Intensity in Double-Slit Interference — Question 10

Question 10

Intensity near a dark fringe gives a quadratic displacement signal

Two equal coherent beams form a dark fringe at $y=y_d$. Near it, write $I(y)=4I_0\sin^2[\pi(y-y_d)/p]$, where p is fringe spacing and $4I_0$ is the bright level. Let $p=2.0\,\mathrm{mm}$.

1. Derive the leading quadratic intensity and find the displacement magnitude that produces one percent of the bright level using the exact expression.

2. Compare with the quadratic estimate and explain the sign ambiguity of the inferred displacement.

Question 10 - Solution

Intensity near a dark fringe gives a quadratic displacement signal

(a) Method and calculation

$$\sin x\simeq x\Longrightarrow\boxed{I\simeq4I_0\pi^2(y-y_d)^2/p^2}.$$$$I/(4I_0)=0.01\Longrightarrow\boxed{|y-y_d|=\frac p\pi\sin^{-1}(0.1)=63.7686\,\mu\mathrm m}.$$

(b) Interpretation and further calculation

$$\boxed{|y-y_d|_{\mathrm{quad}}=0.1p/\pi=63.6620\,\mu\mathrm m}.$$$$I(y_d+\eta)=I(y_d-\eta)\Longrightarrow\boxed{\eta=\pm63.7686\,\mu\mathrm m\ \text{locally}}.$$

Physical check

The quadratic estimate is slightly smaller than the exact displacement. A dark-fringe intensity cannot determine which side of the minimum the detector occupies.