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Next: Two-Dimensional Fourier Optics Up: Wave Optics Previous: Single-Slit Diffraction

Multi-Slit Diffraction

Suppose that the opaque screen in our interference/diffraction apparatus contains $ N$ identical, equally spaced, parallel slits of finite width. Let the slit spacing be $ d$ , and the slit width $ \delta$ , where $ \delta < d$ . It follows that the aperture function for the screen is written

$\displaystyle F(x) = \frac{1}{N}\sum_{n=1,N} F_2(x-x_n),$ (1051)

where

$\displaystyle x_n = [n-(N+1)/2]\,d,$ (1052)

and

$\displaystyle F_2(x)=\left\{ \begin{array}{ccc} 1/\delta&\mbox{\hspace{0.5cm}}&\vert x\vert\leq \delta/2 [0.5ex] 0 &&\vert x\vert> \delta/2 \end{array}\right..$ (1053)

We recognize $ F_2(x)$ as the aperture function for a single slit, of finite width $ \delta$ , that is centered on $ \theta =0$ . [See Equation (1046).]

Assuming normal incidence (i.e., $ \theta_0=0$ ), the interference/diffraction function, which is the Fourier transform of the aperture function, takes the form [see Equation (1044)]

$\displaystyle \bar{F}(\theta) = \int_{-\infty}^{\infty} F(x) \cos(k \sin\theta x) dx.$ (1054)

Hence,

$\displaystyle \bar{F}(\theta)$ $\displaystyle = \frac{1}{N}\sum_{n=1,N}\int_{-\infty}^\infty F_2(x-x_n) \cos(k \sin\theta x) dx$    
  $\displaystyle =\frac{1}{N}\sum_{n=1,N} \left[\cos(k \sin\theta x_n) \int_{-\infty}^\infty F_2(x') \cos(k \sin\theta x') dx'\right.$    
  $\displaystyle    \left. - \sin(k \sin\theta x_n) \int_{-\infty}^\infty F_2(x') \sin(k \sin\theta x') dx'\right]$    
  $\displaystyle =\left[\frac{1}{N}\sum_{n=1,N} \cos(k\,\sin\theta\,x_n)\right] \int_{-\infty}^\infty F_2(x')\,\cos(k\,\sin\theta\,x')\,dx',$ (1055)

where $ x'=x-x_n$ . Here, we have made use of the result $ \int_{-\infty}^\infty F_2(x') \sin(\alpha x') dx'=0$ , for any $ \alpha$ , which follows because $ F_2(x')$ is even in $ x'$ , whereas $ \sin(\alpha x')$ is odd. We have also employed the trigonometric identity $ \cos(x-y)\equiv \cos x \cos y-\sin x \sin y$ . (See Appendix B.) The previous expression reduces to

$\displaystyle \bar{F}(\theta) = \bar{F}_1(\theta)\,\bar{F}_2(\theta).$ (1056)

Here [cf., Equation (1030)],

$\displaystyle \bar{F}_1(\theta)$ $\displaystyle = \int_{-\infty}^\infty F_1(x) \cos(k \sin\theta x) dx=\frac{1}{N}\sum_{n=1,N}\cos(k \sin\theta x_n)$    
  $\displaystyle =\frac{1}{N}\,\frac{\sin[\pi\,N\,(d/\lambda)\,\sin\theta]}{\sin [\pi\,(d/\lambda)\,\sin\theta]},$ (1057)

is the interference/diffraction function for $ N$ identical parallel slits of negligible width that are equally spaced a distance $ d$ apart, and

$\displaystyle F_1(x) = \frac{1}{N} \sum_{n=1,N} \delta(x-x_n),$ (1058)

is the corresponding aperture function. Furthermore [cf., Equation (1047)],

$\displaystyle \bar{F}_2(\theta) = \int_{-\infty}^\infty F_2(x) \cos(k \sin\theta x) dx = {\rm sinc}[\pi (\delta/\lambda) \sin\theta],$ (1059)

is the interference/diffraction function for a single slit of width $ \delta$ .

Figure 73: Multi-slit far-field interference pattern calculated for $ N=10$ , $ d/\lambda = 10$ , and $ \delta /\lambda =2$ , assuming normal incidence.
\begin{figure}
\epsfysize =3in
\centerline{\epsffile{Chapter10/fig11.eps}}
\end{figure}

We conclude, from the preceding analysis, that the interference/diffraction function for $ N$ identical, equally spaced, parallel slits of finite width is the product of the interference/diffraction function for $ N$ identical, equally spaced, parallel slits of negligible width, $ \bar{F}_1(\theta)$ , and the interference/diffraction function for a single slit of finite width, $ \bar{F}_2(\theta)$ . We have already encountered both of these functions. The former function (see Figure 69, which shows $ [\bar{F}_1(\theta)]^{\,2}$ ) consists of a series of sharp maxima of equal amplitude located at [see Equation (1031)]

$\displaystyle \theta_j = \sin^{-1}\left(j\,\frac{\lambda}{d}\right),$ (1060)

where $ j$ is an integer. The latter function (see Figure 71, which shows $ [\bar{F}_2(\theta-\theta_0)]^{\,2}$ ) is of order unity for $ \vert\theta\vert\lesssim \sin^{-1}(\lambda/\delta)$ , and much less than unity for $ \vert\theta\vert\gtrsim \sin^{-1}(\lambda/\delta)$ . It follows that the interference/diffraction pattern associated with $ N$ identical, equally spaced, parallel slits of finite width, which is given by

$\displaystyle {\cal I}(\theta) \propto \left[\bar{F}_1(\theta)\,\bar{F}_2(\thet...
...pto\left[\bar{F}_1(\theta)\right]^{\,2}\, \left[\bar{F}_2(\theta)\right]^{\,2},$ (1061)

is similar to that for $ N$ identical, equally spaced, parallel slits of negligible width, $ [\bar{F}_1(\theta)]^{\,2}$ , except that the intensities of the various maxima in the pattern are modulated by $ [\bar{F}_2(\theta)]^{\,2}$ . Hence, those maxima lying in the angular range $ \vert\theta\vert< \sin^{-1}(\lambda/\delta)$ are of similar intensity, whereas those lying in the range $ \vert\theta\vert> \sin^{-1}(\lambda/\delta)$ are of negligible intensity. This is illustrated in Figure 73, which shows the multi-slit interference/diffraction pattern calculated for $ N=10$ , $ d/\lambda = 10$ , and $ \delta /\lambda =2$ . As expected, the maxima lying in the angular range $ \vert\theta\vert< \sin^{-1}(0.5) = \pi/6$ have relatively large intensities, whereas those lying in the range $ \vert\theta\vert> \pi/6$ have negligibly small intensities.


next up previous
Next: Two-Dimensional Fourier Optics Up: Wave Optics Previous: Single-Slit Diffraction
Richard Fitzpatrick 2013-04-08