Chapter 23, Problem 36P

To determine

The critical angle for the interface between water and Lucite.

The medium from which ray should enter for total internal reflection to occur.

Answer to Problem 36P

The critical angle is $\text{61}{\text{.7}}^{\text{o}}$.

The medium from which ray should enter for total internal reflection to occur: Lucite

Explanation of Solution

Given info:

Refractive index for water is $\text{1}\text{.33}$ andLucite is $\text{1}\text{.51}$ .

Formula used:

According to Snell’s laws

  ${\text{n}}_{\text{1}}{\text{sin(θ}}_{\text{1}}{\text{)=n}}_{\text{2}}{\text{sin(θ}}_{\text{2}}\text{)}\text{}$

Critical angle as ${\text{θ}}_{\text{1}}$ , ${\text{θ}}_{\text{2}}$ for water and Lucite and indices of refraction as $\text{1}\text{.33}$ , $\text{1}\text{.51}$ respectively.

Calculation:

According to the question,

  ${\text{n}}_{\text{1}}{\text{sin(θ}}_{\text{1}}{\text{)=n}}_{\text{2}}{\text{sin(θ}}_{\text{2}}\text{)}\text{}$

  $\left(\text{1}\text{.51}\right){\text{sin(θ}}_{\text{1}}\text{) = }\left(\text{1}\text{.33}\right)\text{sin}\left({\text{90}}^{\text{o}}\right)$

  $\begin{array}{l}{\text{θ}}_{\text{1}}\text{ is of course the critical angle}\text{.}\hfill \\ {\text{θ}}_{\text{1}}{\text{ = sin}}^{{}_{\text{-1}}}\left(\text{1}\text{.33/1}\text{.51}\right)\text{ = 61}{\text{.7}}^{\text{o}}\hfill \end{array}$

  $\text{= 61}{\text{.7}}^{\text{o}}$

Also, light must start in the Lucite because it has higher index.

Conclusion:

The critical angle for the interface between water and Lucite is $\text{ 61}{\text{.7}}^{\text{o}}$ .