91Ó°ÊÓ

$$\begin{gathered} n_1=1.6 \\ {n}_2=1.4 \\ {n}_3=1.2 \end{gathered}$$

Using Snell’s law of refraction, we are able to calculate the value for $\mathrm{θ}$.

Formula:

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

From the diagram, we can say that

$$\begin{gathered} {\mathrm{θ}}_2=90°-{\mathrm{θ}}_C \\ \sin {\mathrm{θ}}_2=\sin \left(90-{\mathrm{θ}}_C\right) \\ \sin {\mathrm{θ}}_2=\cos {\mathrm{θ}}_C \\ \cos {\mathrm{θ}}_C=\sqrt{1-{\sin }^2{\mathrm{θ}}_C} \end{gathered}$$

We apply Snell’s law for material 1 and 2, and we get

$n_1\sin \mathrm{θ}=n_2\sin {\mathrm{θ}}_2$

We have

$\sin {\mathrm{θ}}_2=\cos {\mathrm{θ}}_C$

And

$$\begin{gathered} \cos {\mathrm{θ}}_C=\sqrt{1-{\sin }^2{\mathrm{θ}}_C} \\ \sin \mathrm{θ}=\frac{n_2}{n_1}×\left(\sqrt{1-{\sin }^2{\mathrm{θ}}_C}\right) \end{gathered}$$

For${\mathrm{θ}}_C,$we have

$n_2\sin {\mathrm{θ}}_C=n_3\sin {\mathrm{θ}}_3$

Here,${\mathrm{θ}}_3=90°$

$$\begin{gathered} \sin {\mathrm{θ}}_C=\frac{n_3}{n_2} \\ \sin \mathrm{θ}=\frac{n_2}{n_1}×\left(\sqrt{1-{\left(\frac{n_3}{n_2}\right)}^2}\right) \\ \sin \mathrm{θ}=\frac{1.4}{1.6}×\left(\sqrt{1-{\left(\frac{1.2}{1.4}\right)}^2}\right) \\ \sin \mathrm{θ}=0.875×0.515 \\ \mathrm{θ}={\sin }^{-1}0.45 \\ \mathrm{θ}=26.78°~26.8° \end{gathered}$$

Increase in angle $\mathrm{θ}$ increases the angle with which the light is travelling from material 2 to material 3.

This will cause the angle to be greater than that of the critical angle.

From this, we can conclude that there will be some transmission of light into the third material.