Maxwell's Rainbow — Question 2

Question 2

The center of a logarithmic spectral interval is a geometric mean

A detector covers vacuum frequencies $f_1=4.0\times10^{14}$ Hz to $f_2=9.0\times10^{14}$ Hz. Define its logarithmic midpoint by equal frequency ratios on either side. Use $c=3.00\times10^8$ m/s.

1. Find the midpoint frequency and corresponding wavelength.

2. Compare that wavelength with the arithmetic mean of the endpoint wavelengths and explain the difference.

Question 2 - Solution

The center of a logarithmic spectral interval is a geometric mean

(a) Method and calculation

$$f_m/f_1=f_2/f_m\Rightarrow f_m=\sqrt{f_1f_2}=\boxed{6.0\times10^{14}\,\mathrm{Hz}}.$$$$\boxed{\lambda_m=c/f_m=500\,\mathrm{nm}}.$$

(b) Interpretation and further calculation

$$\lambda_1=750\,\mathrm{nm},\quad\lambda_2=333.333\,\mathrm{nm}.$$$$\boxed{(\lambda_1+\lambda_2)/2=541.667\,\mathrm{nm}\ne\lambda_m}.$$

Physical check

Reciprocal conversion preserves a geometric midpoint, not an arithmetic midpoint.