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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 \(p ?\)

Short Answer

Expert verified
The pressure on the absorbing surface is \( \frac{p}{4} \).

Step by step solution

01

Understanding Radiation Pressure on Reflecting Surface

When radiation falls on a perfectly reflecting surface, the radiation pressure is given as \( p = \frac{2I}{c} \), where \( I \) is the intensity of the radiation and \( c \) is the speed of light. This is because the radiation reflects, effectively doubling the change in momentum.
02

Considering Perfect Absorbing Surface

For a perfectly absorbing surface, the radiation pressure is different. The pressure is given by \( \frac{I}{c} \), since absorption results in a single change of momentum. This is half of the pressure compared to a perfectly reflecting surface for the same area.
03

Adjust for Surface Area

The problem states that the absorbing surface has twice the area that the reflecting surface had. When spreading the same intensity over twice the area, the intensity \( I \) per unit area becomes \( \frac{I}{2} \).
04

Calculate Pressure on Absorbing Surface

Substitute the adjusted intensity into the pressure formula for absorbing surfaces: \( p_{\text{absorb}} = \frac{\frac{I}{2}}{c} = \frac{I}{2c} \).
05

Express Absorbing Pressure in Terms of Initial Pressure

Given \( p = \frac{2I}{c} \) from the reflecting surface, rearrange this expression to find \( \frac{I}{c} = \frac{p}{2} \). Substitute into \( p_{\text{absorb}} = \frac{I}{2c} \) to get: \( p_{\text{absorb}} = \frac{p}{4} \).

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Key Concepts

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

Reflecting Surface
When light or any form of radiation falls upon a perfectly reflecting surface, it bounces back rather than being absorbed. This bouncing action involves a change in momentum, which in turn gives rise to radiation pressure. The formula for this pressure is expressed as:
  • \[ p = \frac{2I}{c} \]
where:
  • \( p \) is the radiation pressure,
  • \( I \) is the intensity of the radiation, and
  • \( c \) is the speed of light.
The key here is the doubling of momentum change. Since the radiation reflects, the direction of momentum effectively reverses, which is why we multiply the intensity term by 2. This doubled change in momentum leads to greater pressure exerted on the reflecting surface.
Absorbing Surface
Unlike a reflecting surface, a perfectly absorbing surface does not bounce back the radiation. Instead, it retains the incoming energy, absorbing the radiation entirely. In this scenario, the formula for calculating the radiation pressure becomes:
  • \[ \frac{I}{c} \]
This indicates a single change in momentum as the radiation is absorbed, compared to the double change on a reflective surface. As a result, the pressure is half that of the reflecting surface for the same radiation intensity and area of exposure.When the absorbing surface is larger—specifically, when the area is doubled as in the problem—the intensity per unit area decreases. This adjustment leads to an even smaller pressure exerted on the surface.
Intensity of Radiation
The intensity of radiation, denoted by \( I \), represents the power per unit area carried by the waves. It plays a central role in determining the radiation pressure either on a reflecting or absorbing surface. Intensity is typically measured in watts per square meter (W/m²). Upon facing different scenarios:
  • For a reflecting surface, the intensity directly influences the effect resulting in double momentum change, hence, a greater pressure.
  • For an absorbing surface, when spread over a larger area, the intensity decreases per unit area.
  • This reduced intensity further translates to lower pressure exerted.
Understanding this concept is crucial in calculating how surfaces interact differently under the same radiation conditions.
Momentum Change
Momentum change is a key factor in the functioning of both reflecting and absorbing surfaces concerning radiation. It's the alteration in the motion state of radiation particles when they interact with a surface, which directly influences radiation pressure. For a reflecting surface:
  • The change is pronounced as the radiation bounces back, causing a reversal of motion equal to twice the incoming momentum.
For an absorbing surface:
  • Only the incoming momentum is considered since the particles do not leave the surface, leading to less momentum change compared to the reflecting surface.
Doubling of momentum change occurs due to the directional reversal in reflection, enhancing pressure. Conversely, lesser pressure is generated due to a single momentum change via absorption.

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Most popular questions from this chapter

A sinusoidal electromagnetic wave has an average intensity of \(100 \mathrm{~W} / \mathrm{m}^{2} .\) By what factor would the electric-field amplitude of the wave have to be increased in order for the wave to have an average intensity of \(10,000 \mathrm{~W} / \mathrm{m}^{2} ?\)

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A beam of polarized light passes through a polarizing filter. When the angle between the polarizing axis of the filter and the direction of polarization of the light is \(30^{\circ}\), the intensity of the emerging beam is \(I\). If you instead want the intensity to be \(I / 2,\) what should be the angle between the filter axis and the original direction of polarization of the light?

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