moving with velocity then the momentum of a particle,
$\overrightarrow{P}=m\overrightarrow{v} \qquad(1)$
According to mass-energy relation
$U=mc^{2}$
Here $U$ - Total energy of the particle
$m=\frac{U}{c^{2}} \qquad(2)$
From equation $(1)$ and equation $(2)$
$\overrightarrow{P}=\frac{U}{c^{2}} \overrightarrow{v} \qquad(3)$
If the electromagnetic wave is propagating along the x-axis then
$\overrightarrow{v}=c \hat{i}$
Put this value in the above equation $(3)$
$\overrightarrow{P}=\frac{U}{c} \hat{i} \qquad(4)$
We know that the equation of energy flow in electromagnetic wave
$\overrightarrow{S}= \frac{1}{\mu_{0} c} E^{2} \hat{n}$
Here wave is propagating along x-axis i.e
$\hat{n}=\hat{i}$
$\overrightarrow{S}= \frac{1}{\mu_{0} c} E^{2} \hat{i} \qquad(5)$
The energy density in plane electromagnetic wave in free space:
$U=\epsilon_{0} E^{2}$
Where $E$ - Magnitude of electric field
$E^{2}=\frac{U}{\epsilon_{0}} \qquad(6)$
Now substitute the value of $E^{2}$ in equation$(5)$
$\overrightarrow{S}= \frac{1}{\mu_{0} c} \frac{U}{\epsilon_{0}} \hat{i} $
$\overrightarrow{S}= \frac{c^{2}}{c} U \hat{i} \qquad (\because \frac{1}{\sqrt{ \mu_{0} \epsilon_{0}}}=c) $
$\overrightarrow{S}= c U \hat{i} $
$U \hat{i}=\frac{\overrightarrow{S}}{c} \qquad(7)$
Now substitute the value of $ U \hat{i} $ in equation $(4)$. Then
$\overrightarrow{P}=\frac{\overrightarrow{S}}{c}$
$\overrightarrow{P}=\frac{(\overrightarrow{E} \times \overrightarrow{B})}{ \mu_{0}c^{2}}$
$\overrightarrow{P}=\epsilon_{0}(\overrightarrow{E} \times \overrightarrow{B})$
This is the equation of "Momentum of electromagnetic 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
Comments
Post a Comment
If you have any doubt. Please let me know.