Question

In Figure

(a), a beam of light in a material1is incident on a boundary at an angle ${\mathbf{\text{θ}}}_{\mathbf{\text{1}}}\mathbf{\text{=40°}}$. Some of the light travels through the material 2, and then some of it emerges into the material 3. The two boundaries between the three materials are parallel. The final direction of the beam depends, in part, on the index of refraction $n_3$of the third material. Figure (b) gives the angle of refraction ${\mathbf{\text{θ}}}_{\mathbf{\text{3}}}$in that material versus ${\mathit{n}}_{\mathbf{3}}$a range of possible${\mathit{n}}_{\mathbf{3}}$values. The vertical axis scale is set by${\mathbf{\text{θ}}}_{\mathbf{\text{3a}}}\mathbf{\text{=30.0°}}$ and ${\mathbf{\text{θ}}}_{\mathbf{\text{3b}}}\mathbf{\text{=50.0°}}$.(a) What is the indexof refraction of material , or is the index impossible to calculate without more information?

(b) What is the index of refraction of material 2, or is the index impossible to calculate without more information?

(c) It ${\mathbf{\text{θ}}}_{\mathbf{\text{1}}}$is changed to $\mathbf{70}\mathbf{°}$and the index of refraction of a material 3 is$\mathbf{\text{2.4}}$ , what is ${\mathbf{\text{θ}}}_{\mathbf{\text{3}}}$?

Figure:

Short answer

1. The index of refraction of a material is$\text{1.6}$
2. It requires more information to find the index of refraction of the material 2.
3. Then ${\text{θ}}_{\text{3}}$, it ${\text{θ}}_{\text{1}}$is changed to ${\text{70}}^{\text{0}}$and the index of refraction of material 3 is ,$\text{2.4}$ is ${\text{39}}^{\text{0}}$.

Step-by-step solution

Step 1: Given

The angle of incidence at the boundary of material 1 is ${\text{θ}}_{\text{1}}{\text{=40}}^{\text{0}}$.

Determining the concept

Use the formula of Snell’s law to each boundary to find the index of refraction of material 1and angle${\theta }_3$, it${\theta }_1$is changed to${70}^0$and the index of refraction of material 3 is$\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{.}$

The formula is as follows:

${\mathit{n}}_{\mathbf{1}}\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_{\mathbf{1}}\mathbf{=}{\mathit{n}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_{\mathbf{2}}$

Where,

${\mathit{\theta }}_{\mathbf{1}}$= angle of incidence,

${\mathit{\theta }}_{\mathbf{2}}$= angle of refraction,

n₁= index of refraction of the incident medium,

n₂= index of refraction of the refractive medium,

Determining the index of refraction of the material .

a.According to Snell’s law,

$\begin{array}{l}{n}_1{\text{sinθ}}_{\text{1}}=n_2{\text{sinθ}}_{\text{2}}\\ n_2{\text{sinθ}}_{\text{2}}=n_3{\text{sinθ}}_{\text{3}}\end{array}$

Therefore,

$n_1{\text{sinθ}}_{\text{1}}=n_3{\text{sinθ}}_{\text{3}}{\text{θ}}_{\text{1}}={\text{θ}}_{\text{3}}$

So,

$\text{When}{n}_1=n_3$

Since.${\text{θ}}_{\text{1}}{\text{=40}}^{\text{0}}$Then,${\text{θ}}_{\text{3}}{\text{=40}}^{\text{0}}$.

Figure 33.50 (b),${\text{θ}}_{\text{3}}{\text{=40}}^{\text{0}}$is at$n_3=\text{1.6}.$

Therefore,

$n_1=\text{1.6}$

Therefore, the index of refraction of a material 1 is $\text{1.6}$.

Determining the index of refraction of material  2 or is the index impossible to calculate without more information.  

b. Part a, $n_2$cancels the manipulation. Therefore, it requiressome more information to determine the refractive index of the material 2.

Determining the θ3if θ1 is changed to  700and the index of refraction of material 3,

c. It is known,

$n_1{\text{sinθ}}_{\text{1}}=n_3{\text{sinθ}}_{\text{3}}$

For,${\text{θ}}_{\text{1}}{\text{=70}}^{\text{0}}$ and ${\text{n}}_{\text{3}}\text{=2.4}$,

$n_1{\text{sinθ}}_{\text{1}}=n_3{\text{sinθ}}_{\text{3}}$

$\begin{array}{l}{\text{θ}}_{\text{3}}{\text{=sin}}^{\text{-1}}\left(\frac{n_1{\text{sin70}}^{\text{0}}}{\text{2.4}}\right)\\ {\text{θ}}_{\text{3}}{\text{=39}}^{\text{0}}\end{array}$

Therefore, it${\text{θ}}_{\text{1}}$ is changed to ${\text{70}}^{\text{0}}$and the index of refraction of material 3 is $\text{2.4}$,${\text{θ}}_{\text{3}}$ is${\text{39}}^{\text{0}}$ .

Using Snell’s law, the angle of incidence and the refraction index of the layers of a multilayered material can be found.