Chapter 18, Problem 45A

To determine

The critical angle for a diamond in air.

Answer to Problem 45A

  $24.4°$

Explanation of Solution

Given:

The light ray travels from diamond into air.

Formula used:

Critical angle is given as,

  ${\theta }_c={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$

  $\begin{array}{l}{n}_1=\text{ index of refraction of incident medium}\\ n_2=\text{ index of refraction of refracting medium}\end{array}$

Calculation:

It is known that,

Index of refraction of diamond, $n{1}_{}=2.42$

Index of refraction of air, $n{2}_{}=1.0$

Then, using the given values in above, the critical angle for diamond will be,

  $\begin{array}{c}{\theta }_c={\sin }^{-1}\left(\frac{1.0}{2.42}\right)\\ =24.4°\end{array}$

Conclusion:

Hence, the critical angle for diamond is $24.4°$ .