## Friday, April 20, 2012

### Frequency versus Wavelength The Optics of Life: A Biologist's Guide to Light in Nature, by Sonke Johnsen.
I am currently reading The Optics of Life: A Biologist’s Guide to Light in Nature, by Sonke Johnsen. I hope to have more to say about this fascinating book when I finish it, but today I want to consider a point made in Chapter 2 (Units and Geometry), which addresses the tricky issue of measuring light intensity as a function of either frequency or wavelength. Johnsen favors using wavelength whenever possible.
However, one critical issue must be discussed before we put frequency away for good. It involves the fact that light spectra are histograms. Suppose you measure the spectrum of daylight, and that the value at 500 nm is 15 photons/cm2/s/nm. That doesn’t mean that there are 15 photons/cm2/s with a wavelength of exactly 500 nm. Instead, it means that, over a 1-nm-wide interval centered on a wavelength of 500 nm, you have 15 photons/cm2/s. The bins in a spectrum don’t have to be 1 nm wide, but they all must have the same width.

Let’s suppose all the bins are 1 nm wide and centered on whole numbers (i.e., one at 400 nm, one at 401 nm, etc.). What happens if we convert these wavelength values to their frequency counterparts? Let’s pick the wavelengths of two neighboring bins and call them λ1 and λ2. The corresponding frequencies ν1 and ν2 are equal to c/λ1 and c/λ2, where c is the speed of light. We know that λ1−λ2 equals 1 nm, but what does ν1−ν2 equal?
ν1−ν2 = … = −c/λ12
…So the width of the frequency bins depends on the wavelengths they correspond to, which means they won’t be equal! In fact, they are quite unequal. Bins at the red end of the spectrum (700 nm) are only about one-third as wide as bins at the blue end (400 nm). This means that a spectrum generated using bins with equal frequency intervals would look different from one with equal wavelength intervals. So which one is correct? Neither or both. The take-home message is that the shape of a spectrum depends on whether you have equal frequency bins or equal wavelength bins.
Johnsen goes on to note that the wavelength at which the spectrum is maximum depends on if you use equal frequency or equal wavelength bins. It does not make sense to say that the spectrum of, say, sunlight peaks at a particular wavelength, unless you specify the type of spectrum you are using. Furthermore, you cannot unambiguously say light is “white” (a uniform spectrum). White light using equal wavelength bins is not white using equal frequency bins. Fortunately, if you integrate the spectrum, you get the same value regardless of if you express it in terms of wavelength or frequency.

Russ Hobbie and I discuss this issue in Chapter 14 (Atoms and Light) of the 4th edition of Intermediate Physics for Medicine and Biology.
Early measurements of the radiation function were done with equipment that made measurements vs. wavelength. It is also possible to measure vs. frequency. To rewrite the radiation function in terms of frequency, let λ1 and λ2 =  λ1 + dλ be two slightly different wavelengths, with power Wλ(λ, T) dλ emitted per unit surface area at wavelengths between λ1 and λ2. The same power must be emitted between frequencies ν1 = c1 and ν2 = c2:

Wν(ν,T) dν = Wλ(λ,T) dλ .     (14.35)

Since ν = c/λ, dν/dλ = − c2, and

|dν| = + c λ2 |dλ| .                 (14.36)

... This transformation is shown in Fig. 14.24. The amount of power per unit area radiated in the 0.5 μm interval between two of the vertical lines in the graph on the lower right is the area under the curve of Wλ between these lines. The graph on the upper right transforms to the corresponding frequency interval. The radiated power, which is the area under the Wν curve between the corresponding frequency lines on the upper left, is the same. We will see this same transformation again when we deal with x rays. Note that the peaks of the two curves are at different frequencies or wavelengths.
Students who prefer visual explanations should see Fig. 14.24, which Russ drew. It is one of my favorite pictures in our book, and provides an illuminating comparison of the two spectra.

One detail I should mention: why in Eq. 14.36 do we use absolute values to eliminate the minus sign introduced by the derivative dν/dλ? Typically, when you integrate a spectrum, you start from the lower frequency and go to the higher frequency (say, zero to infinity), and you start from the shorter wavelength and go to the longer wavelength (again, zero to infinity). However, zero frequency corresponds to an infinite wavelength, and an infinite frequency corresponds to zero wavelength. So, really one case should be integrated forward (zero to infinity) and the other backwards (infinity to zero). If we keep the convention of always integrating from zero to infinity in both cases, we introduce an extra minus sign, which cancels the minus sign introduced by dν/dλ.

Sometimes it helps to have an elementary example to illustrate these ideas. Therefore, I have developed a new homework problem that introduces an extremely simple spectrum for which you can do the math fairly easily, thereby allowing you to focus on the physical interpretation. Enjoy.
Section 14.7

Problem 23 ½ Let Wν(ν) = A ν (νο - ν) for ν less than νο, and Wν(ν) = 0 otherwise.
(a) Plot Wν(ν) versus ν.
(b) Calculate the frequency corresponding to the maximum of Wν(ν), called νmax.
(c) Let λο = c/νο and λmax = c/νmax. Write λmax in terms of λο.
(d) Integrate Wν(ν) over all ν to find Wtot.
(e) Use Eqs. 14.35 and 14.36 to calculate Wλ(λ).
(f) Plot Wλ(λ) versus λ.
(g) Calculate the wavelength corresponding to the maximum of Wλ(λ), called λ*max, in terms of λο.
(h) Compare λmax and λ*max. Are they the same or different? If λο is 400 nm, calculate λmax and λ*max? What part of the electromagnetic spectrum is each of these in?

(i) Integrate Wλ(λ) over all λ to find W*tot. Compare Wtot and W*tot. Are they the same or different?