Let’s use phasors to find the intensity of the diffraction pattern of a single slit. FIGURE 33.17a shows a slit of width $a$ with $N$ point sources of Huygens’ wavelets, each separated by distance $a\text{/}N$. (We’ll soon consider the entire wave front by letting $N\to \text{∞}$.) We need to calculate the superposition of these $N$ wavelets at a point on a distant screen, so far away in comparison with the slit width $a$ that the directions are all essentially parallel at angle $\theta$.

FIGURE 33.17 Phasor diagrams for single-slit diffraction.

In the double-slit experiment, two waves from slits separated by distance $d$ had a phase difference $\text{Δ}\varphi =2\pi d\text{sin}\theta \text{/}\text{λ}$. By exactly the same reasoning, the phase difference between two adjacent wavelets separated by $a\text{/}N$ is $\text{Δ}{\varphi }_{\text{adj}}=2\pi \left(a\text{/}N\right)\text{sin}\theta \text{/}\text{λ}$. This is the phase difference for every pair of adjacent wavelets.

FIGURE 33.17b analyzes the diffraction at $\theta =0$, the center of the diffraction pattern. Here all the wavelets travel straight ahead, and the phase difference between adjacent wavelets is $\text{Δ}{\varphi }_{\text{adj}}=0$. Consequently, the phasor diagram shows $N$ phasors in a straight line with amplitude $E_0=Ne$.

FIGURE 33.17c is the phasor diagram for superposition at an arbitrary point on the screen with $\theta \ne 0$. All $N$ phasors have the same length $e$—so the length of the chain of phasors is still $E_0=Ne$—but each is rotated by angle $\text{Δ}{\varphi }_{\text{adj}}$ with respect to the preceding phasor. The angle of the last phasor, after $N$ rotations, is

Notice that $\beta$ is independent of $N$. It is the phase difference between a wavelet originating at the top edge of the slit and one originating at the bottom edge, distance $a$ away.

$E$ is the amplitude of the superposition of the $N$ wavelets. To determine $E$, let $N\to \text{∞}$. This makes our calculation exact, because now we’re considering every point on the wave front. It also makes our calculation easier, because now the chain of phasors, with length $E_0$, is simply the arc of a circle.

FIGURE 33.18 shows the geometry. The triangle at the upper right is a right triangle, so $\alpha +\beta =90\text{°}$. But $\alpha$ and the angle subtending the arc also add up to 90°, so the angle subtending the arc must be $\beta$. We’ve divided it into two angles, each $\beta \text{/}2$, in order to create two right triangles along $E$. You can see that $E=2R\text{sin}\left(\beta \text{/}2\right)$. In addition, the arc length, if $\beta$ is in radians, is $E_0=\beta R$. Eliminating $R$, we find that the amplitude of the superposition is

FIGURE 33.18 Calculating the superposition.

The diffraction-pattern intensity is proportional to $E^2$, thus

where $I_0$ (proportional to $E_0{}^2$) is the intensity at the center of the central maximum, $\theta =0$. (Recall, from l’Hôpital’s rule, that $\text{sin}x\text{/}x\to 1$ as $x\to 0$.) FIGURE 33.19 is a graph of Equation 33.26. You can see that it is, indeed, exactly what we observe for single-slit diffraction—a bright central maximum flanked by weaker secondary maxima. The minima occur where the numerator of Equation 33.26 is zero. This requires $\text{sin}{\theta }_p=p\text{λ}\text{/}a$ for $p=1\text{,}2\text{,}3\text{,}\dots$, which is exactly the result for the dark fringes that we found previously.

FIGURE 33.19 The single-slit diffraction pattern.

We usually measure positions on a screen rather than angles. For a screen at distance $L$, a point on the screen at distance $y$ from the center of the pattern is at angle $\theta =\text{tan}{}^{-1}\left(y\text{/}L\right)$. For very small angles, which is also typical, the small-angle approximation $\text{tan}\theta \approx \text{sin}\theta \approx \theta$ allows us to write the intensity at position $y$ as

In this case the minima are at $y_p=p\text{λ}L/a$, also as we found previously.

You might think that the maxima would occur where the numerator in Equation 33.27 is 1. However, $y$ also appears in the denominator, and that affects the maxima. Setting the derivative of Equation 33.27 to zero, to locate the maxima, leads to a transcendental equation, one that cannot be solved algebraically. It can be solved numerically, leading to the result that, for small angles, the first two maxima occur at $y_{\text{max}1}=1.43\text{λ}L/a$ and $y_{\text{max}2}=2.46\text{λ}L/a$.