Chapter 34: Q. 29 Excercises And Problems (page 991)

An air bubble inside an $8.0 - c m$ -diameter plastic ball is localid="1651400303536" $2.0 cm$ 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

Expert verified

As a result, the picture of the bubble appears to be $1.5 c m$ 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 $s^{'}$ from it.

$\frac{n_{1}}{s} + \frac{n_{2}}{s^{'}} = \frac{n_{2} - n_{1}}{R}$

$n_{1}$ 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,

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

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

$= - 4.0 cm$

$\frac{n_{1}}{s} + \frac{n_{2}}{s^{l}} = \frac{n_{2} - n_{1}}{R}$ rearrange for $s^{'}$

$s^{'} = \frac{n_{2}}{\frac{n_{2} - n_{1}}{R} - \frac{n_{1}}{s}}$

Calculation for image of the bubble's position

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

$s' = \frac{n_{2}}{\frac{n_{2} - n_{1}}{R} - \frac{n_{1}}{s}}$

Substitute $1$ for role="math" localid="1651400853986" $n_{2},$ $1.59$ forrole="math" localid="1651400866937" $n_{1},$ $- 4.0 cm$ for $R$ and $2.0 cm$ for $s$

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

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

$= - 1.5 cm$

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An air bubble inside an $8.0 - c m$ -diameter plastic ball is localid="1651400303536" $2.0 cm$ 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

Step by step solution

Relationship between the refractive indices

Expression for s'

Calculation for image of the bubble's position

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An air bubble inside an 8.0-cm-diameter plastic ball is localid="1651400303536" 2.0cmfrom 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

Step by step solution

Relationship between the refractive indices

Expression for s'

Calculation for image of the bubble's position

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Wave Optics

Circular Motion and Gravitation

Electrostatics

Thermodynamics

Nuclear Physics

Kinematics Physics

Study anywhere. Anytime. Across all devices.

An air bubble inside an $8.0 - c m$ -diameter plastic ball is localid="1651400303536" $2.0 cm$ 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?