Chapter 18.1, Problem 4PP

To determine

The index of refraction of the material of the block.

Answer to Problem 4PP

  $1.50$

Explanation of Solution

Given:

The angle of incidence of the light at the block, ${\theta }_1=31°$

The angle of refraction of the light in the block, ${\theta }_2=27°$

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 water, $n{1}_{}=1.33$

Then, using Snell’s law of refraction, index of refraction of the material of the block will be,

  $\begin{array}{c}{n}_1\sin {\theta }_1=n_2\sin {\theta }_2\\ \left(1.33\right)\sin 31°=n_2\sin 27°\\ n_2=1.50\end{array}$

Conclusion:

Hence, the index of refraction of the material of the block is $1.50$ .