Derivation of Einstein Coefficient Relation→
Let us consider the $N_{1}$ and $N_{2}$ is the mean population of lower energy state and upper energy state respectively. If the energy density of incident light is $\rho(\nu)$ then
The rate of transition of number of atoms due to absorption process:
$R_{abs}=B_{12} \: \rho(v) \: N_{1} \qquad(1)$
The above equation shows the number of atoms absorbing the photon per second per unit volume
Where $B_{12}$= Einstein Absorption Coefficent
The rate of transition of number of atoms due to sponteneous emission process:
$R_{sp}=A_{21} \: N_{2} \qquad(2)$
The above equation shows the number of atoms emitting the photon per second per unit volume due to spontaneous emission
Where $A_{21}$= Einstein Spontaneous Emission Coefficient
The rate of transition of the number of atoms due to stimulated emission process:
$R_{st}=B_{21} \: \rho(v) \: N_{2} \qquad(3)$
The above equation shows the number of atoms emitting the photon per second per unit volume due to stimulated emission
Where $B_{21}$= Einstein Stimulated Emission Coefficient
Under the thermal equilibrium, the mean population $N_{1}$ and $N_{2}$ in lower and upper energy states respectively must remain constant. This condition requires that the transition of the number of atoms from $E_{2}$ to $E_{1}$ must be equal to the transition of the number of atoms from $E_{1}$ to $E_{2}$. Thus
$\left.\begin{matrix}The \: number \: of \: atoms \: absorbing \\ photons \: per \: second \: per \: unit \: volume
\end{matrix}\right\} \\ = \left.\begin{matrix} The \: number \: of \: atoms \: emitting \\ photons \: per \: second \: per \: unit \: volume
\end{matrix}\right\}$
i.e $R_{abs}= R_{sp}+R_{st}$
$B_{12} \: \rho(v) \: N_{1}= A_{21} \: N_{2} + B_{21} \: \rho(v) \: N_{2}$
$B_{12} \: \rho(v) \: N_{1} - B_{21} \: \rho(v) \: N_{2} = A_{21} \: N_{2} $
$ \rho(v) (B_{12} \: N_{1} - B_{21} \: N_{2} ) = A_{21} \: N_{2} $
$\rho(v)=\frac{A_{21} \: N_{2}}{(B_{12} \: N_{1} - B_{21} \: N_{2} )} \qquad(4)$
We know that
$\frac{N_{1}}{N_{2}}=e^{\frac{(E_{2}-E_{1})}{kT}}$
$\frac{N_{1}}{N_{2}}=e^{\frac{h\nu}{kT}}$
Now substitute the value of $\frac{N_{1}}{N_{2}}$ in equation $(4)$
$\rho(v)=\frac{A_{21}}{B_{12}} \left [ \frac{1}{e^{\frac{h\nu}{kT}}- \frac{B_{21}}{B_{12}}} \right ] \qquad(5)$
According to Planck's Radiation Law
$\rho(v)=\frac{8\pi h \nu^{3}}{c^{3}} \left [ \frac{1}{e^{\frac{h\nu}{kT}}- 1} \right ] \qquad(6)$
Now comparing the equation $(5)$ and equation $(6)$
$\frac{B_{21}}{B_{12}}=1$ and $\frac{A_{21}}{B_{12}}=\frac{8\pi h \nu^{3}}{c^{3}}$
From the above equation, we get
$B_{21}=B_{12}$
$B_{12}=B_{21}=\frac{c^{3}}{8\pi h \nu^{3}}A_{21}$
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.