Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Spectra, Temperature, and Energy
Name: Wayne
Status: other
Grade: 4-5
Location: N/A
Country: N/A
Date: 1/24/2005

My daughter asked me a question regarding a possible science fair project she might like to attempt and I find myself without an answer.

The project of interest is to recreate Sir William Herschel's Infrared experiment of the year 1800.

Essentially, Herschel divided the sun's light, using a prism, into the spectrum of visible light so he could then measure the temperature of each specific region of colour. He noticed that the temperature progressively increased from the "Blue" colour to the "Red" colour. A further observation was that the region just beyond the visible "Red" colour was yet the hottest area. Herschel discovered Infrared. My understanding of the "Electromagnetic spectrum" is that as we move from the longer wavelengths (low frequencies such as radio waves) through Infrared, through visible light, through UV to the shorter wavelengths (higher frequencies such as X-rays) we find an increase in energy. Her question; Why does the Herschel experiment find the Infrared region of the spectrum, with its longer wavelengths, to be at a higher temperature (energy) than any visible colour of light which has shorter wavelengths? Remember, Infrared comes before visible colour as we move from long to short wavelengths. (or low to high frequencies).

You are correct that the energy per photon of electromagnetic radiation is proportional to the frequency: (E = h x f ) and that the frequency and wavelength inversely proportional ( l = c / f ) where 'E' is the energy per photon, 'h' is Planck's constant, 'f' is the frequency, 'l' is the wavelength, and 'c' is the speed of light. But that is only one factor that enters into Herschel's experiment. Let us trace the light path from the Sun to the temperature measuring device (i.e. the temperature measuring device may be a regular thermometer or it could be some solid state probe such as a thermocouple).

The intensity of light that leaves the Sun depends upon the wavelength of the light. While it is approximately constant above the atmosphere a substantial portion of the ultraviolet, fortunately, is absorbed in the upper atmosphere. The atmosphere is essentially transparent in the visible region, and in the infrared at wavelengths just longer than the visible red part of the solar output. You can find a comparison of the solar output vs. the radiation at sea level at the web site:

among others -- search Google for "solar spectrum" or "solar radiation" and/or "solar radiation at sea level". For reference the visible range is about 400 to 700 nanometers. So a lot of the ultraviolet radiation never gets to the earth's surface.

The prism used to disperse the incoming radiation may (or may not) absorb the wavelengths of the incident light depending upon the material of construction. Common glass absorbs most of the ultraviolet and infrared. The absorption spectrum of glass you can search for because it depends upon the type of glass. One site is:

(Caution: on this site the units are wave numbers ( cm^-1) which is commonly used in the infrared)

Last but not least, the "thermometer" (I use quotation marks because the "thermometer" could be a variety of detectors, not just a common thermometer.) must respond to the radiation that hits it. So, for example, if you used a blackened target you would find that the temperature of the thermometer would respond to visible light because the visible light is absorbed by the blackened target and converted to heat (infrared radiation), as well as the incident infrared and some of the ultraviolet radiation as well.

To summarize: Objects in the light path that absorb radiation will not allow that radiation to get to the thermometer. Objects that reflect radiation do not convert the radiation into heat (which is equivalent to infrared radiation). Radiation not absorbed by the thermometer will not be recorded as an increase in temperature.

Vince Calder

As you move into higher frequencies, the energy in the wave form does increase.

However, just because there is a higher energy content does not mean more energy can be transferred as heat. Through the visible spectrum, for example, there are a number of wavelengths which are very good for exciting solar cells and producing electricity. As you move even further up the scale, you will come to X-rays, which are so high energy they pass directly through most softobjects. (little energy being transferred) Even higher frequency than that are microwaves. A microwave oven cooks the food by exciting one specific molecule which is very common, water. So the short answer would be that Infra-red can be largely absorbed as heat, while visible light tends to just bounce off.

Ryan Belscamper

Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (, or at Argonne's Educational Programs

Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory