**INTENSITY IN THE SINGLE-SLIT PAITERN**

We can derive an expression for the intensity distribution for the single-slit diffraction pattern by the same phasor-addition method that we used in Section 37-4 to obtain Eqs. (37-10) and (37-14) for the two-slit interference pattern. We again imagine a plane wave front at the slit subdivided into a large number of strips. We superpose the contributions of the Huygens wavelets from all the strips at a point P on a distant screen at an angle 8 from the normal to the slit plane. To do this, we use a phasor to represent the sinusoidally varying if field from each individual strip. The magnitude of the vector sum of the phasors at each point P is the amplitude Ep of the total if field at that point. The intensity at P is proportional to E/ At the point 0 shown in Figure 38-6a, corresponding to the center of the pattern where 8 = 0, there are negligible path differences for x » a; the phasors are all essentially in phase (that is, have the same direction). In Fig. 38-7a we draw the phasors at time t = 0 and denote the resultant amplitude at 0 by Eo’ In this illustration we have divided the slit into 14 strips.

Now consider wavelets arriving from different strips at point P in Fig. 38-6a, at an angle 8 from point O. Because of the differences in path length, there are now phase differences between wavelets coming from adjacent strips; the corresponding phasor diagram is shown in Fig. 38-7b. The vector sum of the phasors is now part of the perimeter of a many-sided polygon, and E”, the amplitude of the resultant electric field at P, is the chord. The angle fJ is the total phase difference between the wave from the top strip of Fig. 38-6a and the wave from the bottom strip; that is, fJ is the phase of the wave at P from the top strip with respect to the wave received at P from the bottom strip. We may imagine dividing the slit into narrower and narrower strips. In the limit that there is an infinite number of infinitesimally narrow strips, the phasor diagram becomes an arc of a circle (Fig. 38-7c), with arc length equal to the length Eo in Fig. 38-7a. The center C of this arc is found by constructing perpendiculars at A and B. From the definition of the radian as a measure of angles, the radius of the arc is EolfJ; the amplitude Ep of the resultant electric field at P is equal to the chord AB, which is 2(Eol fJ) sin (fJ12) We then have We can express the phase difference fJ in terms of geometric quantities, as we did for the two-slit pattern. From Eq. (37-11) the phase difference is 21C1)’times the path difference. Figure 38-6 shows that the path difference between the ray from the top of the slit and the ray from the middle of the slit is (a/2) sin 8. The path difference between the rays from the top of the slit and the bottom of the slit is twice this, so

This agrees with our previous result, Eq. (38-2). Note again that P = 0 (corresponding to 8 = 0) is not a minimum. Equation (38-5) is indeterminate at p = 0, but we can evaluate the limit as p ~ 0 using L’Hopital’s rule. We find that at p = 0, 1=/0, as we should expect.