G

2 J Ilk ) dk + J Ilk ) explikx ) dk + J Ilk ) expl-ikx ) dk

If we let

JIlk)dk

be the total power, and we define I(-k) = I(k), this simplifies to

the waves as E0 exp[i(kx — wt)], where k = 2v/\ and w = 2-nv, and E0 is the electric field amplitude of the wave. So, the two waves that will be recom-bined can be written as

Taking the total electric field, E = Ea + E2, and the intensity, I = EE*, we can write (see Problem 4.34)

The integral in this expression is the Fourier transform of I(x). So by measuring I(x), we are also measuring the Fourier transform . This means that we can find I(k), power as a function of wavelength, from the inverse Fourier transfrom of I(x), which is q f I(x) exp [ikx ] dk

In a real measurement, we don't measure I(x) for all values of x. There are two limitations. One is the total range over which we move the mirror. This limits our ability to do the integral from minus to plus infinity. The other is that we can only move the mirror in finite steps. This means that we only measure I(x) at those positions, so this limits our ability to approximate the inverse transform integral as a sum. The closer together we measure I(x), the shorter wavelengths (higher frequencies) we are sensitive to. The greater the largest value of x at which we measure I(x), the more information we have on the longer wavelengths (lower frequencies), so this sets the limit in the frequency resolution in the computed spectrum.

0 0

Post a comment