Light of (angular) frequency w passes from medium , through a slab (thickness $\mathbf{d}$) of medium $\mathbf{2}$, and into medium $\mathbf{3}$(for instance, from water through glass into air, as in Fig. 9.27). Show that the transmission coefficient for normal incidence is given by

localid="1658907323874" ${\mathbf{T}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\frac{1}{\mathbf{4}{\mathbf{n}}_1{\mathbf{n}}_3}\mathbf{[}{\mathbf{(}{\mathbf{n}}_{\mathbf{1}}\mathbf{+}{\mathbf{n}}_{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{(}{\mathbf{n}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{-}{\mathbf{n}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{)}\mathbf{(}{\mathbf{n}}_{\mathbf{3}}^{\mathbf{2}}\mathbf{-}{\mathbf{n}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{)}}{{\mathbf{n}}_{\mathbf{2}}^{\mathbf{2}}}{\mathbf{sin}}^{\mathbf{2}}\left(\frac{n_2\mathrm{\omega d}}{c}\right)\mathbf{]}$

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.

$\mathbf{T}\mathbf{=}\frac{{\mathbf{\epsilon }}_3{\mathbf{\upsilon }}_3}{{\mathbf{\epsilon }}_1{\mathbf{\upsilon }}_1}$ …… (1)

Here, ${\mathbf{\epsilon }}_3$relative permittivity of medium 3, ${\mathbf{\upsilon }}_3$wave velocity at medium 3, ${\mathbf{\epsilon }}_1$relative permittivity of medium 1 and${\mathbf{\upsilon }}_1$ 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.

$\begin{array}{c}{\overrightarrow{\mathrm{E}}}_1={\mathrm{E}}_{01}{\mathrm{e}}^{\mathrm{i}({\mathrm{k}}_1\mathrm{z}-\mathrm{\omega t})}\widehat{\mathrm{x}}\\ {\overrightarrow{\mathrm{B}}}_1=\frac{{\mathrm{E}}_{01}}{{\mathrm{\upsilon }}_1}{\mathrm{e}}^{\mathrm{i}({\mathrm{k}}_1\mathrm{z}-\mathrm{\omega t})}\widehat{\mathrm{z}}\times \widehat{\mathrm{x}}\\ =\frac{{\mathrm{E}}_{01}}{{\mathrm{\upsilon }}_1}{\mathrm{e}}^{\mathrm{i}({\mathrm{k}}_1\mathrm{z}-\mathrm{\omega t})}\widehat{\mathrm{y}}\\ {\overrightarrow{\mathrm{E}}}_{\mathrm{R}}={\mathrm{E}}_{0\mathrm{R}}{\mathrm{e}}^{\mathrm{i}(-{\mathrm{k}}_1\mathrm{z}-\mathrm{\omega t})}\widehat{\mathrm{x}}\end{array}$

Solve further as

$\begin{array}{c}{\overrightarrow{B}}_R=\frac{E_{0R}{e}^{i(-k_{1}z-\omega t)}}{{\upsilon }_1}(-\widehat{z})\times \widehat{x}\\ =-\frac{E_{0R}}{{\upsilon }_1}{e}^{i(-k_{1}z-\omega t)}\widehat{y}\end{array}$

Now fields on material $2$:

$\begin{array}{c}{\overrightarrow{E}}_r=E_{0r}{e}^{i(k_{2}z-\omega t)}\widehat{x}\\ {\overrightarrow{B}}_r=\frac{E_{0r}}{{\upsilon }_2}{e}^{i(k_{2}z-\omega t)}\widehat{y}\\ {\overrightarrow{E}}_{\iota }=E_{0\iota }{e}^{i(-k_{2}z-\omega t)}\widehat{x}\\ {\overrightarrow{B}}_{\iota }=-\frac{E_{0\iota }}{{\upsilon }_2}{e}^{i(-k_{2}z-\omega t)}\widehat{y}\end{array}$

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

$\begin{array}{l}{\overrightarrow{E}}_T=E_{0T}{e}^{i(k_{3}z-\omega t)}\widehat{x}\\ {\overrightarrow{B}}_T=\frac{E_{0T}}{{\upsilon }_3}{e}^{i(k_{3}z-\omega t)}\widehat{y}\end{array}$

Apply boundary conditions on both planes. The initial one, ${\overrightarrow{\mathrm{E}}}_{\parallel ,1}={\overrightarrow{\mathrm{E}}}_{\parallel ,2}$, 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, ${\overrightarrow{B}}_{\parallel ,1}/{\mu }_1={\overrightarrow{B}}_{\parallel ,2}/{\mu }_2$, gives:

$\frac{1}{{\mu }_1{\upsilon }_1}(E_{0l}-E_{0R})=\frac{1}{{\mu }_2{\upsilon }_2}(E_{0r}-E_{0l})$ …... (4)

And

$\frac{1}{{\mu }_3{\upsilon }_3}{E}_{0T}{e}^{i{k}_{3}d}=\frac{1}{{\mu }_2{\upsilon }_2}\left(E_{0r}{e}^{i{k}_{2}d-E_{0l}{e}^{-i{k}_{2}d}}\right)$ …… (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:

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

Here, ${\beta }_{12}={\mu }_1{\upsilon }_1/{\mu }_2{\upsilon }_2$. Next add (3) and (5):

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

Subtract (3) and (5):

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

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

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

Now:

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

Since, ${\beta }_{12}{\beta }_{23}={\beta }_{13}$

Determine the transmission coefficient for normal incidence.

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

$\begin{array}{c}\mathrm{T}=\frac{{\mathrm{\epsilon }}_3{\mathrm{\upsilon }}_3}{{\mathrm{\epsilon }}_1{\mathrm{\upsilon }}_1}\frac{4}{{\cos }^2\left({\mathrm{k}}_{2\mathrm{d}}\right){(1+{\mathrm{\beta }}_{13})}^2+{\sin }^2\left({\mathrm{k}}_2\mathrm{d}\right){({\mathrm{\beta }}_{12}+{\mathrm{\beta }}_{23})}^2}\\ =\frac{{\mathrm{\epsilon }}_3{\mathrm{\upsilon }}_3}{{\mathrm{\epsilon }}_1{\mathrm{\upsilon }}_1}\frac{4}{{(1+{\mathrm{\beta }}_{13})}^2+{\sin }^2\left({\mathrm{k}}_2\mathrm{d}\right)({({\mathrm{\beta }}_{12}+{\mathrm{\beta }}_{23})}^2-{(1+{\mathrm{\beta }}_{13})}^2)}\end{array}$

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

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

Therefore, the value of transmission coefficient for normal incidence is

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