Question

In Fig.33-52a, a beam of light in material1is incident on a boundary at an angle of${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{30}\mathbf{°}$. The extent of refraction of the light into material2depends, in part, on the index of refraction${\mathit{n}}_{\mathbf{2}}$of material 2. Fig.33-52bgives the angle of refraction${\mathit{\theta }}_{\mathbf{2}}$versus${\mathit{n}}_{\mathbf{2}}$for a range of possible${\mathit{n}}_{\mathbf{2}}$values. The vertical axis scale is set by${\mathit{\theta }}_{\mathbf{2}\mathbf{a}}\mathbf{=}\mathbf{20}{\mathbf{.0}}^{\mathbf{°}}$and${\mathit{\theta }}_{\mathbf{2}\mathbf{b}}\mathbf{=}\mathbf{40}{\mathbf{.0}}^{\mathbf{°}}$

(a) What is the index of refraction of material 1?

(b) If the incident angle is changed to${\mathbf{60}}^{\mathbf{°}}$and material 2has ${\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.4}$, then what is angle ${\mathit{\theta }}_{\mathbf{2}}$?

Short answer

1. The index of refraction of material 1 is $1.7$.
2. The angle , if incident angle changed to ${60}^0$and material 2 has $n_2=2.4$is${38}^{\circ }$.

Step-by-step solution

Given data

The angle of incidence at the boundary of material 1 is,${\theta }_1={30}^{\circ }$.

Understanding the concept

We can use the formula of Snell’s law to each boundary to find the index of refraction of material 1and angle${\theta }_2$; if${\theta }_1$is changed to${60}^{\circ }$and the index of the refraction of material 3 is2.4.

Formula:

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

(a) Calculate The index of refraction of the material 1

According to Snell’s law,

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

Therefore,$n_1=n_2$if ${\theta }_1={\theta }_2$.

Since, ${\theta }_1={30}^{\circ }$,${\theta }_2={30}^{\circ }$

From figure33-52(b), we can conclude that,

${\theta }_2={30}^{\circ }$is at$n_2=1.7$

Therefore,

$n_1=1.7$

Therefore, the index of the refraction of material 1 is$1.7$

(b) Calculate the angleθ2 ; if the incident angle changed to 60∘, and material 2 has n2=2.4

We have,

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

For,${\theta }_1={60}^{\circ }$ and$n_2=2.4$,

Then we can calculate the angle${\theta }_2$as,

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{\left(1.7\right)\sin {60}^{°}}{2.4}\right)\\ {\theta }_2={37.83}^{°}\\ ~{38}^{°}\end{array}$

Therefore, angle${\theta }_2$, if the incident angle changed to${60}^{\circ }$and material 2 has$n_2=2.4$is${38}^{°}$.