Question

An air bubble inside an$8.0-cm-$-diameter plastic ball is $2.0cm$from the surface. As you look at the ball with the bubble turned toward you, how far beneath the surface does the bubble appear to be?

Short answer

The air bubble seems to be $1.06cm$below the surface of the plastic ball.

Step-by-step solution

Step: 1 Refractions:

Refraction from a spherical surface of radius $R$can produce an image. The distances between the item and also the picture from such a surface area linked by

localid="1649161561334" $\frac{n_1}{s}+\frac{n_2}{s^{\mathrm{\prime }}}=\frac{n_2-n_1}{R}$

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 polystyrene plastic into the air and model the air bubble as a point source. The rays originating from the air bubble are refracted away from the normal at the surface and diverge outward because $n_1>n_2$, where$n_2=1.00$is the index of refraction of air. The air bubble seems closer to the surface than it is because the rays refract away from the normal. The image point within the sphere may be seen by extending the outgoing rays backward.

Step: 3 Substituing vales:

The object surface is concave by

$\begin{array}{l}\frac{n_1}{s}+\frac{n_2}{s^{\mathrm{\prime }}}=\frac{n_2-n_1}{R}\\ s^{\mathrm{\prime }}=\frac{n_2}{\frac{n_2-n_1}{R}-\frac{n_1}{s}}\\ s^{\mathrm{\prime }}=\frac{1.00}{\frac{1.59-1.00}{-4.0\mathrm{cm}}-\frac{1.59}{2.0\mathrm{cm}}}\\ s^{\mathrm{\prime }}=-1.06\mathrm{cm}.\end{array}$