Interference and diffraction are manifestations of superposition. Mathematically, the superposition of waves at a fixed point in space ($r_1$ and $r_2$ constant) involves sums such as $e\text{cos}\left(\omega t\right)+e\text{cos}(\omega t+\text{Δ}\varphi )$, where $e$ is the electric field amplitude of each wave and $\text{Δ}\varphi$ is the phase difference between them due to the fact that the waves have traveled different distances. Interestingly, we can use geometry to compute the sums that are relevant to interference and diffraction.

FIGURE 33.16a shows two vectors, each with amplitude $e$, rotating in the $xy$-plane with angular frequency $\omega$. At any instant, the angle from the $x$-axis is the vector’s phase, $\omega t$ or $\omega t+\text{Δ}\varphi$. A rotating vector that encodes amplitude and phase information is called a phasor, and Figure 33.16a is a phasor diagram. Notice two key features. First, the vectors rotate together, keeping a fixed angle $\text{Δ}\varphi$ between them. Second, the projections of the phasors onto the $x$-axis are $e\text{cos}\left(\omega t\right)$ and $e\text{cos}(\omega t+\text{Δ}\varphi )$, exactly what we add in a superposition calculation.

FIGURE 33.16 Phasor diagrams for double-slit interference.

To see how this works, let’s return to Young’s double-slit experiment. The bright and dark interference fringes arise from the superposition of two waves, one from each slit, with a phase difference $\text{Δ}\varphi =2\pi \text{Δ}r\text{/}\text{λ}$ due to the path-length difference $\text{Δ}r$. FIGURE 33.16b represents each wave as a phasor with amplitude $e$. We’re interested only in their superposition, not the rapid oscillation at frequency $\omega$, so we can draw the first phasor horizontally. If we add the phasors as vectors, using the tip-to-tail method, the magnitude $\mathit{E}$ of their vector sum is the electric field amplitude of the superposition of the two waves.

We can use geometry and trigonometry to determine $E$. We have an isosceles triangle whose large angle, complementary to $\text{Δ}\varphi$, is $180\text{°}-\text{Δ}\varphi$. Consequently, the two smaller, equal angles are each $\text{Δ}\varphi \text{/}2$, and thus the base of the isosceles triangle has length $E=2e\text{cos}\left(\text{Δ}\varphi \text{/}2\right)$. The figure shows a triangle for which $\text{cos}\left(\text{Δ}\varphi \text{/}2\right)$ is positive, but $\text{cos}\left(\text{Δ}\varphi \text{/}2\right)$ can be negative at some points in the double-slit pattern. Amplitude, however, must be a positive number, so in general

Equation 33.23 is identical to Equation 33.10, which we found previously to be the amplitude of the double-slit interference pattern. The intensity of the interference pattern is proportional to $E^2$.