Materials in Electronics/Wave-Particle Duality/The Two-Slit Experiment/Ideal

The Experimental Set-up
The two slit experiment is theoretically very simple to set up and demonstrates a very interesting property of waves, even outside the area of quantum physics. A coherent monochromatic wave source is directed at two slits in an otherwise impenetrable screen. Because the slits will diffract the waves, this produces two light sources which are related very closely:
 * They both have identical wavelengths (as the original source is monochromatic)
 * They are both coherent (the waves are in phase)

This is shown conceptually in the diagram below. This is a very simple diagram, but it is useful to describe the concept.



Let us now specify some parameters for the set up. We will produce our waves using a light source (such as a laser) with a wavelength of 800nm. This is just outside the red end of the visible spectrum, and is not quite visible to the human eye. However, it is a convenient number to use as it is easily divided. If you do this experiment in real life, it is likely that the light will be around 700nm (deep red). We will shine our light through two slits (which we will assume to be so narrow that the diffraction is perfect), with a separation of 0.1mm, which, as we will see, gives a significant effect and is not too difficult to achieve practically. We will then place our screen at a distance of 5 metres from the slits. So, our parameters are:


 * $$\lambda = 400 \times 10^{-9} \mbox{ m}\,$$
 * $$d = 0.1 \times 10^{-3} \mbox{ m}\,$$
 * $$L = 5 \mbox{ m} \,$$

Assuming that our medium has a refractive index that is very close to 1 (such as air), we also know the speed of the wave, c, as this is a universal constant:


 * $$c = 3 \times 10^{8} \mbox{ m s}^{-1}$$

We can therefore work out the frequency, f, and the angular frequency, &omega;:
 * $$f = \frac{c}{\lambda}=375 \times 10^{12} \mbox{ Hz}$$
 * $$\omega = 2 \pi f =750\pi \times 10^{12} \mbox{ rad s}^{-1}$$

The wavenumber (number of cycles per metre), k, can also be found:
 * $$k = \frac{2 \pi}{\lambda}=2.5\pi \times 10^{6} \mbox{ m}^{-1}$$.

The Equation for a Single Slit
Let us first consider the waves emanating from just one of the slits. If we assume that the slit is very long (many times the wavelength) and very narrow (much smaller than the wavelength), and that the light from the laser is a plane wave (so the phase at every point on the slit is identical), then we can say that the diffracted wave is cylindrical:



This can be proved directly from the wave equation. This means that we can just consider a "slice" of the wavefront and not worry about the effect of vertical distance (i.e. in the direction of the slits), as it does not affect the wave. In real life, a laser produces a tight point of light, so the wave produced is a very thin cylindrical wave (because it is not diffracted vertically by the long slit).

Let us now consider the equation for a progressive wave. The disturbance, u produced by the wave at a distance x from the source and time t, is given by the following equation:


 * $$u \left( x,t \right)=A \left(x,t \right) \sin \left( k x - \omega t + \phi \right)$$

where &phi; is a phase offset, which we can set to zero from here on, as all the waves we are dealing with are in phase. A(x,t) is the amplitude envelope of the wave. For simplicity, we will assume that the wave does not decrease in amplitude significantly in this experiment. This is an acceptable simplification, because we are concerned with only the phase relation of the waves. In real life, the amplitude of a cylindrical wave varies inversely with distance (not the square of distance as a spherical wave does).

Below is a plot of a progressive wave in one dimension, over a whole period. This wave does not decay with distance.



Now, because we know we are dealing with a cylindrical wave, we only need to consider a 2D slice of the whole system. We set the x and y axes as shown in the image at the top of the page. The distance, r, to a point at (x,y) in this case is given by the Pythagorean Theorem:


 * $$r=\sqrt{x^2+y^2}$$

Thus, the equation our progressive wave in 2D is now (remember we are ignoring the amplitude and phase offset):


 * $$u \left( x,y,t \right)=\sin \left( k \sqrt{x^2+y^2} - \omega t\right)$$

The image below shows this wave at t=0. Notice that the wave is not smooth at the origin (you can see a flipped-over - actually 180° out of phase - version of the function here). If this were animated, it would appear to flow outwards.



We can center one of these point sources at any point (x0,y0) using the following equation:


 * $$u \left( x,y,t \right)=\sin \left( k \sqrt{(x-x_0)^2+(y-y_0)^2} - \omega t\right)$$

Modeling the Experiment
We can now construct a model of our two-slit experiment. If we place the sources at $$\left(\pm \frac{d}{2},0 \right)$$, we get our slit separation of d. The equation for the disturbance at any point in space and time is therefore given by the following equation:


 * $$u(x,y,t)=

\sin \left( k \sqrt{ \left( x-\frac{d}{2} \right)^2 + y^2} - \omega t \right)+ \sin \left( k \sqrt{ \left( x+\frac{d}{2} \right)^2 + y^2} - \omega t \right) $$

Our slit sepation is 0.1mm and the wavelength is 800nm. This means that the slits are 125 wavelengths apart, so we will be able to make out the diffraction pattern quite easily on a reasonably sized graph. Below is a density plot of the diffraction pattern for our setup. Black corresponds to the minimum amplitude and white corresponds to the maximum, while mid-grey is zero. We are now only concerned with positive y values, as the screen with the slits in it forms a barrier at y=0.



We can clearly see the criss-crossing pattern, but we cannot immediately see any "fringing" effects. This is because the fringing is a result of the wave's average value varying with position. In fact, the image does contain faint evidence of fringing. You may notice that there are approximately radial "bands" of wave where it changes from white to black and back again, and bands where it stays around mid grey. These will stay in position with time. Bands that run across the image are an artifact of the cyclic and progressive nature of the wave, and these will shift with time. The fringing becomes clearly visible when the wave is averaged over a period of time. The image below shows a plot of the same area, but instead of the wave frozen in time as b, it is a plot composed of many samples of the wave over time. The light area correspond to places where the amplitude of the resulting wave is high, and the black areas are where the wave has a zero amplitude.



You can see the fringes here, but they are very close together, and would not be easily discernible to the naked eye. Also, the fringes are not perfectly radial (coming from a single point), so we cannot easily work out their spacing without plotting this graph. If we view the system (averaged as before) from much further away as below, we can see a much simpler pattern:



The resulting wave now appears to be a sequence of radial lines of alternating high and low intensity that originate from the slits (which are imperceptibly separated at this scale). As we find in the next section, this makes a very useful approximation accurate. Also, the fringes are much further apart, so we can measure the fringe separation much more easily. Beware of the apparent curved lines in the bottom corner - there are no curved lines here, but the regular grid of pixels combined with the sloping lines causes false patterns to appear. For more information on related phenomena, see moiré and heterodynes and beat frequencies.

We can plot a graph of the wave at y=5m. The image below is the central maximum, over a one period of the wave:



You can't see this oscillation in real life because of the immensely high frequency of visible light (hundreds of terahertz). However, we can perceive the irradiance at the screen. (see this page for a derivation. Now, the irradiance, I, at a point on the screen is given by:


 * $$I=\epsilon v \langle \mathbf{E}^2 \rangle$$

where $$\langle \mathbf{E}^2 \rangle$$ means the time-average of the magnitude of the electric field squared. This is an elementary result of fields theory and this is not the place to prove it. For the moment, as we are dealing with relative intensities, we will leave out the constants and set


 * $$I=\langle \mathbf{E}^2 \rangle$$

From the principle of superposition,


 * $$\mathbf{E}=\mathbf{E}_1+\mathbf{E}_2$$

where E1 and E2 are the components of the field resulting from each slit. We can now write:


 * $$\mathbf{E}^2=\left( \mathbf{E}_1+\mathbf{E}_2 \right)^2$$


 * $$\mathbf{E}^2=\mathbf{E}_1^2+\mathbf{E}_2^2+2 \mathbf{E}_1+\mathbf{E}_2$$

Time-averaging both sides of this equation gives us:

The image below is the central 5 maxima at t=0. You can easily see the regular-looking maxima where the wave has a high amplitude, and the minima, where the wave's amplitude is very small. You should also be able to see that the variation in intensity is similar to the absolute value of a cosine wave.



You can compare this to the fringes seen in real life (the upper pattern result from two slits, the bottom pattern from five slits):



Let us now find a relationship between the physical properties of the system and the distance between the fringes.

An Approximation for Fringe Spacing
If we make certain approximations, it is simple to find an expression for the fringe spacing in this system. Consider the diagram on the right. If shows two rays of light, one from each slit, both going the same point on the screen, which is marked by a black dot. Because the distance to the screen is much greater than the distance between the slits, the rays are essentially parallel.

Now, the lower ray has further to go than the upper ray, and we will call the difference in path length &Delta;l. This is marked in red on the diagram. For perfect constructive interference, the path difference must be a whole number of wavelengths:


 * $$\Delta l=n \lambda, \quad n=(0,1,2,\ldots)$$

Because the rays are parallel, we can see that the two angles marked &theta; are identical. This means that we can determine the path difference from the angle to the point using trigonometry:


 * $$\sin \theta = \frac{\Delta l}{d}$$

We can also work out the distance of the point from the centre of the screen, s, using &theta;:


 * $$\tan \theta = \frac{s}{L}$$

For small angles, we can make the following approximation:


 * $$\sin \theta \approx \tan \theta$$.

Therefore, we can write


 * $$\frac{\Delta l}{d}=\frac{s}{L}$$

And substituting in our value for maximum constructive interference, we get an expression for the distance from the centre line to the nth bright fringe:


 * $$\frac{n \lambda}{d}=\frac{s}{L}$$

And rearranging gives:


 * {|style=" border: solid 2px #D6D6FF; padding: 1em;" valign="top"


 * align="left"|$$s=\frac{n \lambda L}{d}, \quad n=(0,1,2,\ldots)$$
 * }

Let's try this for our theoretical example from earlier. To find the basic fringe spacing, just set n=1 (this is the spacing between the central fringe and the first fringe to one side):


 * $$s=\frac{\lambda L}{d}$$


 * $$s=\frac{800 \times 10^{-9} \times 5}{0.1 \times 10^{-3}}$$


 * $$s=0.04 \mbox{ m}\,$$

This is almost exactly what we saw (it is only an approximation, so it is not technically exact). The following conditions must apply for this to work:
 * The distance from the slits to the screen, L must be much larger than the slit separation, d.
 * The angle, &theta;, to the fringe from the slits must be small
 * As before, the slits must be narrow enough to refract almost perfectly