Lectures on Physics has been derived from Benjamin Crowell's Light and Matter series of free introductory textbooks on physics. See the editorial for more information.... 
Home Optics Wave Optics DoubleSlit Diffraction  
Search the VIAS Library  Index  
DoubleSlit Diffraction
Let's now analyze doubleslit diffraction, (a), using Huygens' principle. The most interesting question is how to compute the angles such as X and Z where the wave intensity is at a maximum, and the inbetween angles like Y where it is minimized. Let us measure all our angles with respect to the vertical center line of the figure, which was the original direction of propagation of the wave. If we assume that the width of the slits is small (on the order of the wavelength of the wave or less), then we can imagine only a single set of Huygens ripples spreading out from each one, (b). The only dimension of the diffracting slits that has any effect on the geometric pattern of the overlapping ripples is then the centertocenter distance, d, between the slits.
We know from our discussion of the scaling of diffraction that there must be some equation that relates an angle like θ_{Z} to the ratio λ/d, λ / d ↔ θ_{Z} . If the equation for θ_{Z} depended on some other expression such as λ+d or λ^{2}/d, then it would change when we scaled λ and d by the same factor, which would violate what we know about the scaling of diffraction. Along the central maximum line, X, we always have positive waves coinciding with positive ones and negative waves coinciding with negative ones. (I have arbitrarily chosen to take a snapshot of the pattern at a moment when the waves emerging from the slit are experiencing a positive peak.) The superposition of the two sets of ripples therefore results in a doubling of the wave amplitude along this line. There is constructive interference. This is easy to explain, because by symmetry, each wave has had to travel an equal number of wavelengths to get from its slit to the center line, (c).
At the point along direction Y shown in the same figure, one wave has traveled ten wavelengths, and is therefore at a positive extreme, but the other has traveled only nine and a half wavelengths, so it at a negative extreme. There is perfect cancellation, so points along this line experience no wave motion. But the distance traveled does not have to be equal in order to get constructive interference. At the point along direction Z, one wave has gone nine wavelengths and the other ten. They are both at a positive extreme.
To summarize, we will have perfect constructive interference at any point where the distance to one slit differs from the distance to the other slit by an integer number of wavelengths. Perfect destructive interference will occur when the number of wavelengths of path length difference equals an integer plus a half.
Now we are ready to find the equation that predicts the angles of the maxima and minima. The waves travel different distances to get to the same point in space, (d). We need to find whether the waves are in phase (in step) or out of phase at this point in order to predict whether there will be constructive interference, destructive interference, or something in between.
One of our basic assumptions in this chapter is that we will only be dealing with the diffracted wave in regions very far away from the object that diffracts it, so the triangle is long and skinny. Most realworld examples with diffraction of light, in fact, would have triangles with even skinner proportions than this one. The two long sides are therefore very nearly parallel, and we are justified in drawing the right triangle shown in figure (e), labeling one leg of the right triangle as the difference in path length , LL', and labeling the acute angle as θ. (In reality this angle is a tiny bit greater than the one labeled θ in the previous figure.) The difference in path length is related to d and θ by the equation
Constructive interference will result in a maximum at angles for which LL' is an integer number of wavelengths,
Here m equals 0 for the central maximum, 1 for the first maximum to its left, +2 for the second maximum on the right, etc. Putting all the ingredients together, we find mλ/d=sin θ, or
Similarly, the condition for a minimum is
That is, the minima are about halfway between the maxima. As expected based on scaling, this equation relates angles to the unitless ratio λ/d. Alternatively, we could say that we have proven the scaling property in the special case of doubleslit diffraction. It was inevitable that the result would have these scaling properties, since the whole proof was geometric, and would have been equally valid when enlarged or reduced on a photocopying machine!
Counterintuitively, this means that a diffracting object with smaller dimensions produces a bigger diffraction pattern, (f ).
Although the equation λ/d = sin θ/m is only valid for a double slit, it is can still be a guide to our thinking even if we are observing diffraction of light by a virus or a flea's leg: it is always true that (1) large values of λ/d lead to a broad diffraction pattern, and (2) diffraction patterns are repetitive. In many cases the equation looks just like λ/d = sin θ/m but with an extra numerical factor thrown in, and with d interpreted as some other dimension of the object, e.g. the diameter of a piece of wire.


Home Optics Wave Optics DoubleSlit Diffraction 