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# Single slit diffraction pattern

The diffraction pattern made by waves passing through a slit of width a,a (larger than lambda,λ) can be understood by imagining a series of point sources all in phase along the width of the slit. The waves moving directly forward are all in phase (they have zero path difference), so they form a large central maximum.

If the wave travels at an angle theta,θ from the normal to the slit, then there is a path difference x,x between the waves produced at the two ends of the slit.

x, equals, a, sine, theta.  x=asinθ

The path difference between the top and middle waves is then

x, prime, equals, start fraction, a, divided by, 2, end fraction, sine, theta.  x′=a/2​sinθ

If the path difference between the top and middle waves is start fraction, lambda, divided by, 2, end fraction,2λ​, then they are exactly out of phase and cancel each other out. This happens to all consecutive pairs of waves (the ones produced by the second source from the top and the second source past the middle etc.) at this angle, so there is no resultant wave at this angle. Thus, a minimum in the diffraction pattern is obtained at

lambda, equals, a, sine, theta.  λ=asinθ

Now the slit can be divided into four equal sections and the pairing of sources to give destructive interference can be repeated for the top two sections, which is identical to the result of pairing off matching sources in the bottom two sections. In the case, we obtain for a minimum (since every pair of waves we consider will destructively interfere due to our choice of geometry and pairing), to givestart fraction, lambda, divided by, 2, end fraction, equals, start fraction, a, divided by, 4, end fraction, sine, theta.

λ/2​=a/4​sinθ

We can then divide the slit aperture into six equal sections, and pair off sources in the top two divisions, then the middle two divisions, and then the bottom two, to give destructive interference for every matched pair. The minima of intensity are obtained at anglesn, lambda, equals, a, sine, theta.

=asinθ

where n,n is an integer left bracket, 1, comma, 2, comma, point, point, point, right bracket,(1,2,…), but not n, equals, 0,n=0. There is a maximum of intensity in the centre of the pattern. This process only gives the positions of the minima, does not work for positions of the maxima, and so does not give the intensities of the maxima.

These results lead to the diffraction pattern minima shown below, which can be represented as a graph of intensity of the diffracted wave against angle. See a level 6 section for an explanation of this diffraction pattern using phasors.