91Ó°ÊÓ

Consider the Light of (angular) frequency w passes from medium $1$, through a slab (thickness$d$ ) of medium$2$ , and into medium $3$ (for instance, from water through glass into air.

Write the formula of transmission coefficient for normal incidence.

$\mathbf{T}\mathbf{=}\frac{{\mathbf{Î\mu }}_3{\mathbf{Ï\dots }}_3}{{\mathbf{Î\mu }}_1{\mathbf{Ï\dots }}_1}$ …… (1)

Here, ${\mathbf{Î\mu }}_3$relative permittivity of medium 3, ${\mathbf{Ï\dots }}_3$wave velocity at medium 3, ${\mathbf{Î\mu }}_1$relative permittivity of medium 1 and${\mathbf{Ï\dots }}_1$ wave velocity at medium 1.

The fields are, in material $1$, if the two planes are $z=0$ and $z=d$, the electric field travels down the z-axis, and the electric field is polarised along the x-axis.

$\begin{array}{c}{\stackrel{→}{\mathrm{E}}}_1={\mathrm{E}}_{01}{\mathrm{e}}^{\mathrm{i}({\mathrm{k}}_1\mathrm{z}−\mathrm{Ó\neg t})}\widehat{\mathrm{x}}\\ {\stackrel{→}{\mathrm{B}}}_1=\frac{{\mathrm{E}}_{01}}{{\mathrm{Ï\dots }}_1}{\mathrm{e}}^{\mathrm{i}({\mathrm{k}}_1\mathrm{z}−\mathrm{Ó\neg t})}\widehat{\mathrm{z}}×\widehat{\mathrm{x}}\\ =\frac{{\mathrm{E}}_{01}}{{\mathrm{Ï\dots }}_1}{\mathrm{e}}^{\mathrm{i}({\mathrm{k}}_1\mathrm{z}−\mathrm{Ó\neg t})}\widehat{\mathrm{y}}\\ {\stackrel{→}{\mathrm{E}}}_{\mathrm{R}}={\mathrm{E}}_{0\mathrm{R}}{\mathrm{e}}^{\mathrm{i}(−{\mathrm{k}}_1\mathrm{z}−\mathrm{Ó\neg t})}\widehat{\mathrm{x}}\end{array}$

Solve further as

$\begin{array}{c}{\stackrel{→}{B}}_R=\frac{E_{0R}{e}^{i(−k_{1}z−\mathrm{Ó\neg }t)}}{{\mathrm{Ï\dots }}_1}(−\widehat{z})×\widehat{x}\\ =−\frac{E_{0R}}{{\mathrm{Ï\dots }}_1}{e}^{i(−k_{1}z−\mathrm{Ó\neg }t)}\widehat{y}\end{array}$

Now fields on material $2$:

$\begin{array}{c}{\stackrel{→}{E}}_r=E_{0r}{e}^{i(k_{2}z−\mathrm{Ó\neg }t)}\widehat{x}\\ {\stackrel{→}{B}}_r=\frac{E_{0r}}{{\mathrm{Ï\dots }}_2}{e}^{i(k_{2}z−\mathrm{Ó\neg }t)}\widehat{y}\\ {\stackrel{→}{E}}_{\mathrm{ι}}=E_{0\mathrm{ι}}{e}^{i(−k_{2}z−\mathrm{Ó\neg }t)}\widehat{x}\\ {\stackrel{→}{B}}_{\mathrm{ι}}=−\frac{E_{0\mathrm{ι}}}{{\mathrm{Ï\dots }}_2}{e}^{i(−k_{2}z−\mathrm{Ó\neg }t)}\widehat{y}\end{array}$

The transmitted wave is the only wave in material 3, where $r$and $l$represent for waves travelling to the right and left, respectively.

$\begin{array}{l}{\stackrel{→}{E}}_T=E_{0T}{e}^{i(k_{3}z−\mathrm{Ó\neg }t)}\widehat{x}\\ {\stackrel{→}{B}}_T=\frac{E_{0T}}{{\mathrm{Ï\dots }}_3}{e}^{i(k_{3}z−\mathrm{Ó\neg }t)}\widehat{y}\end{array}$

Apply boundary conditions on both planes. The initial one, ${\stackrel{→}{\mathrm{E}}}_{âˆ\yen ,1}={\stackrel{→}{\mathrm{E}}}_{âˆ\yen ,2}$, provides:

${\mathrm{E}}_{01}+{\mathrm{E}}_{0\mathrm{R}}={\mathrm{E}}_{0\mathrm{R}}+{\mathrm{E}}_{0\mathrm{l}}$ …… (2)

And

$E_{0T}{e}^{i{k}_{3}d}=E_{0r}{e}^{i{k}_{2}d}+E_{0l}{e}^{−i{k}_{2}d}$ …… (3)

And the second one, ${\stackrel{→}{B}}_{âˆ\yen ,1}/{\mathrm{μ}}_1={\stackrel{→}{B}}_{âˆ\yen ,2}/{\mathrm{μ}}_2$, gives:

$\frac{1}{{\mathrm{μ}}_1{\mathrm{Ï\dots }}_1}(E_{0l}−E_{0R})=\frac{1}{{\mathrm{μ}}_2{\mathrm{Ï\dots }}_2}(E_{0r}−E_{0l})$ …... (4)

And

$\frac{1}{{\mathrm{μ}}_3{\mathrm{Ï\dots }}_3}{E}_{0T}{e}^{i{k}_{3}d}=\frac{1}{{\mathrm{μ}}_2{\mathrm{Ï\dots }}_2}\left(E_{0r}{e}^{i{k}_{2}d−E_{0l}{e}^{−i{k}_{2}d}}\right)$ …… (5)

Therefore, we need to express $E_{0T}$ in terms of $E_{0I}$ and have $4$equations with $4$unknowns. To (2) and (4), add:

$\begin{array}{c}2{\mathrm{E}}_{0\mathrm{l}}={\mathrm{E}}_{0\mathrm{r}}+{\mathrm{E}}_{0\mathrm{l}}+{\mathrm{β}}_{12}({\mathrm{E}}_{0\mathrm{r}}−{\mathrm{E}}_{0\mathrm{l}})\\ ={\mathrm{E}}_{0\mathrm{r}}(1+{\mathrm{β}}_{12})+{\mathrm{E}}_{0\mathrm{l}}(1−{\mathrm{β}}_{12})\end{array}$ …… (6)

Here, ${\mathrm{β}}_{12}={\mathrm{μ}}_1{\mathrm{Ï\dots }}_1/{\mathrm{μ}}_2{\mathrm{Ï\dots }}_2$. Next add (3) and (5):

${\mathrm{E}}_{0\mathrm{T}}{\mathrm{e}}^{{\mathrm{ik}}_3\mathrm{d}}({\mathrm{β}}_{23}+1)−2{\mathrm{E}}_{0\mathrm{r}}{\mathrm{e}}^{{\mathrm{ik}}_2\mathrm{d}}$ …… (7)

Subtract (3) and (5):

${\mathrm{E}}_{0\mathrm{T}}{\mathrm{e}}^{{\mathrm{ik}}_3\mathrm{d}}(1−{\mathrm{β}}_{23})=2{\mathrm{E}}_{0\mathrm{l}}{\mathrm{e}}^{−{\mathrm{ik}}_2\mathrm{d}}$ …… (8)

Now put (7) and (8) into (6):

$\begin{array}{c}2{\mathrm{E}}_{0\mathrm{I}}={\mathrm{E}}_{0\mathrm{r}}(1+{\mathrm{β}}_{12})+{\mathrm{E}}_{0\mathrm{l}}(1−{\mathrm{β}}_{12})\\ =(1+{\mathrm{β}}_{12})\frac{1}{2}{\mathrm{E}}_{0\mathrm{T}}{\mathrm{e}}^{\mathrm{id}({\mathrm{k}}_3−{\mathrm{k}}_2)}(1+{\mathrm{β}}_{23})+(1−{\mathrm{β}}_{12})\frac{1}{2}{\mathrm{E}}_{0\mathrm{T}}{\mathrm{e}}^{\mathrm{id}({\mathrm{k}}_3+{\mathrm{k}}_2)}(1−{\mathrm{β}}_{23})\\ \frac{{\mathrm{E}}_{0\mathrm{T}}}{{\mathrm{E}}_{0\mathrm{l}}}=\frac{4{\mathrm{e}}^{{\mathrm{idk}}_3}}{[(1+{\mathrm{β}}_{12})(1+{\mathrm{β}}_{23}){\mathrm{e}}^{−{\mathrm{idk}}_2}+(1−{\mathrm{β}}_{12})(1−{\mathrm{β}}_{23}){\mathrm{e}}^{{\mathrm{idk}}_2}]}\end{array}$

Now:

$\begin{array}{c}(1+{\mathrm{β}}_{12})(1+{\mathrm{β}}_{23}){\mathrm{e}}^{−{\mathrm{idk}}_2}+(1−{\mathrm{β}}_{12})(1−{\mathrm{β}}_{23}){\mathrm{e}}^{{\mathrm{idk}}_2}\\ ={\mathrm{e}}^{−{\mathrm{idk}}_2}(1+{\mathrm{β}}_{12}+{\mathrm{β}}_{23}+{\mathrm{β}}_{12}{\mathrm{β}}_{23})+{\mathrm{e}}^{{\mathrm{idk}}_2}(1−{\mathrm{β}}_{12}−{\mathrm{β}}_{23}+{\mathrm{β}}_{12}{\mathrm{β}}_{23})\\ =2\cos \left({\mathrm{k}}_2\mathrm{d}\right)−2\mathrm{isin}\left({\mathrm{k}}_{2\mathrm{d}}\right){\mathrm{β}}_{12}−2\mathrm{isin}\left({\mathrm{k}}_2\mathrm{d}\right)+2{\mathrm{β}}_{12}{\mathrm{β}}_{23}\cos \left({\mathrm{k}}_2\mathrm{d}\right)\\ =2\cos \left({\mathrm{k}}_2\mathrm{d}\right)(1+{\mathrm{β}}_{13})−2\mathrm{isin}\left({\mathrm{k}}_2\mathrm{d}\right)({\mathrm{β}}_{12}+{\mathrm{β}}_{23})\end{array}$

Since, ${\mathrm{β}}_{12}{\mathrm{β}}_{23}={\mathrm{β}}_{13}$

Determine the transmission coefficient for normal incidence.

Substitute $\frac{4}{{\cos }^2\left({\mathrm{k}}_{2\mathrm{d}}\right){(1+{\mathrm{β}}_{13})}^2+{\sin }^2\left({\mathrm{k}}_2\mathrm{d}\right){({\mathrm{β}}_{12}+{\mathrm{β}}_{23})}^2}$for ${\left|\frac{E_{0T}}{E_{0I}}\right|}^2$into equation (1).

$\begin{array}{c}\mathrm{T}=\frac{{\mathrm{Î\mu }}_3{\mathrm{Ï\dots }}_3}{{\mathrm{Î\mu }}_1{\mathrm{Ï\dots }}_1}\frac{4}{{\cos }^2\left({\mathrm{k}}_{2\mathrm{d}}\right){(1+{\mathrm{β}}_{13})}^2+{\sin }^2\left({\mathrm{k}}_2\mathrm{d}\right){({\mathrm{β}}_{12}+{\mathrm{β}}_{23})}^2}\\ =\frac{{\mathrm{Î\mu }}_3{\mathrm{Ï\dots }}_3}{{\mathrm{Î\mu }}_1{\mathrm{Ï\dots }}_1}\frac{4}{{(1+{\mathrm{β}}_{13})}^2+{\sin }^2\left({\mathrm{k}}_2\mathrm{d}\right)({({\mathrm{β}}_{12}+{\mathrm{β}}_{23})}^2−{(1+{\mathrm{β}}_{13})}^2)}\end{array}$

We accept that ${\mathrm{μ}}_1={\mathrm{μ}}_2={\mathrm{μ}}_3={\mathrm{μ}}_0$and hence $\mathrm{n}=\sqrt{\mathrm{c}}$. Also, ${\mathrm{B}}_{\mathrm{ij}}={\mathrm{Ï\dots }}_{\mathrm{i}}/{\mathrm{Ï\dots }}_{\mathrm{j}}={\mathrm{n}}_{\mathrm{j}}/{\mathrm{n}}_{\mathrm{i}}$, so:

$\begin{array}{c}\mathrm{T}=\frac{{\mathrm{n}}_3^2}{{\mathrm{n}}_1^2}\frac{1}{{\mathrm{β}}_{13}}\frac{4}{{(1+{\mathrm{β}}_{13})}^2+{\sin }^2\left({\mathrm{k}}_2\mathrm{d}\right)({\mathrm{β}}_{12}^2+{\mathrm{β}}_{23}^2+2{\mathrm{β}}_{12}{\mathrm{β}}_{23}−1−2{\mathrm{β}}_{13}−{\mathrm{β}}_{13}^2)}\\ =\frac{{\mathrm{n}}_3}{{\mathrm{n}}_1}\frac{4}{{(1+{\mathrm{n}}_3/{\mathrm{n}}_1)}^2+{\sin }^2{\mathrm{k}}_2\mathrm{d}({({\mathrm{n}}_3/{\mathrm{n}}_2)}^2+{({\mathrm{n}}_2/{\mathrm{n}}_1)}^2−{({\mathrm{n}}_3/{\mathrm{n}}_1)}^2−1)}\\ =\frac{4{\mathrm{n}}_1{\mathrm{n}}_3}{{({\mathrm{n}}_1+{\mathrm{n}}_3)}^2+{\sin }^2{\mathrm{k}}_2\mathrm{d}({\mathrm{n}}_1^2{({\mathrm{n}}_3/{\mathrm{n}}_2)}^2+{\mathrm{n}}_2^2−{\mathrm{n}}_3^2−{\mathrm{n}}_1^2)}\\ =\frac{4{\mathrm{n}}_1{\mathrm{n}}_3}{{({\mathrm{n}}_1+{\mathrm{n}}_3)}^2+{\sin }^2\left(\frac{{\mathrm{Ó\neg n}}_2\mathrm{d}}{\mathrm{c}}\right)\frac{({\mathrm{n}}_1^2−{\mathrm{n}}_2^2)({\mathrm{n}}_3^2−{\mathrm{n}}_2^2)}{{\mathrm{n}}_2^2}}\end{array}$

Therefore, the value of transmission coefficient for normal incidence is

${\text{T}}^{−1}=\frac{\text{1}}{{\text{4n}}_{\text{1}}{\text{n}}_{\text{3}}}\left[{({\text{n}}_{\text{1}}+{\text{n}}_{\text{3}})}^{\text{2}}+\frac{({\text{n}}_{\text{1}}^{\text{2}}−{\text{n}}_{\text{2}}^{\text{2}})({\text{n}}_{\text{3}}^{\text{2}}−{\text{n}}_{\text{2}}^{\text{2}})}{{\text{n}}_{\text{2}}^{\text{2}}}{\text{sin}}^{\text{2}}\left(\frac{{\text{n}}_{\text{2}}\mathrm{Ó\neg }\text{d}}{\text{c}}\right)\right]$