Chapter 1.4, Problem 47PPS

a.

To determine

To determine the type of angle.

Answer to Problem 47PPS

m $\angle 1$ is an obtuse angle.

Explanation of Solution

Given information:

  

Evaluation:

m $\angle 1$ is the angle between the incident ray and the surface of water. m $\angle 1$ is an obtuse angle because it is greater than $90°$ .

b.

To determine

To determine the type of angle.

Answer to Problem 47PPS

m $\angle 2$ is an acute angle.

Explanation of Solution

Given information:

  

Evaluation:

m $\angle 2$ is the angle between the refracted ray and the surface of water. m $\angle 1$ is an acute angle because it is less than $90°$ .

c.

To determine

To determine how many degrees the path of light changes when it enters water.

Answer to Problem 47PPS

  ${\theta }_2^{}={\sin }^{-1}(\frac{n_1^{}\sin {\theta }_1^{}}{n_2^{}})$

Explanation of Solution

Given information:

  

Formula used:

Snell’s Law:

  

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

  $n_1^{}$  and $n_2^{}$ are the two different mediums that will impact the refraction.?1 is the angle of incidence; ?2 is the angle of refraction.

Evaluation:

According to Snell’s Law:

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

Therefore, to find ${\theta }_2^{}$

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

Taking inverse of sin,

  ${\theta }_2^{}={\sin }^{-1}(\frac{n_1^{}\sin {\theta }_1^{}}{n_2^{}})$