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Chapter 33: Problem 22

A black, totally absorbing piece of cardboard of area \(A=2.0 \mathrm{~cm}^{2}\) intercepts light with an intensity of \(10 \mathrm{~W} / \mathrm{m}^{2}\) from a camera strobe light. What radiation pressure is produced on the cardboard by the light?

Short Answer

Expert verified
Radiation pressure on the cardboard is approximately \( 3.33 \times 10^{-8} \, \text{N/m}^2 \).

Step by step solution

01

Identify the Equation

The key principle here is that the radiation pressure, when light is totally absorbed by a surface, is given by the equation: \( P = \frac{I}{c} \), where \( P \) is the radiation pressure, \( I \) is the intensity of the light, and \( c \) is the speed of light in a vacuum \( (c = 3 \times 10^8 \, \text{m/s}) \).
02

Check Unit Consistency

Make sure all the units are consistent. Here, intensity (\( I = 10 \, \text{W/m}^2 \)) is already in the correct units to use in the equation, and \( c = 3 \times 10^8 \, \text{m/s} \) is standard in SI units.
03

Apply the Equation

Substitute the given values into the equation: \( P = \frac{10 \, \text{W/m}^2}{3 \times 10^8 \, \text{m/s}} \).
04

Calculate Radiation Pressure

Perform the division: \( P = \frac{10}{3 \times 10^8} \). Simplifying this gives \( P \approx 3.33 \times 10^{-8} \, \text{N/m}^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intensity of Light
The intensity of light is a crucial factor when analyzing how energy interacts with surfaces. Think of light intensity as the amount of energy the light delivers per unit area per second. This is measured in watts per square meter (W/m虏).
It's like measuring how much sunlight heats your skin during a hot summer day, except here we're dealing with scientific values. In our exercise, the intensity is given as 10 W/m虏. This means every square meter of the surface being lit by the camera strobe light receives 10 watts of energy every second. - Intensity depends on several factors: - The power of the light source - The distance from the light source - The angle at which light hits the surface
Understanding intensity helps in estimating the impact of light on surfaces, with higher intensities typically exerting greater effects.
Absorption
Absorption is the process through which a surface takes in light energy. This means that when light hits a surface, some or all of it can be absorbed rather than being reflected. In the case of our exercise, the cardboard is described as black and totally absorbing, which implies it absorbs all the light energy that hits it. - Some key points about absorption: - Dark surfaces generally absorb more light compared to lighter ones. - The energy absorbed can cause an increase in temperature or convert into other forms of energy. - Absorption directly influences radiation pressure. In practical terms, since the cardboard absorbs all the light, the radiation pressure experienced is more straightforward to calculate鈥攖he entire intensity of the light contributes to the pressure on the surface.
Speed of Light
The speed of light is the rate at which light travels through a vacuum, and it is the ultimate speed limit in the universe. The speed of light is denoted by the symbol "c" and has a value of approximately 3 x 10鈦� meters per second (m/s).This value plays a vital role in calculating phenomena involving light, such as radiation pressure, as in our exercise. Calculating radiation pressure involves dividing the light intensity by this speed.
  • The consistency of the speed of light value across calculations is crucial for reliable results.
  • This speed remains the same, regardless of the observer's movement or the light's source.
  • It is a fundamental constant in physics, pivotal to theories like relativity.
In our specific context, the speed of light allows us to understand how fast the energy is being transferred and helps achieve the calculated radiation pressure value using the formula \( P = \frac{I}{c} \). Every instance the light applies force, it moves at this incredibly brisk speed.

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Most popular questions from this chapter

High-power lasers are used to compress a plasma (a gas of charged particles) by radiation pressure. A laser generating radiation pulses with peak power \(1.5 \times 10^{3} \mathrm{MW}\) is focused onto \(1.0 \mathrm{~mm}^{2}\) of high-electron-density plasma. Find the pressure exerted on the plasma if the plasma reflects all the light beams directly back along their paths.

The magnetic component of a polarized wave of light is given by \(B_{x}=(4.00 \mu \mathrm{T}) \sin \left[k y+\left(2.00 \times 10^{15} \mathrm{~s}^{-1}\right) t\right] .\) (a) In which direction does the wave travel, (b) parallel to which axis is it polarized, and (c) what is its intensity? (d) Write an expression for the electric field of the wave, including a value for the angular wave number. (e) What is the wavelength? (f) In which region of the electromagnetic spectrum is this electromagnetic wave?

As a comet swings around the Sun, ice on the comet's surface vaporizes, releasing trapped dust particles and ions. The ions, because they are electrically charged, are forced by the electrically charged solar wind into a straight ion tail that points radially away from the Sun (Fig. \(33-39\) ). The (electrically neutral) dust particles are pushed radially outward from the Sun by the radiation force on them from sunlight. Assume that the dust particles are spherical, have density \(3.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3},\) and are totally absorbing. (a) What radius must a particle have in order to follow a straight path, like path 2 in the figure? (b) If its radius is larger, does its path curve away from the Sun (like path 1 ) or toward the Sun (like path 3 )?

Prove, for a plane electromagnetic wave that is normally incident on a flat surface, that the radiation pressure on the surface is equal to the energy density in the incident beam. (This relation between pressure and energy density holds no matter what fraction of the incident energy is reflected.)

Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of \(1.4 \mathrm{~kW} / \mathrm{m}^{2}\). (a) Assuming that Earth (and its atmosphere) behaves like a flat disk perpendicular to the Sun's rays and that all the incident energy is absorbed, calculate the force on Earth due to radiation pressure. (b) For comparison, calculate the force due to the Sun's gravitational attraction.
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