91Ó°ÊÓ

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:

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

Where,

${\mathrm{θ}}_1$= angle of incidence.

${\mathrm{θ}}_2$= 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,${\mathrm{θ}}_1$corresponding angles of refraction${\mathrm{θ}}_2$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${\mathrm{θ}}_1$corresponding angles of refraction${\mathrm{θ}}_2$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,

${\mathrm{θ}}_1={90}^{∘}$

And,

$\begin{array}{l}{\mathrm{θ}}_2=\frac{3}{4}×{\mathrm{θ}}_1\\ {\mathrm{θ}}_2={68.25}^{°}\end{array}$

Now use Snell’s law:

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

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,

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

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

Similarly, from the graph for material 2,

${\mathrm{θ}}_1={90}^{∘}$

Therefore,

$\begin{array}{c}{\mathrm{θ}}_2=0.5×90\\ {\mathrm{θ}}_2={45}^{°}\end{array}$

Now use Snell’s law:

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

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,

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

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