Intensity in Double-Slit Interference — Question 4

Question 4

Quadrature bias converts a tiny phase change into a linear intensity signal

Equal coherent beams give $I(\delta)=2I_0(1+\cos\delta)$. Bias the interferometer at $\delta_0=\pi/2$ and impose a small optical path change $\Delta\mathcal L$, giving $\epsilon=2\pi\Delta\mathcal L/\lambda$. Let $\lambda=600\,\mathrm{nm}$.

1. Linearize the intensity about the bias and calculate the change for $\Delta\mathcal L=3.0\,\mathrm{nm}$.

2. Find the leading relative error of this linear response and compare the first-order sensitivity with biasing at a bright fringe.

Question 4 - Solution

Quadrature bias converts a tiny phase change into a linear intensity signal

(a) Method and calculation

$$I(\pi/2+\epsilon)=2I_0(1-\sin\epsilon)\simeq2I_0(1-\epsilon).$$$$\epsilon=0.01\pi\Longrightarrow\boxed{\Delta I\simeq-0.062832I_0}.$$

(b) Interpretation and further calculation

$$\sin\epsilon=\epsilon-\epsilon^3/6+\cdots\Longrightarrow\boxed{\frac{|\epsilon-\sin\epsilon|}{|\epsilon|}\simeq\epsilon^2/6=1.6449\times10^{-4}}.$$$$\frac{dI}{d\delta}=-2I_0\sin\delta:\quad\boxed{I^\prime(\pi/2)=-2I_0,\quad I^\prime(0)=0}.$$

Physical check

Quadrature provides a first-order signal, while a bright fringe responds only at second order to a sufficiently small phase displacement.