Q20E - University Physics with Modern Physics

Question

A person is lying on a diving board 3.00 m above the surface of the water in a swimming pool. She looks at a penny that is on the bottom of the pool directly below her. To her, the penny appears to be a distance of 7.00 m from her. What is the depth of the water at this point?

Step-by-Step Solution

Answer

The depth of water from this point is 5.332m

Step 1: Important Concepts

The object image relationship for a spherical reflecting surface is 

$\frac{{\mathbf{n}}_{\mathbf{1}}}{\mathbf{s}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{2}}}{\mathbf{s}\mathbf{\text{'}}}\mathbf{=}\frac{{\mathbf{n}}_{\mathbf{2}}\mathbf{-}{\mathbf{n}}_{\mathbf{1}}}{\mathbf{R}}$     

Where ${\mathit{n}}_{\mathbf{2}}$ and ${\mathit{n}}_{\mathbf{1}}$ are the refraction index of the surfaces 

s is th object distance from the vertex of the spherical surface and 

$\mathit{s}\mathbf{\text{'}}$ is the image distance from the vertex of the spherical surface and R is the radius of curvature of the spherical surface

Step 2: Find the object distance

Here we have 

                                                            $n_2=1$                                                  Since it is air.

                                                             $n_1=1.33$  

The image is located at 

                                                              $$\begin{gathered} s\text{'}=-\left(7-3\right)m \\ s\text{'}=-4m \end{gathered}$$                   

It is negative since the image is on the opposite side of the outgoing light.

 

We know since the surface of water-air interaction is planar hence $R=\infty$

 

Input this information in the formula 

$$\begin{gathered} \frac{n_1}{s}+\frac{n_2}{s\text{'}}=\frac{n_2-n_1}{\infty } \\ \frac{n_1}{s}+\frac{n_2}{s\text{'}}=0 \\ \frac{n_1}{s}=-\frac{n_2}{s\text{'}} \end{gathered}$$ 

                                                            

 Rearranging we have 

$$\begin{gathered} s=\frac{n_{1}s\text{'}}{n_2} \\ =\frac{1.33\times -4m}{-1} \\ s=5.332m \end{gathered}$$                                                     

The depth of water from this point is 5.332m