Why does William Herschel's experiment show red light as warmer than blue if blue light is higher energy?

Here is an explanation of Herschel's experiment http://www.ipac.caltech.edu/outreach/Edu/Herschel/backyard.html

In short, Herschel placed multiple thermometers on the light separated by a prism. This led to the discovery of infrared light as he placed his control thermometer outside of the visible spectrum on the red side.

You may see similar apparatus to his here:

enter image description here

As you can see, the red thermometer is reading a higher temperature.

My guess is that blue is scattered more, therefor the thermometer for red absorbs more light.

  • 4
    $\begingroup$ I think you should make your post self-contained, i.e. explain the experimental setup you're considering so that we don't have to follow a link, just to be able to understand your question. $\endgroup$
    – Danu
    Jan 9, 2016 at 21:02
  • 1
    $\begingroup$ See physics.stackexchange.com/questions/59456/…. $\endgroup$
    – HDE 226868
    Jan 9, 2016 at 21:05

3 Answers 3


The main reason is due to the fact that the prism refracts light in such a way that the "blue" part is more spread than "red" part. So that overall the energy hitting the thermometer is greater in the infrared and red part than on the blue part of the spectrum.

Edit: I have just seen your edit. You're right. There you can see the details.

Here's a quote from the website:

The answer turns out to be the experimental design, and a failure to correct for refraction. In Herschel's setup, sunlight is refracted by a prism. The index of refraction of course must vary with wavelength so that the sunlight would be dispersed into its various colors. If the index of refraction varied linearly with wavelength, Herschel would not have needed to correct for that variation, since the wavelengths would be uniformly spaced along his table. However, since the index of refraction varies non-linearly with wavelength, the wavelengths will not be uniformly spaced along Herschel's measuring table. The actual spacing of the wavelengths versus distance along his table for an incidence angle of 45° from air into glass shows that the infrared region is much more highly concentrated than optical wavelengths. (The plot shows the spacing along the spectrum divided by the distance from the prism. Hence to get the actual spacing in cm or inches, multiply by the distance from the prism in cm or inches.) The relative concentration factor is shown normalized to 0.60 micron. The net result is that Herschel's observed "temperature" should then peak in the infrared.

The website also provides some graphs and in one you can see that the red part (~650 nm) of the spectrum is about 3 times more concentrated than the blue one (~450 nm), while the infrared part is even more concentrated.


Each individual blue photon is higher energy than a red photon, but the distribution of intensity in sunlight (or any source) is not equal for all wavelengths. "This light source emits in blue wavelengths and also in red wavelengths" is not enough to conclude that the total energy delivered by the blue will be more than that by the red.

  • $\begingroup$ typo error at the end ? not enough to conclude that the individual energy delivered by the blue will be less than that by the red. $\endgroup$
    – user46925
    Jan 13, 2016 at 20:17

It also depends on the absorptivity=emissivity spectrum of the thermometer . It may be more reflective in the UV .


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