Question

In Fig. 33-47a, a light ray in an underlying material is incident at an angleon a boundary with water, and some of the light refracts into the water. There are two choices of the underlying material. For each, the angle of refractionversus the incident angleis given in Fig. 33-47b.The horizontal axis scale is set by${\mathit{\theta }}_{\mathbf{1}\mathbf{s}}\mathbf{=}\mathbf{90}\mathbf{°}$.Without calculation, determine whether the index of refraction of

(a) material1 and

(b) material2 is greater or less than the index of water$\mathbf{(}\mathbf{n}\mathbf{=}\mathbf{1}\mathbf{.33}\mathbf{)}$.What is the index of refraction of

(c) material 1 And

(d) material 2?

Short answer

1. This implies that a light ray travels from a denser medium to a rarer medium.

Hence.$n_2>n$

2. This implies that a light ray travels from a denser medium to a rarer medium.

Hence.$n_2>n$

3.The value of n₁ from the graph of material 1 is $n_1=\text{1.9}$.

4.The value of n₂ from the graph of material 2 is$n_1=\text{1.432}$ .

Step-by-step solution

Given data

The index of refraction for water is,$n=\text{1.33}.$

Determining the concept

Using the equation of Snell’s law and comparing the values of incident angle and refraction angle, whether the refractive index is greater or smaller can be determined. To calculate the actual value, take the values of angles from the graph.

The formula is as follows:

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

Where,

${\theta }_1$= angle of incidence.

${\theta }_2$= angle of refraction.

n₁ = index of refraction of the incident medium.

n₂ = index of refraction of the refractive medium.

(a) Determining without calculation, whether the index of refraction of a material  is greater or less than the index of water(n=1.33)

From the graph, it is observed that when light travels from material 1 to water, the angle of incidence${\theta }_1$is less than the angle of refraction${\text{θ}}_{\text{2}}$in water.

This implies that a light ray travels from a denser medium to a rarer medium.

Hence$n_1>n.$

(b) Determining without calculation, whether the index of refraction of a material  is greater or less than the index of water(n=1.33)

From the graph, it is observed that when light travels from material 2 to water, the angle of incidence ${\text{θ}}_{\text{1}}$ is less than the angle of refraction ${\text{θ}}_{\text{2}}$ in water.

This implies that a light ray travels from a denser medium to a rarer medium.

Hence $n_2>n$.

 Step 5: (c) Determining what is the index of refraction of material 1

From the graph of material 1,

${\theta }_2=90°$

And

${\theta }_1=45°$

Now use Snell’s law:

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

Here$n_1$is the refractive index of material 1,$n_2$is the refractive index of water.

Substitute the values in the above expression, and we get,

$\begin{array}{c}{n}_1\sin {90}^{\circ }=1.33\sin {90}^{\circ }\\ n_1=1.9\end{array}$

Hence,the value of n₁ from the graph of material 1 is$n_1=\text{1.9}$ .

(d) Determining what is the index of refraction of material 2

From the graph of material 2,

${\theta }_2={90}^{\circ }$

And,

${\theta }_1={68.25}^{°}$

Now use Snell’s law:

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

Here$n_1$is the refractive index of material 2,$n_2$is the refractive index of water.

Substitute the values in the above expression, and we get,

$\begin{array}{c}{n}_1\sin {68.25}^{\circ }=1.33\sin {90}^{\circ }\\ n_1=1.432\end{array}$

Hence, the value of n₁ from the graph of material 2 is $n_1=\text{1.432}$.