Ask Dr. SETI ®

Colorblind in Kansas

The Doctor Responds:
Your question is not at all trivial, Colorblind. It touches on a basic principle which was discussed in a previous column, that of finding the radiation peak of a Planck blackbody curve. So, let's run the numbers.

We start by considering the Sun a 5780 Kelvin thermal blackbody. From Wien's Law, we compute:

Lambda (peak) = A / T_eq

where A is Wien's Constant, 2.898 x 10^-3 meters times Kelvins. Solving for peak wavelength:

Lambda (peak) = (2.898 x 10^-3 m ^. K) / (5780 K)

Note that the Kelvins in the numerator cancels the Kelvins in the denominator, so wavelength thus comes out simply in meters.

Solving the above equation, we find that the peak wavelength of the Sun's thermal emission is on the order of 5.01 x 10^-7 m, or 501 nm. Converting to frequency, we divide into the speed of light (3 x 10⁸ m/s), for a result of 599 THz.

Now, the optical spectrum extends from red, a wavelength of 750 nm (frequency = 400 THz) all the way to violet, at a wavelength of 400 nm (frequency = 750 THz). Note the interesting symmetry there. In any case, our sunlight peak occurs about 29% up the optical spectrum from the red end, for an equivalent color of ... yellow!

So, in fact, solar radiation peaks just where our Sun-adapted eyes say it does. This is no surprise, since our eyes evolved to see by sunlight. In a few billion years, when our Sun has cooled and expanded into a red giant, our eyes will need to adapt to this longer wavelength. Of course, by then, the frequency response of our eyes will be the least of our worries.