Q21E. - University Physics with Modern Physics

Question

A Spherical Fish Bowl. A small tropical fish is at the centre of a water-filled, spherical fish bowl 28.0 cm in diameter.
(a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?

Step-by-Step Solution

Answer

A. Image is formed at the centre of the bowl and magnification is 1.33

B. $s\text{'}$ is larger than the diameter, hence focal point outside the bowl

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 ${\mathit{n}}_{\mathbf{1}}$ and ${\mathit{n}}_{\mathbf{2}}$ 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

Magnification is given by 

$\mathit{m}\mathbf{=}\mathbf{-}\frac{{\mathbf{n}}_{\mathbf{1}}\mathbf{s}\mathbf{\text{'}}}{{\mathbf{n}}_{\mathbf{2}}\mathbf{s}}$                                                 

Step 2: Find the object distance

Here 

$R=-14,n_1=1.33,n_2=1 and s=14$                                       

Hence solve for $s\text{'}$

$\frac{1.33}{14}+\frac{1}{s\text{'}}=\frac{1.33-1}{-14}$ 

                                                

We get

$s\text{'}=-14cm$                                                 

The image is formed 14cmm to the left of the bowl surface.

 

Magnification is 

$$\begin{gathered} m=-\frac{n_{1}s\text{'}}{n_{2}s}=-\frac{1.33\left(-14\right)}{14} \\ m=1.33 \end{gathered}$$               

Step 3: Find focal point

To find focal point use $s=\infty$ and solve for s’

Use $R=-14,n_2=1.33,n_1=1$

$$\begin{gathered} \frac{1}{\infty }+\frac{1.33}{s\text{'}}=\frac{1.33-1}{-14} \\ s\text{'}=56cm \end{gathered}$$ 

 

S’ is larger than the diameter, hence the focal point is outside the bowl