Thermography QOD 10-13-2008

When striking an object, IR energy is similar to visible light with some exceptions.


…reflected and absorbed by materials that are in its path.

Kevin, you edited your question.

The more highly polished some surfaces are, the more IR energy the surface will reflect.

The question has not changed. I just merely moved the "IR Radiation can be:" to from the body of the post to up in the poll area.


OK, I just jumped the gun a bit early.

Thanks for your time in coming up with the QOD.

I’m surprised that no one answered this one correctly, but there were only 13 people who answered.

Anyway, the answer is **“Reflected, Transmitted, Emitted”

**Visible light can be “absorbed”, while IR radiation can be “emitted”

E + R + T = 1.0

I’ll have another on on Monday.


How does your answer line up with this:

**Infrared Interactions **

The quantum energy of infrared photons is in the range 0.001 to 1.7 eV which is in the range of energies separating the quantum states of molecular vibrations. Infrared is absorbed more strongly than microwaves, but less strongly than visible light. The result of infrared absorption is heating of the tissue since it increases molecular vibrational activity. Infrared radiation does penetrate the skin further than visible light and can thus be used for photographic imaging of subcutaneous blood vessels

Well, I would say it doesn’t line up with that statement. I’m no Physicist, but my question is basic IR Theory. More importantly, the Thermography QOD is designed for “Building applications” and home inspectors.

“**The result of infrared absorption is heating of the tissue since it increases molecular vibrational activity.”

**The above statement is outside the scope of my knowledge. However, I do know that knowing the correct answer to a question based on ones training and experience verses looking up a page of information on the Internet can be quite different. I think you are comparing apples to oranges, IMHO.

Emittance is how well an object radiates IR energy when compared to a blackbody at the same wavelength and temperature. For most applications, Transmittance is usually zero, so Reflectance and Emittance are variables.

Absorbtion is not a factor in determining E R and T.


Not trying to argue just understand.

I think the reason many chose absorbed is because we all feel the IR radiation of the sun as it is absorbed by our skin.

Can you provide a source of information that supports your question?

I think the question was not very clear.


IR radiation can also be “Absorbed” especially on black body objects. If a material absorbs all visible light (dull black surface), it is likely to absorb some of the infrared range as well.

The way I understand it is:

Emissivity + Reflectivity + Transmissivity = 1.0

and E = emitter and/or absorber

Correct: That which is absorbed is emitted. If it passes through or is reflected it can not be emitted. This is why T-Reflect options (which are not available on all cameras) is necessary to determine true temp vs. apparent temps.

The object always wants to go back to a state of rest (so to speak). As you add more energy it absorbs and then emits more.

David, not to nit pick but I would use the term gray body as a true black body does not exist. Some T or R will always occur.

Keep up the good work. It stimulates the thought process.

Thanks for everyone’s responses!!

The whole premise to this question was to compare visible light to Infrared in order to get you thinking about the differences.

I think what we all are saying is somewhat correct, but the best answer to my question is what I have already stated.

An infrared camera uses colors to represent the thermal world it sees, just like we see colors visibly.

The BIG difference is colors in an Infrared image express both “reflection” and “Emission.” When we are looking through our cameras at an object, “absorption” is not a factor.

This is directly form Level I Infraspection and FLIR/ITC Training manuals.


Black body is used in all the training manuals, so why would I go ahead and change it to your standards?


It is not my standard and I’m not asking you to change. Maybe it was very late at night but your question was not very clear.

In my training:
Blackbody-a perfect absorber and emitter of radiation (e = 1). This is usually referenced during calibration and camera adjustment for emissivity.

Gray body-which also can be referred to as “real body” (objects with e<1). This references objects in the field that are not true black bodies.

These terms are simply used to differentiate between the emissivity of the object in the field in reference to a true blackbody (Which does not exist).

The only reason I’m bringing this up is to clarify the questions you’re asking. It may not confuse everyone but it is confusing when you discuss a blackbody that is not a blackbody. That’s all.

For some reason 99.9% of the people answered your question incorrectly. That is the only reason I’m pointing this out.
It’s not about your question or your training[FONT=Tahoma], it’s about the phrasing of the question and your expected answer.

As standardization is the issue at hand, the use of these terms should probably be standardized to prevent misunderstanding. [size=2]But they are obviously not[FONT=Tahoma]. [size=2]Larry [FONT=Tahoma]Kage and I sat next to one another in one class and probably perceive things a little differently than those from other programs. We need to bridge the gap so there is understanding about what a “real body” is if the terminology is encountered.


I get your point and it is a valid one. I think what the other David was trying to communicate is that there is really no perfect emitter. All real world objects have “E” values of less than 1.0.



Please read the thread. It wasn’t David Valley who asked the original question, it was me.



Sorry for the topic change. But, I have been meaning to ask you a question. I noticed in your signature, you have the title “[/size][/size][/FONT][/FONT][/size][/FONT]ITC Certified Infrared Thermologist Cert#1958” listed.

Does ITC offer certification in Thermology?

Thermology is the medical science that derives diagnostic indications from highly detailed and sensitive infrared images of the human body.

Do you do that type of work?

Please don’t take offense to my question, just asking for my own curiosity.