Why did Herschel's infrared thermometer show higher temperature than visible red light?

[b612_l_prince]

Hi there, I have a question about his experiment.

Herschel discovered IR by placing the blacken thermometer next to the red light (out from the prism), since he noticed the TREND OF INCREASING TEMPERATURE proceeded from BLUE (lowest temperature) TO RED (highest temperature). So he wanted to find out what the temperature would be just outside visible red light.

So my question is, we always say when frequency of a light is higher (wavelength is shorter), the energy that light carries is higher. However, in his experiment, Herschel found the highest temperature rise is indeed in Infra Red region... shortest wavelength?!

Is there anyone could give me some clues on this "contradiction"? What mistake have I made in having such a contradicting conclusion?

Thanks a lot.


Charlotte

P.S. The details of the experiment can be found at http://coolcosmos.ipac.caltech.edu/cosmic_classroom/classroom_activities/herschel_bio.html

 

[chroot]

While each individual blue photon contains more energy than each individual red photon, the sun makes a whole lot more red photons than blue photons.

- Warren

 

[b612_l_prince]

thanks warren

thanks Warren for letting me know. I appreciate it. =)

 

[lord_91106]

A more correct answer

The Infrared thermometer is heated the most because the infrared light
is more concentrated onto it. The IR photons are spatially more concentrated by the prism.
Note - Herschel's prism cut-off longward of 2 microns so the the near-IR from 1-2 um is doing the heating.

See
http://home.znet.com/schester/calculations/herschel/index.html