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Chapter 5: Problem 8

Calculate the radiation pressure of sunlight when the incoming energy/(s \(\mathrm{m}^{2}\) ) is \(1.4 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2}\) and the radiation is completely absorbed. Compare this value with the atmospheric pressure. What is the force on a surface of \(1 \mathrm{~m}^{2}\) ? What is the force if the light is completely reflected? Hint: The radiation pressure is the momentum transferred per unit of time and surface area.

Short Answer

Expert verified
Radiation pressure is \(4.67 \times 10^{-6} \, \mathrm{N/m^2}\). Force on 1 m虏 is \(4.67 \times 10^{-6} \, \mathrm{N}\) for absorbed, \(9.34 \times 10^{-6} \, \mathrm{N}\) for reflected.

Step by step solution

01

Understand Radiation Pressure

Radiation pressure is the pressure exerted by electromagnetic radiation on a surface due to the momentum transfer from absorbed or reflected photons. In the case of complete absorption, the radiation pressure can be calculated using the relation:\[P = \frac{E}{c}\]where \(P\) is the radiation pressure, \(E\) is the energy per unit area, and \(c\) is the speed of light in vacuum, approximately \(3 \times 10^8 \) m/s.
02

Calculate Radiation Pressure for Absorbed Light

Given that the incoming energy is \(1.4 \times 10^{3} \, \mathrm{W/m^2}\), substitute \(E = 1.4 \times 10^{3} \, \mathrm{W/m^2}\) and \(c = 3 \times 10^8\, \mathrm{m/s}\) into the radiation pressure formula:\[P = \frac{1.4 \times 10^{3}}{3 \times 10^8} \, \mathrm{N/m^2} = 4.67 \times 10^{-6} \, \mathrm{N/m^2}\]This calculation gives the radiation pressure when the light is completely absorbed.
03

Compare with Atmospheric Pressure

The atmospheric pressure at sea level is approximately \(1.01 \times 10^5 \, \mathrm{N/m^2}\). Compared to atmospheric pressure, the radiation pressure \(4.67 \times 10^{-6} \, \mathrm{N/m^2}\) is extremely small.
04

Calculate Force on 1 m虏 Surface for Absorbed Light

The force \(F\) on a surface is given by the product of the pressure \(P\) and the area \(A\):\[F = P \times A = 4.67 \times 10^{-6} \, \mathrm{N/m^2} \times 1 \, \mathrm{m^2} = 4.67 \times 10^{-6} \, \mathrm{N}\]This is the force on a surface area of 1 m虏 when the light is completely absorbed.
05

Calculate Force for Completely Reflected Light

If the light is completely reflected, the radiation pressure doubles because the momentum change is twice as large. Therefore, the pressure is:\[P_{reflected} = 2 \times 4.67 \times 10^{-6} \, \mathrm{N/m^2} = 9.34 \times 10^{-6} \, \mathrm{N/m^2}\]The force on a 1 m虏 surface is then:\[F_{reflected} = 9.34 \times 10^{-6} \, \mathrm{N/m^2} \times 1 \, \mathrm{m^2} = 9.34 \times 10^{-6} \, \mathrm{N}\]Thus, this is the force for completely reflected light on a 1 m虏 surface.

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Key Concepts

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

Electromagnetic Radiation
Electromagnetic radiation is a form of energy that travels through space as waves. These waves consist of oscillating electric and magnetic fields. One of the most familiar forms of electromagnetic radiation is sunlight.
The energy carried by these waves can interact with matter, leading to phenomena such as light absorption, reflection, and scattering.
  • Light has a dual nature, functioning as both waves and particles known as photons.
  • Photons are the carriers of electromagnetic radiation and can exert pressure and transfer momentum when interacting with surfaces.
The interaction of photons with surfaces is what gives rise to radiation pressure, which can be observed when electromagnetic waves exert a force on objects.
Momentum Transfer
Momentum transfer is a key concept in understanding how radiation pressure is exerted. When photons hit a surface, they transfer momentum, which is a property of moving objects and is related to both their mass and velocity.
The momentum of a photon is given by the equation \( p = \frac{E}{c} \), where \(E\) is the energy of the photon and \(c\) is the speed of light.
- In the process of absorption, photons transfer their momentum entirely to the surface.- If the light reflects, the momentum transfer is doubled since the direction of momentum changes.The transfer of momentum from photons to surfaces results in radiation pressure, which although extremely small compared to everyday forces, has significant implications in space science and technology.
Speed of Light
The speed of light, denoted by \(c\), is one of the fundamental constants of nature. It is the speed at which electromagnetic waves propagate through the vacuum of space, approximately \(3 \times 10^8\) meters per second.
This value is not just a measure of how fast light travels, but it also plays a crucial role in the calculations involving electromagnetic phenomena.
  • In the context of radiation pressure, \(c\) is used to determine the momentum of photons.
  • The equation \( p = \frac{E}{c} \) highlights how the speed of light links energy and momentum.
The invariance of the speed of light underpins many foundational principles in physics, including Einstein's theory of relativity, which has transformed our understanding of space, time, and energy.
Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of the air above a given point. It is typically measured in Newtons per square meter (\(\mathrm{N/m^2}\)), and at sea level, it averages around \(1.01 \times 10^5\, \mathrm{N/m^2}\).
This pressure plays a crucial role in weather patterns and affects our daily lives. In the context of radiation pressure, comparing the two highlights the vast difference in magnitude.
  • Radiation pressure for absorbed sunlight is approximately \(4.67 \times 10^{-6} \mathrm{N/m^2}\), a tiny fraction of atmospheric pressure.
  • This comparison emphasizes why radiation pressure has minimal impact on Earth's weather systems but can be a significant force in space, where there is no atmospheric pressure.
Understanding the difference in these pressures helps in fields like meteorology and aerospace, where forces of different scales are continually assessed.

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Most popular questions from this chapter

A photon with \(10^{4} \mathrm{eV}\) energy collides with a free electron at rest and is scattered through an angle of \(60^{\circ}\). Calculate a) the change in energy, frequency and wavelength of the photon, and b) the kinetic energy, momentum and direction of the electron after the collision.

A pulsed laser with a repetition rate of \(20 \mathrm{~Hz}\) produces light pulses with a length of 30 ps (picoseconds). Its laser medium is Nd: YAG and the wavelength is \(1064 \mathrm{~nm}\). The light pulses are passed through a frequency-tripler (efficiency \(10 \%\), i.e. only \(1 / 10\) of the incident energy is available at the output of the frequency-tripler). An output pulse energy of \(1 \mathrm{~mJ}\) is required at the output wavelength \(\lambda\) (to be calculated). a) How many photons \(n\) are contained in an output pulse? b) How high is the peak power \(P\) of the laser system at the output of the tripler? c) What is the average light power \(I\) of the pulsed laser before the tripler, if only \(0.1 \%\) of the electrical power is converted into light power?

X-rays with a wavelength of \(1 \AA\) are scattered on graphite. The scattered radiation is observed perpendicular to the direction of the incident \(x\) -rays. a) How large is the Compton shift \(\Delta \lambda\) ? b) How large is the kinetic energy of the ejected electron? c) What fraction of its original energy does the photon lose? d) How large is the corresponding fraction of energy lost by a photon with a wavelength \(\lambda=0.1 \AA\) if it is deflected through \(90^{\circ}\) by Compton scattering? The electron should be considered at rest before the collision, and the binding energy should be neglected.

A person can perceive yellow light with the naked eye when the power being delivered to the retina is \(1.8 \times 10^{-18} \mathrm{~W}\). The wavelength of yellow light is about \(6000 \dot{A}\). At this power, how many photons fall on the retina each second?

Express the relativistic mass of a photon in terms of \(h, \lambda\) and \(c\).
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