Question

A goldfish lives in a $50-cm$-diameter spherical fish bowl. The fish sees a cat watching it. If the cat's face is$20cm$ from the edge of the bowl, how far from the edge does the fish see it as being? (You can ignore the thin glass wall of the bowl.)

Short answer

The cat seems to be $36cm$away from the edge of the bowl.

Step-by-step solution

Step: 1 Refraction:

The refraction image from surface by

$\frac{n_1}{s}+\frac{n_2}{s^{\mathrm{\prime }}}=\frac{n_2-n_1}{R}$

When the light strikes a media with an index of refraction of $n_1$and is refracted by a medium with an index of refraction of $n_2$. If the surface is convex toward the item, the radius of curvature $R$is positive; otherwise, it is negative.

Step: 2 Paraxial rays:

Consider the paraxial rays that refract from the air into the water using the cat as a point source. The rays originating from the air bubble are refracted toward the normal at the surface because $n_2>n_1$ where$n_1=1.00$ is the index of refraction of air, and the cat's face seems farther from the edge of the bowl than it actually is.

Step: 3 Finding values:

The convex towards cat is

$\begin{array}{l}\frac{n_1}{s}+\frac{n_2}{s^{\mathrm{\prime }}}=\frac{n_2-n_1}{R}\\ s^{\mathrm{\prime }}=\frac{n_2}{\frac{n_2-n_1}{R}-\frac{n_1}{s}}\\ s^{\mathrm{\prime }}=\frac{1.33}{\frac{1.33-1.00}{25\mathrm{cm}}-\frac{1.00}{20\mathrm{cm}}}\\ s^{\mathrm{\prime }}=-36\mathrm{cm}\\ s^{\mathrm{\prime }}=36cm.\end{array}$