Question

If a diver who is underwater shines a flashlight upward toward the surface, at an angle of $\mathbf{29}\mathbf{°}$ from the normal, at what angle does the light emerge from the water? The indices of refraction are: water,$\mathbf{1}\mathbf{.}\mathbf{33}$ ; air,$\mathbf{1}\mathbf{.}\mathbf{00029}$

Short answer

The angle at which the light emerges from the water is $21.38°$.

Step-by-step solution

Identification of given data

Refractive index of water,$n_2=1.33$

Refractive index of air,$n_1=1.00029$

The angle of incident,${\theta }_1=29°$

Determine the formulas to calculate the angle at which the light emerges from the water.

The Snell’s law defines the relationship between the angle of incident and angle of refraction when light of wave passes through two mediums.

${\mathit{n}}_{\mathbf{1}}\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_{\mathbf{1}}\mathbf{=}{\mathit{n}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_{\mathbf{2}}$ …(i)

Here,${\mathit{\theta }}_{\mathbf{2}}$ is the angle of refection.

Determining the angle at which the light emerges from the water

Calculate the angle at which the light emerges from water.

Substitute $29°for{\theta }_1$, $1.33for{n}_2$and $1.00029for{n}_1$into equation (i)

$$\begin{gathered} 1.00029\sin 29°=1.33\sin {\theta }_2 \\ \sin {\theta }_2=\frac{1.00029}{1.33}\sin 29° \\ \sin {\theta }_2=0.752\times 0.4848 \\ \sin {\theta }_2=0.3645 \end{gathered}$$

Solve further as,

$$\begin{gathered} {\theta }_2={\sin }^{-1}\left(0.3645\right) \\ {\theta }_2=21.38° \end{gathered}$$

Hence the angle at which the light emerges from the water is $21.38°$.