Question

Question: In Figure, light initially in material1refracts into material 2 , crosses that material, and is then incident at the critical angle on the interface between materials 2 and 3 . The indexes of refraction are${\mathit{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{,}{\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{,}\mathbf{\text{and}}{\mathit{n}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}$ .(a) What is angle$\mathit{\theta }$ ?(b) If$\mathit{\theta }$ is increased, is there refraction of light into material 3 ?

Figure:

Short answer

Answer

1. 26.8⁰ is the angle.
2. If $\theta$ is increased, there will be some transmission of light into material 3.

Step-by-step solution

Given 

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

Understanding the concept

Using Snell’s law of refraction, we are able to calculate the value for $\theta$.

Formula:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

(a) Calculate the angle θ 

From the diagram, we can say that

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

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

$n_1\sin \theta =n_2\sin {\theta }_2$

We have

$\sin {\theta }_2=\cos {\theta }_C$

And

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

For${\theta }_C,$we have

$n_2\sin {\theta }_C=n_3\sin {\theta }_3$

Here,${\theta }_3=90°$

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

(b) If θ  is increased, find out if there is refraction of light into material 3

Increase in angle $\theta$ 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.