Equations 33.6 and 33.9 locate the positions of maximum and zero intensity. To complete our analysis we need to calculate the light intensity at every point on the screen. In 《 CHAPTER 17, where interference was introduced, you learned that the net amplitude $E$ of two superimposed waves is

where, for light waves, $e$ is the electric field amplitude of each individual wave. Because the sources (i.e., the two slits) are in phase, the phase difference $\text{Δ}\varphi$ at the point where the two waves are combined is due only to the path-length difference: $\text{Δ}\varphi =2\pi (\text{Δ}r/\text{λ})$. Using Equation 33.2 for $\text{Δ}r$, along with the small-angle approximation and Equation 33.5 for $y$, we find the phase difference at position $y$ on the screen to be

Substituting Equation 33.11 into Equation 33.10, we find the wave amplitude at position $y$ to be

The light intensity, which is what we see, is proportional to the square of the amplitude. The intensity of a single slit, with amplitude $e$, is $I_1=C{e}^2$, where $C$ is a proportionality constant. For the double slit, the intensity at position $y$ on the screen is

Replacing $C{e}^2$ with $I_1$, we see that the intensity of an ideal double-slit interference pattern at position $y$ is

We’ve said “ideal” because we’ve assumed that $e$, the electric field amplitude of each wave, is constant across the screen.

FIGURE 33.6a is a graph of the ideal double-slit intensity versus position $y$. Notice the unusual orientation of the graph, with the intensity increasing toward the left so that the $y$-axis can match the experimental layout. You can see that the intensity oscillates between dark fringes $(I_{\text{double}}=0)$ and bright fringes $(I_{\text{double}}=4I_1)$. The maxima occur at points where $y_m=m\text{λ}L/d$. This is what we found earlier for the positions of the bright fringes, so Equation 33.14 is consistent with our initial analysis.

FIGURE 33.6 Intensity of the interference fringes in a double-slit experiment.

FIGURE 33.6 Full Alternative Text

One curious feature is that the light intensity at the maxima is $I=4I_1$, four times the intensity of the light from each slit alone. You might think that two slits would make the light twice as intense as one slit, but interference leads to a different result. Mathematically, two slits make the amplitude twice as big at points of constructive interference $(E=2e)$, so the intensity increases by a factor of $2^2=4$. Physically, this is conservation of energy. The line labeled $2I_1$ in Figure 33.6a is the uniform intensity that two slits would produce if the waves did not interfere. Interference does not change the amount of light energy coming through the two slits, but it does redistribute the light energy on the viewing screen. You can see that the average intensity of the oscillating curve is $2I_1$, but the intensity of the bright fringes gets pushed up from $2I_1$ to $4I_1$ in order for the intensity of the dark fringes to drop from $2I_1$ to 0.

Equation 33.14 predicts that, ideally, all interference fringes are equally bright, but you saw in Figure 33.4b that the fringes decrease in brightness as you move away from the center. The erroneous prediction stems from our assumption that the amplitude $e$ of the wave from each slit is constant across the screen. A more detailed calculation, which we will do in Section 33.5, must consider the varying intensity of the light that has diffracted through each of the slits. We’ll find that Equation 33.14 is still correct if $I_1$ slowly decreases as $y$ increases.

FIGURE 33.6b summarizes this analysis by graphing the light intensity (Equation 33.14) with $I_1$ slowly decreasing as $y$ increases. Comparing this graph to the image, you can see that the wave model of light has provided an excellent description of Young’s double-slit interference experiment.