Question

A light ray strikes the air-water surface at an angle of incidence of ${30}^{\circ }$.The refractive index for water is $\frac{4}{3}.$Find the angle of refraction when the light ray goes (a) from air to water.

Answer 1

Step 1: Assuming the quantities

Let, the angle of refraction of light from air to water be ${\mathrm{r}}_1$

Let, the refractive index for air, ${\mathrm{\mu }}_1=1$

Step 2: Explanation of the quantities

The air-water surface at an angle of incidence, $\mathrm{i}={30}^{\circ }$

The refractive index for water, ${\mathrm{\mu }}_2=\frac{4}{3}$

Step 3: Calculation of the angle of refraction

From Snell's law

$$\begin{gathered} \frac{\sin \mathrm{i}}{\sin {\mathrm{r}}_1}=\frac{{\mathrm{\mu }}_2}{{\mathrm{\mu }}_1} \\ \frac{\sin \mathrm{i}}{\sin {\mathrm{r}}_1}=\frac{{\mathrm{\mu }}_2}{{\mathrm{\mu }}_1} \end{gathered}$$

The angle of refraction when the light ray goes from air to water is:

$$\begin{gathered} \frac{\sin {30}^{\circ }}{\sin {\mathrm{r}}_1}=\frac{{\displaystyle \frac{4}{3}}}{1} \\ \frac{\sin {30}^{\circ }}{\sin {\mathrm{r}}_1}=\frac{4}{3} \\ 3\times \sin {30}^{\circ }=4\times \sin {\mathrm{r}}_1 \\ \sin {\mathrm{r}}_1=\frac{3}{4}\times \sin {30}^{\circ } \\ \sin {\mathrm{r}}_1=\frac{3}{4}\times \frac{1}{2} \\ {\mathrm{r}}_1={\sin }^{-1}\left(\frac{3}{8}\right) \\ {\mathrm{r}}_1=22.{02}^{\circ } \end{gathered}$$

Thus, $22.{02}^{\circ }$is the angle of refraction when the light ray goes from air to water.