Problem 46 - University Physics with Modern Physics

Question

The plane of a flat surface is perpendicular to the propagation direction of an electromagnetic wave of intensity \(1 .\) The surface absorbs a fraction \(w\) of the incident intensity, where \(0 \leq w \leq 1,\) and reflects the rest. (a) Show that the radiation pressure on the surface equals \((2-w) I / c .\) (b) Show that this expression gives the correct results for a surface that is (i) totally absorbing and (ii) totally reflective. (c) For an incident intensity of \(1.40 \mathrm{kW} / \mathrm{m}^{2},\) what is the radiation pressure for 90\(\%\) absorption? For 90\(\%\) reflection?

Step-by-Step Solution

Answer

Radiation pressure is \( \frac{(2-w)I}{c} \). Pressure is \(5.6 \times 10^{-6} \, \text{Pa}\) for 90% absorption and \(8.4 \times 10^{-6} \, \text{Pa}\) for 90% reflection.

Step 1: Understand Radiation Pressure

Radiation pressure is the pressure exerted by electromagnetic radiation on a surface. When radiation is totally absorbed, the pressure exerted is given by \( \frac{I}{c} \). If radiation is totally reflected, the pressure is \( \frac{2I}{c} \).

Step 2: Express Intensity in Terms of Absorption and Reflection

Let \( I \) be the intensity of the incident wave, and \( w \) be the fractional absorption by the surface. Then, the absorbed intensity is \( wI \), and the reflected intensity is \((1-w)I \).

Step 3: Compute Pressure from Absorbed and Reflected Components

For the absorbed component, radiation pressure is \( \frac{wI}{c} \). For the reflected component, the effective pressure is \( \frac{2(1-w)I}{c} \) since reflection doubles the momentum change. Therefore, the total radiation pressure is \( \frac{wI}{c} + \frac{2(1-w)I}{c} \) which simplifies to \( \frac{(2-w)I}{c} \).

Step 4: Verify Total Absorption and Reflection

For total absorption \((w = 1)\), the expression becomes \( \frac{(2-1)I}{c} = \frac{I}{c} \), which is correct for total absorption. For total reflection \((w = 0)\), the expression becomes \( \frac{(2-0)I}{c} = \frac{2I}{c} \), which corresponds to total reflection.

Step 5: Calculate Radiation Pressure for Given Intensities and Absorptions

For an intensity of \(1.40 \ \text{kW/m}^2\) and \(w = 0.9\) absorption, the radiation pressure is \( \frac{(2-0.9)(1.40 \times 10^3)}{3 \times 10^8} = 5.6 \times 10^{-6} \, \text{Pa} \). For \(w = 0.1\) absorption (90% reflection), the pressure is \( \frac{(2-0.1)(1.40 \times 10^3)}{3 \times 10^8} = 8.4 \times 10^{-6} \, \text{Pa} \).

Key Concepts

Intensity Absorption

Intensity absorption refers to how much of the electromagnetic wave's energy is absorbed by a particular surface. When an electromagnetic wave hits a surface, some of the energy gets absorbed, and the amount absorbed depends on the nature of the surface. In the given problem, the absorption fraction is denoted by \( w \), where \( 0 \leq w \leq 1 \).
Here, \( w \) represents the portion of the incident radiation intensity \( I \) that gets absorbed by the surface. For example, if \( w = 0.9 \), it means that 90% of the incident intensity is absorbed. The absorbed intensity can be calculated using the formula \( wI \).

• If a surface is totally absorbing, \( w = 1 \), meaning all the incident intensity is absorbed, and the absorbed intensity is equal to \( I \).
• If a surface is totally reflective, \( w = 0 \), meaning no intensity is absorbed, and the absorbed intensity is zero.

Understanding intensity absorption helps in assessing how much energy remains with the surface after an electromagnetic wave interaction.

Electromagnetic Wave

An electromagnetic wave is a type of wave composed of oscillating electric and magnetic fields that propagate through space. These waves travel at the speed of light \( c \) and carry energy and momentum.
In the context of radiation pressure, electromagnetic waves exert pressure on surfaces they strike. This pressure depends on whether the wave is absorbed or reflected by the surface. Such interactions are common and can be observed with systems that interact with light, such as solar panels or radiation detectors.
Electromagnetic waves are characterized by their wavelength and frequency, and they cover a spectrum ranging from radio waves to gamma rays. The energy carried by these waves is what causes momentum transfer, resulting in radiation pressure. In studying such waves, one often evaluates factors such as intensity (energy per unit area per unit time) and how surfaces modify this when interacting with different media.

Reflective Surfaces

Reflective surfaces are materials that return a portion of the incoming electromagnetic waves instead of absorbing them. When a wave is reflected, it "bounces off" the surface and goes back into the medium from which it came.
In our problem, reflection is measured by the amount \((1-w) \) indicating the fraction of the incident intensity that is not absorbed and instead reflected by the surface. Reflective surfaces are crucial in determining the radiation pressure because they cause a change in momentum that affects the pressure's magnitude.
The radiation pressure due to reflection is given by \( \frac{2(1-w)I}{c} \), indicating that the pressure doubles in the context of reflection. This doubling occurs due to the change in direction of momentum as the wave interacts with the surface. Surfaces that are highly reflective (such as mirrors or shiny metals) can significantly impact radiation pressure readings in optical and physical systems.

Absorption Coefficient

The absorption coefficient is a measure of how much of an incoming electromagnetic wave is absorbed by a medium per unit distance. In the context of our exercise, it’s more general than the fraction \( w \) used for a flat surface, but it relates to how \( w \) impacts the material's interactions with radiation.
Absorption coefficients are key in fields like optics and materials science for designing devices that either maximize absorption (like solar cells) or minimize it (like protective coatings or reflective materials). It reflects properties of the medium, such as thickness, density, and molecular composition.
By using the absorption fraction \(w\), we assume a simplified integration of such coefficients over the path of light through a given medium. Ultimately, being aware of the absorption coefficient allows engineers and scientists to predict or control how much radiation is retained in a system.