Chapter 4, Problem 13PS

a.

To determine

Find ${\theta }_2$ .

Answer to Problem 13PS

  ${\theta }_2=\boxed{40.5°}$

Explanation of Solution

Given information:

When you stand in shallow water and look at an object below the surface of the water, the object will look farther away from you than it really is. This is because when light rays pass between air and water, the water refracts, or bends, the light rays. The index of refraction for water is $1.333$ . This is the ratio of the sine of ${\theta }_1$ and the sine of ${\theta }_2$ (see figure).

  

While standing in water that is $2$ feet deep, you look at a rock at angle ${\theta }_1=60°$ (measured from a line perpendicular to the surface of the water). Find ${\theta }_2$ .

Calculation:

Here, we will consider the following figure:

  

We are given the ratio:

  $\frac{\sin {\theta }_1}{\sin {\theta }_2}=1.333$

Now, for ${\theta }_1=60°$ :

  $\begin{array}{l}\frac{\sin 60°}{\sin {\theta }_2}=1.333\\ \sin {\theta }_2=\frac{\sin 60°}{1.333}\\ \sin {\theta }_2=\frac{\sqrt{3}}{2\left(1.333\right)}\\ \sin {\theta }_2=0.65\\ {\theta }_2={\sin }^{-1}\left(0.65\right)\\ {\theta }_2=\boxed{40.5°}\end{array}$

Hence, ${\theta }_2=\boxed{40.5°}$ .

b.

To determine

Find the distances $x$ and $y$ .

Answer to Problem 13PS

  $\boxed{x=1.71ft}$ , $\boxed{y=3.46ft}$

Explanation of Solution

Given information:

When you stand in shallow water and look at an object below the surface of the water, the object will look farther away from you than it really is. This is because when light rays pass between air and water, the water refracts, or bends, the light rays. The index of refraction for water is $1.333$ . This is the ratio of the sine of ${\theta }_1$ and the sine of ${\theta }_2$ (see figure).

  

Find the distances $x$ and $y$ .

Calculation:

Here, we will find $x$ :

  $\begin{array}{l}\tan {\theta }_2=\frac{x}{2}\\ x=2\tan 40.5°\\ x=\boxed{1.71feet}\end{array}$

Now, we will find $y$ :

  $\begin{array}{l}\tan {\theta }_1=\frac{y}{2}\\ y=2\tan 60°\\ y=\boxed{3.46}\end{array}$

c.

To determine

Find the distance $d$ between where the rock is and where it appears to be.

Answer to Problem 13PS

  $\boxed{‘d’willdecrease}$

Explanation of Solution

Given information:

When you stand in shallow water and look at an object below the surface of the water, the object will look farther away from you than it really is. This is because when light rays pass between air and water, the water refracts, or bends, the light rays. The index of refraction for water is $1.333$ . This is the ratio of the sine of ${\theta }_1$ and the sine of ${\theta }_2$ (see figure).

  

Find the distance $d$ between where the rock is and where it appears to be.

Calculation:

Here, as we move close to the rock, $d$ must get smaller and smaller. The angles ${\theta }_1$ and ${\theta }_2$ will decrease along with the distance $y$ , so $d$ will $\boxed{Decrease}$ .