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Chapter 23: Problem 21

\(\bullet\) The intensity at a certain distance from a bright light source is 6.00 \(\mathrm{W} / \mathrm{m}^{2} .\) Find the radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing surface and (b) a totally reflecting surface.

Short Answer

Expert verified
Absorbing: \(2 \times 10^{-8} \, \text{Pa}\), \(1.97 \times 10^{-13} \, \text{atm}\); Reflecting: \(4 \times 10^{-8} \, \text{Pa}\), \(3.95 \times 10^{-13} \, \text{atm}\).

Step by step solution

01

Understand the Concept of Radiation Pressure

Radiation pressure is the force exerted by electromagnetic radiation on a surface per unit area. It can be calculated using the formula for a totally absorbing surface as \( P = \frac{I}{c} \) and for a totally reflecting surface as \( P = \frac{2I}{c} \), where \( I \) is the intensity and \( c \) is the speed of light in a vacuum (approximately \( 3 \times 10^8 \) meters per second).
02

Calculate Radiation Pressure for Totally Absorbing Surface

For a totally absorbing surface, input the given intensity into the formula for radiation pressure: \( P = \frac{I}{c} = \frac{6.00}{3 \times 10^8} \). Solving this gives \( P = 2 \times 10^{-8} \, \text{Pa}\).
03

Calculate Radiation Pressure for Totally Reflecting Surface

For a totally reflecting surface, the formula becomes \( P = \frac{2I}{c} \). Inputting the given intensity: \( P = \frac{2 \times 6.00}{3 \times 10^8} \). Solving this gives \( P = 4 \times 10^{-8} \, \text{Pa}\).
04

Convert Pressure from Pascals to Atmospheres

Since 1 atmosphere is approximately equivalent to \( 101325 \, \text{Pa} \), convert the computed pressures: - For the absorbing surface: \( 2 \times 10^{-8} \, \text{Pa} \approx \frac{2 \times 10^{-8}}{101325} \approx 1.97 \times 10^{-13} \, \text{atm} \).- For the reflecting surface: \( 4 \times 10^{-8} \, \text{Pa} \approx \frac{4 \times 10^{-8}}{101325} \approx 3.95 \times 10^{-13} \, \text{atm} \).

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Key Concepts

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

Intensity
Intensity describes how much power, in watts, is distributed over a specific area, in square meters, when light strikes a surface. It represents the energy flow of an electromagnetic wave per unit surface area.
Light sources vary in brightness, so different distances from a source will have varying intensities. For instance, if the source is a bright lamp emitting light uniformly, the intensity is measured in watts per square meter (W/m虏).
  • Intensity is a crucial factor when calculating radiation pressure.
  • For this particular problem, we have an intensity of 6.00 W/m虏.
Understanding intensity helps us determine how much force the electromagnetic wave can apply over an area.
Absorbing Surface
An absorbing surface takes in all the electromagnetic radiation that hits it without reflecting any. When a surface absorbs the light completely, the energy of the light is transferred to the surface, often causing an increase in thermal energy or temperature. In this scenario:
  • Radiation pressure on an absorbing surface is calculated using the formula:
\[P = \frac{I}{c}\]
  • Where \( I \) is the intensity of the light and \( c \) is the speed of light \( \approx 3 \times 10^8 \) m/s.
  • For a given intensity of 6.00 W/m虏, the pressure works out as \( 2 \times 10^{-8} \) Pa.
This value represents the force applied per unit area on the absorbing surface by the electromagnetic waves.
Reflecting Surface
Reflecting surfaces bounce back all the electromagnetic radiation that strikes them, similar to how a mirror reflects visible light. This reflection results in twice the interaction force compared to an absorbing surface, since the radiation's momentum changes direction upon reflection.
To compute the radiation pressure on such a surface, the formula used is:
  • \[P = \frac{2I}{c}\]
Where \( I \) and \( c \) have the same meanings as before. For a reflecting surface with our given intensity:
  • The pressure calculated is \( 4 \times 10^{-8} \) Pa.
  • This represents the force per unit area exerted.
This model demonstrates how reflective properties amplify the impact of radiation.
Pressure Conversion
Converting pressure from one unit to another makes it comprehensible in different contexts. In many scientific tasks, you convert pressure into atmospheres (atm), as it gives a more recognizable scale.
One atmosphere is equivalent to 101325 Pascals (Pa). To convert the pressure calculated in Pascals to atmospheres, divide by 101325.
  • For an absorbing surface:
  • \[ \frac{2 \times 10^{-8}}{101325} \approx 1.97 \times 10^{-13} \text{ atm} \]
  • For a reflecting surface:
  • \[ \frac{4 \times 10^{-8}}{101325} \approx 3.95 \times 10^{-13} \text{ atm} \]
Understanding this conversion helps in grasping the relative comparison between the tiny pressures exerted by radiation and normal atmospheric pressure.
Electromagnetic Radiation
Electromagnetic radiation encompasses light and other forms of energy waves traveling through space. It includes visible light, radio waves, microwaves, infrared, ultraviolet, X-rays, and gamma rays. This radiation consists of oscillating electric and magnetic fields propagating through a vacuum or a medium. It carries energy, momentum, and can exert pressure upon interacting with surfaces.
  • In this context, electromagnetic radiation affects how we calculate pressure on surfaces.
  • The speed of light \( c \) plays a vital role in determining the radiation pressure.
Understanding electromagnetic radiation allows us to see the broader implications of its interaction with different materials and its ability to impart force.

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Most popular questions from this chapter

\(\bullet\) Two ideal polarizing filters are oriented so that they transmit the maximum amount of light when unpolarized light is shone on them. To what fraction of its maximum value \(I_{0}\) is the intensity of the transmitted light reduced when the second filter is rotated through (a) \(22.5^{\circ},\) (b) \(45.0^{\circ},\) and (c) \(67.5^{\circ} ?\)

\(\bullet$$\bullet\) A ray of light traveling in a block of glass \((n=1.52)\) is incident on the top surface at an angle of \(57.2^{\circ}\) with respect to the normal in the glass. If a layer of oil is placed on the top surface of the glass, the ray is totally reflected. What is the maximum possible index of refraction of the oil?

\(\bullet$$\bullet\) A light beam is directed parallel to the axis of a hollow cylindrical tube. When the tube contains only air, it takes the light 8.72 ns to travel the length of the tube, but when the tube is filled with a transparent jelly, it takes the light 2.04 ns longer to travel its length. What is the refractive index of this jelly?

\(\bullet$$\bullet\) Radiation falling on a perfectly reflecting surface produces an average pressure \(p .\) If radiation of the same intensity falls on a perfectly absorbing surface and is spread over twice the area, what is the pressure at that surface in terms of \(p ?\)

\(\bullet\) Visible light. The wavelength of visible light ranges from 400 nm to 700 nm. Find the corresponding ranges of this light's (a) frequency, (b) angular frequency, (c) wave number.
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