33.4.2 Analyzing Single-Slit Diffraction

FIGURE 33.13a shows a wave front passing through a narrow slit of width $a$ . According to Huygens’ principle, each point on the wave front can be thought of as the source of a spherical wavelet. These wavelets overlap and interfere, producing the diffraction pattern seen on the viewing screen. The full mathematical analysis, using every point on the wave front, is a fairly difficult problem in calculus. We’ll be satisfied with a geometrical analysis based on just a few wavelets.

The figure labeled (“a”) illustrates a diffraction pattern of a plane wave front passing through a slit.

The figure labeled (b) illustrates a plane wave front passing through a slit.

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FIGURE 33.13b shows the paths of several wavelets that travel straight ahead to the central point on the screen. (The screen is very far to the right in this magnified view of the slit.) The paths are very nearly parallel to each other, thus all the wavelets travel the same distance and arrive at the screen in phase with each other. The constructive interference between these wavelets produces the central maximum of the diffraction pattern at $θ = 0$ .

The situation is different at points away from the center. Wavelets 1 and 2 in FIGURE 33.13c start from points that are distance $a / 2$ apart. If the angle is such that $Δ r_{12}$ , the extra distance traveled by wavelet 2, happens to be $λ / 2$ , then wavelets 1 and 2 arrive out of phase and interfere destructively. But if $Δ r_{12}$ is $λ / 2$ , then the difference $Δ r_{34}$ between paths 3 and 4 and the difference $Δ r_{56}$ between paths 5 and 6 are also $λ / 2$ . Those pairs of wavelets also interfere destructively. The superposition of all the wavelets produces perfect destructive interference.

Figure 33.13c shows six wavelets, but our conclusion is valid for any number of wavelets. The key idea is that every point on the wave front can be paired with another point distance $a / 2$ away. If the path-length difference is $λ / 2$ , the wavelets from these two points arrive at the screen out of phase and interfere destructively. When we sum the displacements of all $N$ wavelets, they will—pair by pair—add to zero. The viewing screen at this position will be dark. This is the main idea of the analysis, one worth thinking about carefully.

You can see from Figure 33.13c that $Δ r_{12} = (a / 2) sin θ$ . This path-length difference will be $λ / 2$ , the condition for destructive interference, if

Δ r_{12} = \frac{a}{2} sin θ_{1} = \frac{λ}{2}

(33.18)

or, equivalently, if $a sin θ_{1} = λ$ .

NOTE Equation 33.18 cannot be satisfied if the slit width $a$ is less than the wave length $λ$ . If a wave passes through an opening smaller than the wavelength, the central maximum of the diffraction pattern expands to where it completely fills the space behind the opening. There are no minima or dark spots at any angle. This situation is uncommon for light waves, because $λ$ is so small, but quite common in the diffraction of sound and water waves.

We can extend this idea to find other angles of perfect destructive interference. Suppose each wavelet is paired with another wavelet from a point $a / 4$ away. If $Δ r$ between these wavelets is $λ / 2$ , then all $N$ wavelets will again cancel in pairs to give complete destructive interference. The angle $θ_{2}$ at which this occurs is found by replacing $a / 2$ in Equation 33.18 with $a / 4$ , leading to the condition $a sin θ_{2} = 2 λ$ . This process can be continued, and we find that the general condition for complete destructive interference is

\begin{matrix} a sin θ_{p} = p λ & p = 1, 2, 3, \dots \end{matrix}

(33.19)

When $θ_{p} ≪ 1 rad,$ which is almost always true for light waves, we can use the small-angle approximation to write

\begin{matrix} θ_{p} = p \frac{λ}{a} & p = 1, 2, 3, \dots & (angles of dark fringes) \end{matrix}

(33.20)

Equation 33.20 gives the angles in radians to the dark minima in the diffraction pattern of Figure 33.11. Notice that $p = 0$ is explicitly excluded. $p = 0$ corresponds to the straight-ahead position at $θ = 0$ , but you saw in Figures 33.11 and 33.13b that $θ = 0$ is the central maximum, not a minimum.

NOTE It is perhaps surprising that Equations 33.19 and 33.20 are mathematically the same as the condition for the $m th$ maximum of the double-slit interference pattern. But the physical meaning here is quite different. Equation 33.20 locates the minima (dark fringes) of the single-slit diffraction pattern.

You might think that we could use this method of pairing wavelets from different points on the wave front to find the maxima in the diffraction pattern. Why not take two points on the wave front that are distance $a / 2$ apart, find the angle at which their wavelets are in phase and interfere constructively, then sum over all points on the wave front? There is a subtle but important distinction. FIGURE 33.14 shows six vector arrows. The arrows in Figure 33.14a are arranged in pairs such that the two members of each pair cancel. The sum of all six vectors is clearly the zero vector $\vec{0}$ , representing destructive interference. This is the procedure we used in Figure 33.13c to arrive at Equation 33.18.

The figure labeled (“a”) consists of three pairs of six vector arrows that illustrate destructive interference.

The figure labeled (b) consists of three pairs of six vector arrows that illustrate destructive interference.

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The arrows in Figure 33.14b are arranged in pairs such that the two members of each pair point in the same direction—constructive interference! Nonetheless, the sum of all six vectors is still $\vec{0}$ . To have $N$ waves interfere constructively requires more than simply having constructive interference between pairs. Each pair must also be in phase with every other pair, a condition not satisfied in Figure 33.14b. Constructive interference by pairs does not necessarily lead to net constructive interference. It turns out that there is no simple formula to locate the maxima of a single-slit diffraction pattern.

Optional Section 33.5 will calculate the full intensity pattern of a single slit. The results are shown graphically in FIGURE 33.15. You can see the bright central maximum at $θ = 0$ , the weaker secondary maxima, and the dark points of destructive interference at the angles given by Equation 33.20. Compare this graph to the image of Figure 33.11 and make sure you see the agreement between the two.

The figure illustrates a position y versus light intensity graph for depicting the full intensity pattern of a single slit.

FIGURE 33.15 Full Alternative Text

EXAMPLE 33.3 Diffraction of a laser through a slit

Light from a helium-neon laser $(λ = 633 nm)$ passes through a narrow slit and is seen on a screen 2.0 m behind the slit. The first minimum in the diffraction pattern is 1.2 cm from the central maximum. How wide is the slit?

MODEL A narrow slit produces a single-slit diffraction pattern. A displacement of only 1.2 cm in a distance of 200 cm means that angle $θ_{1}$ is certainly a small angle.

VISUALIZE The intensity pattern will look like Figure 33.15.

SOLVE We can use the small-angle approximation to find that the angle to the first minimum is

θ_{1} = \frac{1.2 cm}{200 c m} = 0.00600 rad = 0.344 °

The first minimum is at angle $θ_{1} = λ / a$ , from which we find that the slit width is

a = \frac{λ}{θ_{1}} = \frac{633 \times 10^{- 9} m}{6.00 \times 10^{- 3} rad} = 1.1 \times 10^{- 4} m = 0.11 mm

REVIEW This is typical of the slit widths used to observe single-slit diffraction. You can see that the small-angle approximation is well satisfied.

33.4.2 Analyzing Single-Slit Diffraction

FIGURE 33.13 Each point on the wave front is a source of spherical wavelets. The superposition of these wavelets produces the diffraction pattern on the screen.

FIGURE 33.14 Destructive interference by pairs leads to net destructive interference, but constructive interference by pairs does not necessarily lead to net constructive interference.

FIGURE 33.15 A graph of the intensity of a single-slit diffraction pattern.

EXAMPLE 33.3 Diffraction of a laser through a slit