91Ó°ÊÓ

Given data:

The refractive index of water is${\mathrm{n}}_1=\frac{4}{3}$.

The refractive index of glass is${\mathrm{n}}_2=\frac{3}{2}$.

The refractive index of air is${\mathrm{n}}_3=1$.

Write the expression for the inversion of the transmission coefficient.

${\mathbf{T}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{n}}_{\mathbf{1}}{\mathbf{n}}_{\mathbf{3}}}\left[{\left(n_1+n_3\right)}^2+\frac{\left({n_1}^2-{n_2}^2\right)\left({n_3}^2-{n_2}^2\right)}{{n_2}^2}{\sin }^2\frac{n_2\mathrm{Ó\neg »å}}{C}\right]$

Substitute $n_1=\frac{4}{3}$, $n_2=\frac{3}{2}$and $n_3=1$in the above expression.

$$\begin{gathered} T^{-1}=\frac{1}{4\left({\displaystyle \frac{4}{3}}\right)1}\left[{\left(\frac{4}{3}\right)}^2+1+\frac{{\left({\displaystyle \frac{4}{3}}\right)}^2-{\left({\displaystyle \frac{3}{2}}\right)}^2\left(r^2-{\left({\displaystyle \frac{3}{2}}\right)}^2\right)}{{\left({\displaystyle \frac{3}{2}}\right)}^2}{\sin }^2\left(\frac{3\mathrm{Ó\neg }d}{2C}\right)\right] \\ {T}^{-1}=\frac{3}{16}\left[\frac{49}{9}+\left(\frac{17}{36}\right)×\left(\frac{5}{4}\right)×\left(\frac{4}{9}\right){\sin }^2\left(\frac{3\mathrm{Ó\neg }d}{2C}\right)\right] \\ {T}^{-1}=\frac{49}{48}+\frac{85}{\left(48\right)\left(36\right)}{\sin }^2\left(\frac{3\mathrm{Ó\neg }d}{2C}\right) \end{gathered}$$........(1)

For transmission coefficient to be minimum data-custom-editor="chemistry" ${\sin }^2\left(\frac{3\mathrm{Ó\neg }d}{2C}\right)=1$.

Substitute data-custom-editor="chemistry" ${\sin }^2\left(\frac{3\mathrm{Ó\neg }d}{2C}\right)=1$in equation (1).

data-custom-editor="chemistry" $$\begin{gathered} {\mathrm{T}}_{\min }=\frac{48}{49+{\displaystyle \frac{85}{36}}} \\ {\mathrm{T}}_{\min }=0.93455 \end{gathered}$$

For transmission coefficient to be maximumdata-custom-editor="chemistry" ${\sin }^2\left(\frac{3\mathrm{Ó\neg }d}{2C}\right)=0$.

Substitute data-custom-editor="chemistry" ${\sin }^2\left(\frac{3\mathrm{Ó\neg }d}{2C}\right)=0$in equation (1).

$$\begin{gathered} {\mathrm{T}}_{\max }=\frac{48}{49} \\ {\mathrm{T}}_{\max }=0.987959 \end{gathered}$$