# Average Pressure Radiation on Perfectly Absorbing Surface

• rugerts
In summary: The question discussed at that link considers two perfectly reflecting surfaces. The question in post #1 of this thread compares a reflecting surface with an absorbing one.In summary, a light source emitting a sinusoidal electromagnetic wave with average pressure p will exert a pressure radiation of (2 * p)/(speed of light) on a perfectly reflecting surface at a distance R, and a pressure radiation of p/(2 * speed of light) on a perfectly absorbing surface at a distance 2R. Using the equations for intensity, surface area of a sphere, and the factor of 1/2 from absorption, the ratio of pressure radiation on the absorbing surface to that on the reflecting surface is 1/4, not 1/8 as previously thought
rugerts

## Homework Statement

A light source radiates a sinusoidal electromagnetic wave uniformly in all directions. This wave exerts an average pressure p on a perfectly reflecting surface a distance R away from it. What average pressure (in terms of p) would this wave exert on a perfectly absorbing surface that was twice as far from the source?

## Homework Equations

pressure radiation of perfect absorber = (Intensity)/(speed of light)

pressure radiation of perfect reflector = (2 * Intensity)/(speed of light)

Intensity = (Power)/(Area)

Surface Area of Sphere = 4*pi*R^2

## The Attempt at a Solution

Since you probably can't see that, here's an imgur link: https://imgur.com/a/uxltWGV

When I try to do the math, I get that it's a factor of 1/8. But, if I try to reason through it, it seems like 1/4 is a reasonable answer since I'm just doubling a square factor.

Thanks for your time.

#### Attachments

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rugerts said:
since I'm just doubling a square factor.
You are doubling the distance factor, which is squared, but you are also getting afactor of 1/2 from the absorption. Where's the puzzle?

haruspex said:
You are doubling the distance factor, which is squared, but you are also getting afactor of 1/2 from the absorption. Where's the puzzle?
The puzzle is that I get an answer of 1/8 and an answer I found online is 1/4.

rugerts said:
The puzzle is that I get an answer of 1/8 and an answer I found online is 1/4.

haruspex said:

haruspex said:
?

rugerts said:
The question discussed at that link considers two perfectly reflecting surfaces. The question in post #1 of this thread compares a reflecting surface with an absorbing one.

rugerts

## 1. What is average pressure radiation on a perfectly absorbing surface?

Average pressure radiation on a perfectly absorbing surface is the amount of force per unit area exerted by electromagnetic radiation on a surface that absorbs all incident radiation. It is a measure of the average intensity of the radiation on the surface.

## 2. How is average pressure radiation calculated?

The average pressure radiation is calculated by taking the integral of the intensity of the incident radiation over the surface, divided by the speed of light. This can also be represented as the energy flux density divided by the speed of light.

## 3. What factors affect the average pressure radiation on a perfectly absorbing surface?

The average pressure radiation is affected by the intensity and wavelength of the incident radiation, as well as the properties of the surface such as its reflectivity and emissivity. The distance between the radiation source and the surface also plays a role in determining the average pressure radiation.

## 4. How does average pressure radiation on a perfectly absorbing surface relate to heat transfer?

The average pressure radiation is a form of heat transfer, as electromagnetic radiation carries thermal energy. The amount of average pressure radiation on a surface can affect the rate at which heat is transferred from the radiation source to the surface, and can also contribute to the overall temperature of the surface.

## 5. Why is it important to consider average pressure radiation on a perfectly absorbing surface in scientific research?

Average pressure radiation on a perfectly absorbing surface is an important concept in fields such as thermodynamics, heat transfer, and materials science. It helps us understand how electromagnetic radiation interacts with different surfaces and how heat is transferred between objects. This knowledge is crucial in many practical applications, such as designing efficient thermal insulation systems and understanding the behavior of materials in high temperature environments.

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