Light as a Wave — Question 1

Question 1

Equal geometric routes can accumulate unequal optical phase

Two coherent rays of vacuum wavelength $600\,\mathrm{nm}$ traverse equal geometric lengths. In path 1, $12\,\mu\mathrm m$ of air is replaced by glass of index 1.50; path 2 remains in air. Use $\phi=k_0\mathcal L-\omega t$, $\mathcal L=\int n\,ds$, and neglect boundary reflection phases.

1. Calculate the additional optical path and unwrapped phase of path 1.

2. Find the smallest positive increase in glass thickness that restores the same relative phase modulo $2\pi$.

Question 1 - Solution

Equal geometric routes can accumulate unequal optical phase

(a) Method and calculation

$$\Delta\mathcal L=(n-1)t=0.50(12\,\mu\mathrm m)=\boxed{6.0\,\mu\mathrm m}.$$$$\Delta\phi=\frac{2\pi\Delta\mathcal L}{\lambda_0}=\boxed{20\pi\ \mathrm{rad}\equiv0\pmod{2\pi}}.$$

(b) Interpretation and further calculation

$$\delta\phi=2\pi\Longrightarrow(n-1)\delta t=\lambda_0.$$$$\boxed{\delta t=\frac{\lambda_0}{n-1}=1.20\,\mu\mathrm m}.$$

Physical check

Equal geometric lengths need not mean equal phase. Here the extra optical path happens to contain exactly ten vacuum wavelengths.