Chapter 18, Problem 52A

To determine

The angle of refraction for red and blue light.

Answer to Problem 52A

  $11.94°$ and $11.77°$ , respectively

Explanation of Solution

Given:

The angle of incidence at the diamond, ${\theta }_1=30.0°$

The diamond’s index of refraction for red light, $n_{2r}=2.41$

The diamond’s index of refraction for blue light, $n_{2b}=2.45$

Formula used:

Snell’s law of refraction,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Where,

  $\begin{array}{l}{n}_1=\text{ index of refraction of first medium}\\ n_2=\text{ index of refraction of second medium}\\ {\theta }_1=\text{ angle of incidence}\\ {\theta }_2=\text{ angle of refraction}\end{array}$

Calculation:

It is known that,

Index of refraction of air, $n{1}_{}=1.0003$

Then, using Snell’s law of refraction, angle of refraction for red light will be,

  $\begin{array}{c}{n}_1\sin {\theta }_1=n_{2r}\sin {\theta }_{2r}\\ \left(1.0003\right)\sin 30.0°=\left(2.41\right)\sin {\theta }_{2r}\\ {\theta }_{2r}={\sin }^{-1}\left(0.207\right)\\ =11.94°\end{array}$

Similarly, angle of refraction for blue light will be,

  $\begin{array}{c}{n}_1\sin {\theta }_1=n_{2b}\sin {\theta }_{2b}\\ \left(1.0003\right)\sin 30.0°=\left(2.45\right)\sin {\theta }_{2b}\\ {\theta }_{2b}={\sin }^{-1}\left(0.204\right)\\ =11.77°\end{array}$

Conclusion:

Hence, the angle of refraction for red light and blue light is $11.94°$ and $11.77°$ , respectively.