Blackbody Radiation and Emissivity Explained

In summary: IR light. The intensity at each wavelength is the maximum possible because a blackbody is a perfect emitter, meaning it emits the maximum amount of energy at every wavelength for a given temperature. This behavior is due to the fact that in a blackbody, all light that enters is absorbed and re-emitted, resulting in a complete spectrum of all wavelengths of IR light.
  • #1
Wannabeagenius
91
0
Hi All,

I'm reading about blackbody radiation and emissivity and I came upon the following statement

"Imagine looking into a small opening of a deep cave. In the visible wavelengths, the opening looks black because the light that enters the cave is not easily reflected back out. However, the cave glows with emitted thermal IR energy. This energy emerges as a complete spectrum of all wavelengths of IR light. The radiance at each infrared wavelength is the maximum amount possible for a given temperature."

Why does the IR energy that emerges contain the complete spectrum of all wavelengths of IR light and why is the intensity at each wavelength the maximum possible? Why does it behave like a perfect blackbody?

Thanks,
Bob
 
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  • #2
Wannabeagenius said:
Why does the IR energy that emerges contain the complete spectrum of all wavelengths of IR light
Classically it actually contains all wavelengths of all light - it just peaks in the IR

and why is the intensity at each wavelength the maximum possible?
Slightly odd wording - the intensity at each wavelength if it's a blackbody is exactly what the blackbody curve predicts it should be

Why does it behave like a perfect blackbody?
If no outside light is reflected back out without being absorbed by the walls and any light you do see has been emitted by the walls then you get a pretty good blackbody.
 
  • #3
light entering the cavity bounced back and forth till its almost completely absorbed

emission is just the reverse of absorption
 

Related to Blackbody Radiation and Emissivity Explained

1. What is blackbody radiation?

Blackbody radiation is the thermal electromagnetic radiation emitted by an object when it is heated to a certain temperature. It is a continuous spectrum of radiation and its intensity and wavelength distribution depend only on the temperature of the object.

2. How does emissivity affect blackbody radiation?

Emissivity is the measure of an object's ability to emit thermal radiation. It is a value between 0 and 1, where 0 represents a perfect reflector and 1 represents a perfect emitter. The higher the emissivity, the more thermal radiation an object will emit at a given temperature.

3. What is the relationship between temperature and blackbody radiation?

The intensity of blackbody radiation increases with temperature. This is described by Planck's law, which states that the energy of a photon emitted by a blackbody is directly proportional to the temperature of the object.

4. How is blackbody radiation used in scientific research?

Blackbody radiation is used in various fields of research, such as astrophysics, thermodynamics, and materials science. It helps scientists understand the properties of different materials, study the temperature and composition of stars and planets, and develop new technologies such as infrared cameras and thermal imaging devices.

5. Can emissivity be changed or controlled?

Yes, the emissivity of an object can be altered by changing its surface properties, such as roughness or reflectivity. This can be achieved through various methods, including coating the object with a material with a desired emissivity, or manipulating the surface through polishing or texturing.

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