Chapter 24, Problem 60P

To determine

To Calculate: The Brewster’s angles.

Explanation of Solution

Given info:

Refractive index of air, $n_{air}=1.00$

Refractive index of water, $n_{water}=1.33$

Formula used:

  $\tan {\text{θ}}_{\text{p}}=\frac{{\text{n}}_{\text{air}}}{n_{\text{water}}}$

Where,

  ${\text{θ}}_{\text{p}}$ is Brewster’s angle.

  $n_{air}$ is refractive index of air.

  $n_{water}$ is refractive index of water.

Calculation:

Substitute the given data into the formula to get the Brewster’s angle, since the light is coming from water to air, $\tan {\text{θ}}_{\text{p}}=\frac{{\text{n}}_{\text{air}}}{n_{\text{water}}}$

  $\tan {\text{θ}}_{\text{p}}=\frac{1.00}{1.33}$

  $\therefore {\text{θ}}_{\text{p}}={\tan }^{-1}\left(0.752\right)$

  $\therefore {\text{θ}}_{\text{p}}=36.9°$

Now, when refraction occur at the critical angle,

  ${\text{n}}_{\text{air}}\sin {\text{θ}}_1=n_{\text{water}}\sin {\text{θ}}_2$

  $\therefore \left(1.00\right)\sin 90°=\left(1.33\right)\sin {\text{θ}}_c$

  $\therefore {\text{θ}}_c={\sin }^{-1}\left(\frac{\left(1.00\right)\sin 90°}{1.33}\right)$

  $\therefore {\text{θ}}_c=48.8°$

According to the question, if light is coming from air to water, so, the Brewster’s angle,

  $\tan {\text{θ}}_{\text{p}}\text{'}=\frac{n_{\text{water}}}{{\text{n}}_{\text{air}}}=\frac{1.33}{1.00}=1.33$

  $\therefore {\text{θ}}_{\text{p}}\text{'}={\tan }^{-1}\left(1.33\right)$

  $\therefore {\text{θ}}_{\text{p}}\text{'}=53.1°$

Hence,

  ${\text{θ}}_{\text{p}}+{\text{θ}}_{\text{p}}\text{'}=36.9°+53.1°=90°$

Conclusion: The required critical angle and the Brewster’s angle above the surface is $48.8°$ and $53.1°$ .