Chapter 4.3, Problem 56E

Solution

To state: The value of $\sin {\theta }_2$ if ${\theta }_1={42}^{\circ }$ and an index of refraction is $1.52$ for a certain piece of crow glass, ${\theta }_1$ is the angle from which a ray enters the glass. The index of refraction is constant $\mu$ that satisfies the equation $\sin {\theta }_1=\mu \sin {\theta }_2$ .

The resultant answer is $\sin {\theta }_2=0.44$ .

Given information:

The given equation is $\sin {\theta }_1=\mu \sin {\theta }_2$ where ${\theta }_1={42}^{\circ }\text{ and }\mu =1.52$ .

Explanation:

Consider the given equation: $\sin {\theta }_1=\mu \sin {\theta }_2$

Solve for $\sin {\theta }_2$ ,

  $\sin {\theta }_2=\frac{\sin {\theta }_1}{\mu }$

Substitute the values ${\theta }_1={42}^{\circ }\text{ and }\mu =1.52$ ,

  $\begin{array}{c}\sin {\theta }_2=\frac{\sin {42}^{\circ }}{1.52}\\ =\frac{0.67}{1.52}\\ =0.44\end{array}$

Therefore, the index of refraction is $\sin {\theta }_2=0.44$ .