In Fig. 33-48a, a light ray in water is incident at an angle${\mathit{\theta }}_{\mathbf{1}}$on a boundary with an underlying material, into which some of the light refracts. There are two choices of the underlying material. For each, the angle of refraction${\mathit{\theta }}_{\mathbf{2}}$versus the incident angleis given in Fig. 33-48b. The vertical axis scale is set by${\mathit{\theta }}_{\mathbf{2}\mathbf{s}}\mathbf{=}\mathbf{90}\mathbf{°}$. Without calculation, determine whether the index of refraction of

(a) material 1 and

(b) material 2 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) materialAnd

(d) material?

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

Here use the formula for Snell’s law, which gives the relation between the angle of reflection and the angle of refraction.

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.

From the graph for material 1, it is observed that for all the incidence angles,${\theta }_1$corresponding angles of refraction${\theta }_2$are smaller. This implies that the refracted ray bends towards the normal. Hence light travels from a rarer medium to a denser medium.

Hence$n_1>n$.

From the graph for material 2, again, it is observed that for all the incidence angles${\theta }_1$corresponding angles of refraction${\theta }_2$are smaller. This implies that the refracted ray bends towards the normal. Hence light travels from a rarer medium to a denser medium.

Hence$n_2>n.$

At the end of the graph,

${\theta }_1={90}^{\circ }$

And,

$\begin{array}{l}{\theta }_2=\frac{3}{4}\times {\theta }_1\\ {\theta }_2={68.25}^{°}\end{array}$

Now use Snell’s law:

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

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

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

.$\begin{array}{c}1.33\sin \left({90}^{\circ }\right)=n_2\sin \left({68.25}^{\circ }\right)\\ n_2=1.432\\ ~1.4\end{array}$

Therefore, the refractive index of material 1 is $\text{1.4}$.

Similarly, from the graph for material 2,

${\theta }_1={90}^{\circ }$

Therefore,

$\begin{array}{c}{\theta }_2=0.5\times 90\\ {\theta }_2={45}^{°}\end{array}$

Now use Snell’s law:

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

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

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

$\begin{array}{c}1.33\sin \left({90}^{\circ }\right)=n_2\sin \left({45}^{\circ }\right)\\ n_2=1.88\\ ~1.9\end{array}$

Therefore, the refractive index of material 2 is$\text{1.9}$ .