3.2 Mathematics of interference

[link] (a) shows how to determine the path length difference $\text{Δ}l$ for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle $\theta$ between the path and a line from the slits to the screen [part (b)] is nearly the same for each path. In other words, $r_1$ and $r_2$ are essentially parallel. The lengths of $r_1$ and $r_2$ differ by $\text{Δ}l$ , as indicated by the two dashed lines in the figure. Simple trigonometry shows

where d is the distance between the slits. Combining this result with [link] , we obtain constructive interference for a double slit when the path length difference is an integral multiple of the wavelength, or

Similarly, to obtain destructive interference for a double slit, the path length difference must be a half-integral multiple of the wavelength, or

where $\lambda$ is the wavelength of the light, d is the distance between slits, and $\theta$ is the angle from the original direction of the beam as discussed above. We call m the order    of the interference. For example, $m=4$ is fourth-order interference.

The equations for double-slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes    ( [link] ). The closer the slits are, the more the bright fringes spread apart. We can see this by examining the equation

$d\text{sin}\theta =m\lambda ,\text{for}m=0,\text{±}1,\text{±}2,\text{±}3\text{…}$ . For fixed $\lambda$ and m , the smaller d is, the larger $\theta$ must be, since $\text{sin}\theta =m\lambda \text{/}d$ . This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance d apart) is small. Small d gives large $\theta$ , hence, a large effect.

Referring back to part (a) of the figure, $\theta$ is typically small enough that $\text{sin}\theta \approx \text{tan}\theta \approx y_m\text{/}D$ , where $y_m$ is the distance from the central maximum to the m th bright fringe and D is the distance between the slit and the screen. [link] may then be written as

or