91Ó°ÊÓ

The index of the refraction for vacuum is.$\text{n}=1$

The table showing values of the angle of incidence${\mathrm{θ}}_1$ and the angle of refraction${\mathrm{θ}}_2$ is given.

Snell’s law gives the relation between the angle of incidence and the angle of refraction. We have to use Snell’s law to find the refractive index of the water.

Formula:

The Snell’s law of refraction,$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$(1)

Here, ${\text{n}}_{\text{1}}$is the refractive index of thevacuum and${\text{n}}_{\text{2}}$is the refractive index of the water.

For, angle of incidence${\mathrm{θ}}_1=10°$ and angle of emergence ,${\mathrm{θ}}_2=8°$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{l}sin(10°)=n_{2}sin(8°)\\ n_2=\frac{sin(10°)}{sin(8°)}\\ n_2=1.245\end{array}$

Forthe angle of incidence${\mathrm{θ}}_1=20°$and angle of emergence,${\mathrm{θ}}_2=15°{30}^{\text{'}}$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{c}sin(20°)=n_{2}sin(15°{30}^{\text{'}})\\ n_2=\frac{\text{sin}(20°)}{\text{sin}(15°{30}^{\text{'}})}\\ n_2=1.29\end{array}$

Forthe angle of incidence${\mathrm{θ}}_1=30°$and angle of emergence,${\mathrm{θ}}_2=22°{30}^{\text{'}}$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{c}sin(30°)=n_{2}sin(22°{30}^{\text{'}})\\ n_2=\frac{sin(30°)}{sin(22°{30}^{\text{'}})}\\ n_2=1.31\end{array}$

Forthe angle of incidence${\mathrm{θ}}_1=40°$and angle of emergence,${\mathrm{θ}}_2=29°$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{c}sin(40°)=n_{2}sin(29°)\\ n_2=\frac{sin(40°)}{sin(29°)}\\ n_2=1.32\end{array}$

Forthe angle of incidence${\mathrm{θ}}_1=50°$and angle of emergence,data-custom-editor="chemistry" ${\mathrm{θ}}_2=35°$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{c}\text{sin}(50°)=n_2\text{sin}(35°)\\ n_2=\frac{\text{sin}(50°)}{\text{sin}(35°)}\\ n_2=1.33\end{array}$

Forthe angle of incidence${\mathrm{θ}}_1=60°$and angle of emergence,${\mathrm{θ}}_2=40°{30}^{\text{'}}$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{c}sin(60°)=n_{2}sin(40°{30}^{\text{'}})\\ n_2=\frac{sin(60°)}{sin40°30\text{'}}\\ =1.3\end{array}$

Forthe angle of incidence${\mathrm{θ}}_1=70°$and angle of emergence,${\mathrm{θ}}_2=45°{30}^{\text{'}}$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{c}sin(70°)=n_{2}sin(45°{30}^{\text{'}})\\ n_2=\frac{sin(70°)}{sin(45°{30}^{\text{'}})}\\ n_2=1.34\end{array}$

Fortheangle of incidence ${\mathrm{θ}}_1=80°$and angle of emergence,${\mathrm{θ}}_2=50°$ the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{l}sin(80°)=n_{2}sin(50°)\\ n_2=\frac{sin(80°)}{sin(50°)}\\ n_2=1.29\end{array}$

Thus, from all these results, we can say that the refractive index for the water is.$1.3$