Waves/Double slit Diffraction

Narrow Slits


Let us now imagine a plane sine wave normally impingent on a screen with two narrow slits spaced by a distance d, as shown in Figure 2.16. Since the slits are narrow relative to the wavelength of the wave impingent on them, the spreading angle of the beams is large and the diffraction pattern from each slit individually is a cylindrical wave spreading out in all directions, as illustrated in figure 2.13. The cylindrical waves from the two slits interfere, resulting in oscillations in wave intensity at the screen on the right side of figure 2.16.

Constructive interference occurs when the difference in the path lengths of the two rays is an integer multiple of the wavelength, λ:


 * $$l_2 - l_1 = m\lambda, \quad m = 0, \pm 1 , \pm 2 , \ldots \,$$

Similarly, destructive interference occurs at:


 * $$l_2 - l_1 = \left( m + \frac{1}{2} \right) \lambda, \quad m = 0, \pm 1 , \pm 2 , \ldots \,$$

If the distance from the slits to the screen, L is much greater than the slit separation, d, the rays l1 and l2 are approximately parallel and we can say that the path length difference is:


 * $$l_2 - l_1 = d \sin \theta\,$$

Substituting these two equations gives the following condition for constructive interference:


 * $$d \sin \theta = m\lambda, \quad m = 0, \pm 1 , \pm 2 , \ldots \,$$

The integer m is called the interference order and is the number of wavelengths by which the two paths differ. At every angle corresponding to integer m, constructive interference will be greatest and a "fringe" will be formed.

The distance from a point on the screen to the centre line, s, is related to θ as follows:
 * $$\tan \theta = \frac{s}{L}$$

When θ is small, we can say that:
 * $$\tan \theta \approx \sin \theta$$

We can therefore equate the expressions for position on the screen and fringe angles:
 * $$\frac{m \lambda}{d} \approx \frac{s}{L}$$

Rearranging gives an expression for the fringe spacing:
 * $$s \approx \frac{m \lambda L}{d} $$

The following conditions have to hold for this to be a good approximation:
 * The distance from the slits to the screen, L, must be much larger than the slit separation, d.
 * The angle, θ, to the fringe from the slits must be small.
 * The slits must be narrow compared to the wavelength, λ.

Example
Let us consider a setup with the following parameters:
 * d = 0.1 mm
 * λ = 2500 nm (infrared)

We expect to see a fringe spacing at 10m of:
 * $$s=\frac{\lambda L}{d}=\frac{2500 \times 10^{-9} \times 10}{0.1 \times 10^{-3}}=0.25m$$

First, let's look at a "snapshot" of the wave at a certain instance in time. The amplitude at a point, (x,y), is given by:
 * $$u(x,y,t)=

a\sin \left( k \sqrt{ \left( x-\frac{d}{2} \right)^2 + y^2} \right)+ a\sin \left( k \sqrt{ \left( x+\frac{d}{2} \right)^2 + y^2} \right) $$ where k is the wavenumber, given by 2π/λ. In this exercise, we will neglect the decreasing amplitude with distance, so the fringes stay bright over all space. The plot below shows a plot of this function. You can clearly see the circular propagation of each source, but an overall fringe structure is difficult to make out.



Now, the human eye and all light-measuring equipment do not detect the individual oscillations of a light wave, they measure intensity. Here, intensity is a function of the phase difference between the two sources, which in turn is directly related to the path length difference.
 * $$\delta = \frac{2\pi}{\lambda}\Delta l$$
 * $$I \approx 4 a^2 \cos^2 \frac{\delta}{2}$$

The plot below shows this intensity function. We see that a very clear fringe structure is revealed:

At this scale, the fringe structure is not clearly radial, it "bends" around the source. However, we can see that if we were to zoom out, the frenges diverge, and the ones near the centre, which are closest to perfectly radial, will be the ones that matter: In this plot, we see the predicted bright fringes at x=0mm, x=±250mm and x=±500mm.