Chapter 33: Q51P (page 1004)

In Fig. 33-51, light is incident at angle $θ_{1} = 40 .1 °$ 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 $n_{1} = 1 .30, n_{2} = 1 .40, n_{3} = 1 .32$ and $n_{4} = 1 .45$ , what is the value of
(a) $θ_{5}$ in the air and
(b) $θ_{4}$ in the bottom material?

Short Answer

Expert verified

The value of $θ_{5}$ in the air is ${56.9}^{0}$ .
The value of $θ_{4}$ in the bottom material is ${35.3}^{0}$ .

Step by step solution

Given

The angle of incidence at the boundary between two materials $θ_{1} = {40.1}^{0}$
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 $θ_{5}$ in the air. Similarly, by applying Snell’s law to each boundary, we get 4 equations and solving them, we can find thevalue of $θ_{4}$ in the bottom material.

Formula:

$n_{1} \sin θ_{1} = n_{2} \sin θ_{2}$

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

According to Snell’s law,

$n_{1} \sin θ_{1} = n_{5} \sin θ_{5}$

We have for the air $n_{5} = 1$ and $θ_{1} = {40.1}^{0}$ ,

$\begin{matrix} θ_{5} = \sin^{- 1} (\frac{n_{1} \sin θ_{1}}{n_{5}}) \\ θ_{5} = \sin^{- 1} (\frac{1.30 \sin {40.1}^{0}}{1}) \\ θ_{5} = {56.86}^{0} \\ ~ 56.9 ° \end{matrix}$

Therefore, the value of $θ_{5}$ in the air is ${56.9}^{0}$

(b) Calculate the value of θ4in the bottom material

We have,

$\begin{matrix} n_{1} \sin θ_{1} = n_{2} \sin θ_{2} = n_{3} \sin θ_{3} = n_{4} \sin θ_{4} \\ n_{1} \sin θ_{1} = n_{4} \sin θ_{4} \\ θ_{4} = \sin^{- 1} (\frac{n_{1} \sin θ_{1}}{n_{4}}) \\ θ_{4} = \sin^{- 1} (\frac{1.30 \sin {40.1}^{0}}{1.45}) \\ θ_{4} = {35.3}^{0} \end{matrix}$