Question

An air bubble inside an $8.0-cm$-diameter plastic ball is localid="1651400303536" $2.0\mathrm{cm}$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

As a result, the picture of the bubble appears to be $1.5cm$beneath the surface.

Step-by-step solution

Relationship between the refractive indices 

To locate the picture, use the relationship between the refractive indices of the two media and the radius of the lens. An image will be created at distance $s$from a spherical refracting surface if an object is positioned at distance$s^{\text{'}}$from it.

$\frac{n_1}{s}+\frac{n_2}{s^{\text{'}}}=\frac{n_2-n_1}{R}$

$n_1$is refractive index of air,$n_2$is refractive index of plastic, and $R$is radius of curvature.

Expression for s'

Radius of plastic ball,

$R=-\frac{\text{Diameter of plastic ball}}{2}$

$=-\frac{8.0\mathrm{cm}}{2}$

$=-4.0\mathrm{cm}$

$\frac{n_1}{s}+\frac{n_2}{s^l}=\frac{n_2-n_1}{R}$rearrange for $s^{\text{'}}$

${\mathrm{s}}^{\text{'}}=\frac{{\mathrm{n}}_2}{\frac{{\mathrm{n}}_2-{\mathrm{n}}_1}{\mathrm{R}}-\frac{{\mathrm{n}}_1}{\mathrm{s}}}$

Calculation for  image of the bubble's position 

The image of the bubble's position is calculated as follows:

$s\text{'}=\frac{n_2}{\frac{n_2-n_1}{R}-\frac{n_1}{s}}$

Substitute$1$for role="math" localid="1651400853986" $n_2,$$1.59$forrole="math" localid="1651400866937" $n_1,$$-4.0\mathrm{cm}$ for $R$and$2.0\mathrm{cm}$ for$s$

$s\text{'}=\frac{1}{\frac{1-1.59}{-4.0\mathrm{cm}}-\frac{1.59}{2.0\mathrm{cm}}}$

$=\frac{1}{0.147{\mathrm{cm}}^{-1}-0.795{\mathrm{cm}}^{-1}}$

$=-1.5\mathrm{cm}$