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14.4Single-slit diffraction
This section deals in detail with the problem of single-slit diffraction and the effect of slit width on the interference pattern produced. An interference pattern is produced because light originating from one part of the slit interferes with light from all the other parts of the slit.
Diffraction by a single rectangular slit
As we saw in SectionÂ 14.3 when a wave of wavelength Î» is incident on an aperture whose opening size is b, an important wave phenomenon called diffraction takes place where the wave spreads out past the aperture. The amount of diffraction is appreciable if the wavelength is of the same order of magnitude as the opening or bigger, Î» â¥ b. However, diffraction is negligible if the wavelength is much smaller than the opening size, Î» << b.
Figure 14.45 shows a slit of width b through which light passes. When a wavefront reaches the slit, the points labelled A1 to A5 and B1 to B5 are all on the same wavefront, so they are in phase. Each point on the wavefront acts as a source of waves. These waves are also in phase, so they can interfere. We see the result of the interference on a screen placed a large distance away.
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Figure 14.45: For a narrow slit each point on the wavefront entering the slit acts as a source of waves. The result of interference between waves depends on the path difference.

In Figure 14.45 we see rays from A1 and B1 travelling at an angle Î¸ relative to the centre line of the slit. The interference between the waves is seen at point P on the screen. The wave from A1 travels a slightly different distance to reach the screen than the wave from B1. The interference seen at P depends on this path difference. From the diagram, the path difference for the waves from A1 and B1 equals the distance B1C1. Since P is far away, lines A1P and B1P are approximately parallel, so triangle A1B1C1 is approximately right angled and angle B1A1C1 equals Î¸. We see that the distance A1B1 is equal to b2, so:
path difference=B1C1=b2sinÎ¸
If the path difference is half a wavelength, the two waves will destructively interfere when they get to P. But we have only considered waves from A1 and B1. What about waves from A2 and B2?
Figure 14.46 shows that for P a long way from the slit, the rays from A1, A2, B1 and B2 are all parallel. The points are all equally spaced along the slit so A1B1 = A2B2.Triangles A1B1C1 and A2B2C2 are both right-angled and angle B1A1C1 = angle B2A2C2. The triangles are therefore congruent, so the path difference B1C1 is equal to the path difference B2C2.
Thus, if we get zero wave at P from the first pair of points, we will get the same from the second pair as well. For every point on the upper half of the slit there is a corresponding point on the lower half, so we see that all the points on the wavefront will result in complete destructive interference if the first pair results in destructive interference.

Figure 14.46: Triangles A1C1B1 and A2C2B2 are congruent.

The path difference between waves arriving at P from A1 and B1 is b2sinÎ¸.
This means that we get a minimum at P if:
b2sinÎ¸=Î»2
âbsinÎ¸=Î»
(By similar arguments we can show that we get additional minima whenever b sin Î¸ = nÎ», n = 1, 2, 3, ..., but we will not deal with those.)
Since the angle is small (in radians) we may approximate sin Î¸ â Î¸ and we have the approximate formula for the angle of the first diffraction minimum:
Î¸=Î»b
WORKED EXAMPLE 14.7
The first diffraction minimum using a single slit of opening 2.0 Âµm is at an angle of 12.5Â°. What is the wavelength of the light used?
Answer
From Î¸=Î»b we get Î» = bÎ¸.
We must, however, use the angle in radians:
12.5Â°=12.5ÃÏ180=0.218Â radians. (Here, the angle is at the limit of what may be considered a âsmallâ angle.)
And so Î» = 2.0 Ã 10â6 Ã 0.218 = 4.4 Ã 10â7 m.
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Intensity in single slit diffraction
The intensity of light observed on a screen some distance from the slit is shown in Figure 14.47 for red light with slit opening b = 1.4 Ã 10â5 m.

Figure 14.47: The single-slit intensity pattern for red light and slit width 1.4 Ã 10â5 m.

The intensity of the first secondary maximum is only about 4.5% of the intensity at the central maximum. The units on the vertical axis are arbitrary.
We see that the first minimum occurs at Î¸ = 0.05 rad.
Since Î¸=Î»b, we may deduce the wavelength of the light used:
Î»=bÎ¸=1.4Ã10â5Ã0.05=7.0Ã10â7Â m
The double arrow measures the angular width of the central maximum.
It equals 2Î»b or 0.10 rad in this case.
Blue light has a shorter wavelength and so with blue light the first minimum occurs at smaller angle, as shown in Figure 14.48.

Figure 14.48: The single-slit intensity pattern for blue light and slit width 1.4 Ã 10â5 m.

If we increase the slit width, more light goes through so the intensity will increase and the angular width will decrease.
The discussion above applies to monochromatic light (light of one specific wavelength). When white light is incident on a slit there will be a separate diffraction pattern observed for each wavelength making up the white light. Figure 14.49 shows the combined patterns due to just four wavelengths. The central maximum is white since all the colours produce maxima there. But as we move away from the centre the fringes appear coloured with blue showing up first.

Figure 14.49: The single-slit intensity pattern for white light. The central bright spot is white but the rest of the pattern is coloured. As we move to larger angles we see blue first and finally red.

Links
Diffraction and interference are the defining phenomena of wave behaviour. Anything that diffracts or interferes must be a wave. In ChapterÂ 22 we will see that electrons diffract and interfere so electrons must have a wave nature as well as a particle nature.
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NATURE OF SCIENCE
The interference patterns seen in diffraction through a single slit provide evidence of the wave nature of light. The pattern we see is very different from the simple geometrical shadow expected if light consisted of particles. In this section, you have seen how summing the different waves leads to the pattern observed and how the width of the slit affects the intensity pattern. In a similar way, the waves diffracted around objects can be summed. Figure 14.50 shows the result of diffraction of light around a small circular object. There is a bright spot at the centre of the disc, where a particle model of light would predict darkness! It was this spot, predicted by Fresnelâs wave theory and observed by FranÃ§ois Arago, that led to the acceptance of the wave theory in the nineteenth century (the debate between Fresnel and Poisson was described previously in this chapter). This is an example of how theory can be used to predict what should be observed. This can then be tested by experiment to give a result supporting the theory. The bright spot in the centre is now called Fresnelâs spot or Aragoâs spot.

Figure 14.50: a The Fresnel spot. b The intensity of light as a function of horizontal distance showing the peak in the middle of the disc.

Effect of the slit width on the two-slit intensity pattern
We know that if the slit width is negligibly small the intensity pattern in the two-slit experiment has peaks of equal intensity, as shown in blue in Figure 14.51. Also shown on the same graph in red is the single slit diffraction pattern for a slit whose width is not negligible. How is the two-slit intensity pattern modified when the slits have a non-zero width?

Figure 14.51: The blue curve is the two-slit intensity pattern in the case of negligible slit width. The red curve is the single slit diffraction pattern.

The effect of the slit width is that the red curve becomes the envelope of the blue curve, i.e. the blue curve is pushed under the red curve. We say that the single slit diffraction pattern modulates the two-slit pattern so that the intensity is given by the blue curve in Figure 14.52.

Figure 14.52: The two-slit intensity pattern (in blue) when the slit width is not negligible.

CHECK YOURSELF 5
In a two-slit experiment the slit separation is d and the slit width is b=d3. Explain why the third order maximum of the two-slit interference pattern will not be observed.
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TEST YOUR UNDERSTANDING
35A single slit of width 1.50 Î¼m is illuminated with light of wavelength 500.0 nm. Determine the angular width of the central maximum.
36In a single-slit diffraction experiment the slit width is 0.12 mm and the wavelength of the light used is 6.00 Ã 10â7 m. Calculate the linear width of the central maximum on a screen 2.00 m from the slit.
37The intensity pattern for single-slit diffraction is shown in Figure 14.53. (The vertical units are arbitrary.) The wavelength of the light used is Î».

Figure 14.53: For question 37.

aFind the width of the slit b in terms of Î».
bOn a copy of the axes draw a graph to show how intensity varies with diffraction angle for a slit with:
iwidth b2 and wavelength Î»
iiwidth b2 and wavelength Î»2.
(You do not need numbers on the intensity axis.)
38Two very narrow, parallel slits separated by a distance of 1.4 Ã 10â5 m are illuminated by coherent, monochromatic light of wavelength 7.0 Ã 10â7 m.
aDescribe what is meant by coherent and monochromatic light.
bDraw a graph to show how the intensity of light observed on a screen far from the slits varies with angle assuming the slit width is negligible.
cBy drawing another graph on the same axes, illustrate the effect on the intensity distribution of using slits of width 4.7 Ã 10â6 m.
