Refresher on Diffraction


Interference and diffraction

The term interference describes wave phenomena originating from a small number of point sources. (By a point source we mean a source of waves having linear dimensions small in comparison to the wavelength of the radiation.) The term diffraction applies to the spreading of waves at apertures and can be thought of as resulting from the interference of many coherent point sources. Sources are coherent when they oscillate with a phase relationship which does not vary in time.

Suppose now instead of a point source we consider an extended source, i.e., a source larger than the wavelength under consideration. Such a source can be considered as a large array of coherent point sources and from our definition of diffraction it should be clear that waves from such a source should exhibit diffraction effects.

In this experiment we wish to gain some feeling for the inherent limitations in optical instruments due to diffraction effects. The light source in this experiment is a helium neon laser. The apertures which we will use are much larger than the wavelength of the laser light.

Interference from two point sources

Figure 1 represents the wave disturbance originating from two point sources, S1 and S2, separated by a distance $d$. By the superposition principle, the intensity observed at the detector will be a maximum (i.e., the waves will be in phase) if the path difference, $\Delta r = r_2-r_1$, is an integral number of wavelengths, i.e. $\Delta r = n\lambda$. But, for a detector far from the sources, $\Delta r = d\sin\theta$. Therefore, the angles at which the intensity will be a maximum, denoted by $\theta_{max}$, obey the following relation:

$n\lambda = d\sin\theta_{max} (\mathrm{for\;n = 0,1,2,\dots)}$ (1)
Figure 1: Interference from two point sources

The intensity observed at the detector will be a minimum if the two sources are out of phase, i.e., $r = \left(n + \frac{1}{2}\right)\lambda$. Therefore, denoting angles where we expect an intensity minimum by $\theta_{min}$, we have

$\left(n+\dfrac{1}{2}\right)\lambda = d\sin\theta_{min} (\mathrm{for\;n = 0,1,2,\dots)}$ (2)

Single slit diffraction

Now, consider an extended source, e.g., a slit, having width $a$ as shown in Fig. 2. We can consider the pattern of diffracted light coming from such an extended source as that resulting from the interference pattern of an infinite number of point sources within the slit, all oscillating in phase.

Figure 2: Single slit diffraction

To find the angles where the intensity will be a minimum, let us divide the extended source of width $a$ into two equal regions: the top and bottom halves of the slit. Now suppose that we are at such an angle $\theta$ that the path difference between a source at the top of the slit and one at the middle of the slit is exactly one-half wavelength; then it is clear that these two will cancel and give no contribution at the detector. Also, it should be clear that the contribution from the entire slit will as well be zero at this angle. This is because for any source in the top half of the slit, we can find one in the bottom half (namely one a distance $a/2$ from the upper source) which will, when added, give zero intensity. Thus, our condition for a minimum intensity becomes

$\Delta r = \dfrac{a}{2}\sin\theta_{min} = \dfrac{n\lambda}{2} (\mathrm{for\;n = 1,2,3,\dots)}$ (3)

(Note that $n=0 is not included.) Thus,

$\Delta r = {a}\sin\theta_{min} = n\lambda (\mathrm{for\;n = 1,2,3,\dots)}$ (4)

The maxima will occur approximately half way between the minima. A single slit diffraction pattern is shown in Fig. 3.

Figure 3: Single slit diffraction pattern

Two slit diffraction

The pattern of transmission through two slits will depend upon both the width of the slits and their spacing. The amplitude from either slit at angle  is governed by the single slit diffraction pattern - whereas the intensity from the two interfering slits will depend upon the interference pattern from two point sources. An intensity pattern similar to that shown in Fig. 4 will result: here the slit separation is taken to be three times the slit width. Note that the two slit pattern of Fig. 4 would lie entirely beneath the single slit pattern of Fig. 3, showing that the larger scale shape of the two slit pattern is still dictated by the widths of the individual slits.

Figure 4: Two-slit diffraction pattern

Three slit diffraction

Consider the interference pattern produced by three slits. Slits 1 and 2 together would act as a double slit and therefore produce the first minimum given by Eq. (4). Slits 2 and 3 alone would produce a minimum at exactly the same angle. However, the pattern from the slits 1 and 3 will actually produce a (secondary) maximum at the above angle. The overall amplitude at this angle will be one-third the amplitude from that of the diffraction envelope, because two of the slits cancel each other out. So, the intensity at this angle (a minimum in the two slit pattern) will be one-ninth of the diffraction envelope and will be a maximum.

Fig. 5 depicts the intensity pattern from such a three-slit arrangement. Note that the intensity goes to zero on each side of the secondary maximum.

Figure 5: Three slit diffraction pattern

Multiple slit diffraction

The point that we wish to stress here is that when we add a third slit to our two slit arrangement, the position of the first minimum moves closer to zero.  As a result, the central maximum becomes narrower. For similar reasons, the maxima in all orders become increasingly narrow as we add more slits. It can be shown that the angular width of the peaks is inversely proportional to N, the number of slits. Thus, the resolution of a slit system, which is its ability to separate two closely spaced wavelengths, is proportional to N.

Diffraction grating

As noted in the foregoing discussion, the resolution of a slit system increases with increasing number of slits. In the limit as the number of slits gets large, the central maximum of the diffraction pattern becomes narrower. For light normally incident on an equally spaced set of slits (as in Fig. 6), the relation among the diffraction angle $\theta$, the slit separation $d$, and the wavelength of the light $\lambda$ is given by

$n\lambda = d\sin\theta_{max} (\mathrm{for\;n = 0,1,2,\dots)}$ (5)

A grating is most commonly used with a spectrometer to make precise measurements of wavelengths. The spectrometer collimates the incoming light (i.e., makes it parallel) and provides for precise diffraction angle measurements.

In order to determine a wavelength using a grating, one must determine the diffraction order $n$, the grating spacing $d$ and the angle through which the light is diffracted.

Figure 6: Diffraction from a set of slits, showing the relationship among wavelength, slit separation and diffraction angle.