Light as a Wave — Question 4

Question 4

An index gradient changes phase even when the endpoints are fixed

A straight ray crosses a length $L=4.0\,\mathrm{mm}$ with refractive index $n(x)=n_0+bx/L$, where $n_0=1.20$ and $b=0.030$. A reference ray crosses the same length in uniform index $n_0$. Take $\lambda_0=600\,\mathrm{nm}$ and assume the straight path is imposed by a guide.

1. Integrate the optical path difference and compute the unwrapped relative phase.

2. Find how much b must change to advance the phase by one cycle with L fixed.

Question 4 - Solution

An index gradient changes phase even when the endpoints are fixed

(a) Method and calculation

$$\Delta\mathcal L=\int_0^L\frac{bx}{L}\,dx=\frac{bL}{2}=\boxed{60.0\,\mu\mathrm m}.$$$$\boxed{\Delta\phi=k_0\Delta\mathcal L=200\pi\ \mathrm{rad}}.$$

(b) Interpretation and further calculation

$$\delta\phi=\frac{2\pi}{\lambda_0}\frac{L\delta b}{2}=2\pi.$$$$\boxed{\delta b=\frac{2\lambda_0}{L}=3.00\times10^{-4}}.$$

Physical check

Using the exit index throughout the guide would double the predicted excess optical path.