Spectra, Temperature, and Energy
My daughter asked me a question regarding a possible
science fair project she might like to attempt and I find myself without
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
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.
As you move into higher frequencies, the energy in the wave form does
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.
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Update: June 2012