Chapter 18, Problem 102A

To determine

The difference in refracted angles of red and blue light.

Answer to Problem 102A

  $0.63°$

Explanation of Solution

Given:

The angle of incidence at the flint glass, ${\theta }_1=45°$

The glass’s index of refraction for red light, $n_{2r}=1.7273$

The glass’s index of refraction for blue light, $n_{2b}=1.7708$

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, $n{1}_{}=1.00$

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

  $\begin{array}{c}{n}_1\sin {\theta }_1=n_{2r}\sin {\theta }_{2r}\\ \left(1.00\right)\sin 45°=\left(1.7273\right)\sin {\theta }_{2r}\\ {\theta }_{2r}={\sin }^{-1}\left(0.409\right)\\ =24.14°\end{array}$

Similarly, angle of refraction for blue light will be,

  $\begin{array}{c}{n}_1\sin {\theta }_1=n_{2b}\sin {\theta }_{2b}\\ \left(1.00\right)\sin 45°=\left(1.7708\right)\sin {\theta }_{2b}\\ {\theta }_{2b}={\sin }^{-1}\left(0.399\right)\\ =23.51°\end{array}$

Then, the difference in refracted angles of red and blue light will be,

  $\begin{array}{c}\theta ={\theta }_{2r}-{\theta }_{2b}\\ =24.14°-23.51°\\ =0.63°\end{array}$

Conclusion:

Hence, the difference in refracted angles of red and blue light is $0.63°$ .