## Abstract

Optical spectra are typically normalized per unit wavelength or per
unit photon energy, yielding two different expressions or
curves. It is advantageous instead to normalize a spectrum to a
constant fractional bandwidth, providing a unique expression
independent of whether the bandwidth is in dimensions of wavelength or
of photon energy. For the Sun, whereas a per-unit-wavelength
spectrum peaks in the green and a per-unit-photon-energy spectrum peaks
in the IR, when the proposed normalization is used, the output peaks in
the red. This approach applies to any spectral source and provides
curves of constant spectral resolving power, as produced by many
spectrometers.

© 2001 Optical Society of America

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### Equations (9)

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(1)
$$\frac{\mathrm{d}{J}_{\mathrm{\omega}}}{\hslash \mathrm{d}\mathrm{\omega}}=\frac{1}{4{\mathrm{\pi}}^{2}{c}^{2}}\frac{{\mathrm{\omega}}^{3}}{exp\left(\hslash \mathrm{\omega}/\mathit{kT}\right)-1}=2\mathrm{\pi}c\frac{1}{{\mathrm{\lambda}}^{3}\left[exp\left(2\mathrm{\pi}\hslash c/\mathit{kT}\mathrm{\lambda}\right)-1\right]},$$
(2)
$$\frac{\hslash \mathrm{\omega}}{\mathit{kT}}=\frac{2\mathrm{\pi}\hslash c}{\mathit{kT}\mathrm{\lambda}}=2.822.$$
(3)
$$\frac{\mathrm{d}{J}_{\mathrm{\lambda}}}{\mathrm{d}\mathrm{\lambda}}=4{\mathrm{\pi}}^{2}\hslash {c}^{2}\frac{1}{{\mathrm{\lambda}}^{5}\left[exp\left(2\mathrm{\pi}\hslash c/\mathit{kT}\mathrm{\lambda}\right)-1\right]}=\frac{\hslash}{8{\mathrm{\pi}}^{3}{c}^{3}}\frac{{\mathrm{\omega}}^{5}}{exp\left(\hslash \mathrm{\omega}/\mathit{kT}\right)-1}.$$
(4)
$$\frac{2\mathrm{\pi}\hslash c}{\mathit{kT}\mathrm{\lambda}}=\frac{\hslash \mathrm{\omega}}{\mathit{kT}}=4.965.$$
(5)
$$J\prime =\frac{\hslash}{4{\mathrm{\pi}}^{2}{c}^{2}}\frac{{\mathrm{\omega}}^{4}}{exp\left(\hslash \mathrm{\omega}/\mathit{kT}\right)-1}=4{\mathrm{\pi}}^{2}\hslash {c}^{2}\frac{1}{{\mathrm{\lambda}}^{4}\left[exp\left(2\mathrm{\pi}\hslash c/\mathit{kT}\mathrm{\lambda}\right)-1\right]}.$$
(6)
$$\frac{2\mathrm{\pi}\hslash c}{\mathit{kT}\mathrm{\lambda}}=\frac{\hslash \mathrm{\omega}}{\mathit{kT}}=3.921.$$
(7)
$$\mathrm{\Delta}\mathrm{\omega}/\mathrm{\omega}=\mathrm{\Delta}\mathrm{\lambda}/\mathrm{\lambda}.$$
(8)
$$\mathrm{\lambda}/\mathrm{\Delta}\mathrm{\lambda}=|m|N,$$
(9)
$$\frac{\mathrm{\lambda}}{\mathrm{\Delta}\mathrm{\lambda}}=t\left|\frac{\mathrm{d}n}{\mathrm{d}\mathrm{\lambda}}\right|,$$