...them
In the case of two slits, the application of Rayleigh's criterion that the maximum of one peak should lie on or below the first minimum of the other gives a dip in intensity between the peaks to 8/$\pi^{2}_{}$ of the maximum. This follows because the separate intensity patterns vary as (sin($\beta$)/$\beta$)2 where $\beta$ = (kd /2sin($\theta$)) , and if we set one peak at the point where $\beta$ = $\pi$ for the other the dip in intensity corresponds to $\beta$ = $\pi$/2 for each peak, giving a ratio of dip to peak intensity of 2(1/($\pi$/2))2 = 8/$\pi^{2}_{}$ .

If we keep to this criterion, and note that in this case the maximum intensity is the central intensity of one peak plus the intensity of the other peak at the same point, for peaks to be resolved we require two values of $\delta$ , $\Delta$$\delta$ apart, which satisfy

$\displaystyle{8 \over \pi^2}$ $\textstyle\le$ $\displaystyle{1+{1 \over 1 + F \sin^2 (\delta_1+\Delta\delta
/2)}}$   
  = $\displaystyle{1 \over 1 + F \sin^2 ((\delta_1+\Delta\delta/2) /2)}$ + $\displaystyle{1 \over 1 + F \sin^2 ((\delta_2-\Delta\delta/2) /2)}$.   
But we know that at the peaks $\delta_{1}^{}$ and $\delta_{2}^{}$ are each equal to a multiple of 2$\pi$ , so we have

$\displaystyle{8 \over \pi^2}$$\displaystyle{1+{1 \over 1 + F \sin^2 (\Delta\delta /2)}}$ = $\displaystyle{2 \over 1 + F \sin^2 (\Delta\delta /4)}$,

giving, for small $\Delta$$\delta$ (expanding sins to first order, obtaining and solving a quadratic in F$\Delta$$\delta$ ),

$\displaystyle\Delta$$\displaystyle\delta$ $\displaystyle\approx$ $\displaystyle{4.2 \over \sqrt{F}}$.

Now, from Equation L29.1,

$\displaystyle\Delta$$\displaystyle\delta$ = - $\displaystyle{2 \pi \Delta \lambda \over \lambda^2}$2dcos($\displaystyle\theta$)

and $\delta$ itself is a large multiple, p , of $\lambda$ , so making the reasonable approximation of neglecting the phase change $\phi$ compared with 2$\pi$p ,

$\displaystyle{\Delta\lambda \over \lambda}$ $\displaystyle\approx$ $\displaystyle{\Delta \delta \over 2 \pi p}$,

which gives a resolving power

$\displaystyle{\lambda \over (\Delta\lambda)_{\rm min}}$ = $\displaystyle{4 \pi d \over 4.2 \lambda}$$\displaystyle\sqrt{F}$,

since p$\lambda$ $\approx$ 2d.
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...power
This differs very little from the result deduced from our interpretation of the Rayleigh criterion.
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Tony Harker
3/18/2002