Question

In Fig. 33-51, light is incident at angle ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{40}\mathbf{.1}\mathbf{°}$on a boundary between two transparent materials. Some of the light travels down through the next three layers of transparent materials, while some of it reflects upward and then escapes into the air. If${\mathit{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.30}\mathbf{,}\mathbf{}{\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.40}\mathbf{,}\mathbf{}{\mathit{n}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.32}$and${\mathit{n}}_{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{.45}$, what is the value of

(a)${\mathit{\theta }}_{\mathbf{5}}$in the air and

(b) ${\mathit{\theta }}_{\mathbf{4}}$in the bottom material?

Short answer

1. The value of ${\theta }_5$ in the air is ${56.9}^0$.
2. The value of ${\theta }_4$ in the bottom material is ${35.3}^0$.

Step-by-step solution

Given

• The angle of incidence at the boundary between two materials${\theta }_1={40.1}^0$
• The refractive indices $n_1=1.30,n_2=1.40,n_3=1.32,n_4=1.45$

Understanding the concept

We can apply Snell’s law to the material of1and air to find thevalue of${\theta }_5$in the air. Similarly, by applying Snell’s law to each boundary, we get 4 equations and solving them, we can find thevalue of${\theta }_4$in the bottom material.

Formula:

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

(a) Calculate the value of θ5  in the air.

According to Snell’s law,

$n_1\sin {\theta }_1=n_5\sin {\theta }_5$

We have for the air$n_5=1$and${\theta }_1={40.1}^0$,

$\begin{array}{c}{\theta }_5={\sin }^{-1}\left(\frac{n_1\sin {\theta }_1}{n_5}\right)\\ {\theta }_5={\sin }^{-1}\left(\frac{1.30\sin {40.1}^0}{1}\right)\\ {\theta }_5={56.86}^0\\ ~56.9°\end{array}$

Therefore, the value of ${\theta }_5$ in the air is ${56.9}^0$

(b) Calculate the value of   θ4in the bottom material

We have,

$\begin{array}{c}{n}_1\sin {\theta }_1=n_2\sin {\theta }_2=n_3\sin {\theta }_3=n_4\sin {\theta }_4\\ n_1\sin {\theta }_1=n_4\sin {\theta }_4\\ {\theta }_4={\sin }^{-1}\left(\frac{n_1\sin {\theta }_1}{n_4}\right)\\ {\theta }_4={\sin }^{-1}\left(\frac{1.30\sin {40.1}^0}{1.45}\right)\\ {\theta }_4={35.3}^0\end{array}$

Therefore, thevalue of${\theta }_4$in the bottom material is${35.3}^0$