Light as a Wave — Question 6

Question 6

Two wavelengths resolve an optical-path ambiguity

An unknown positive optical path difference $D$ produces an integer number of phase cycles at both vacuum wavelengths $500\,\mathrm{nm}$ and $600\,\mathrm{nm}$. Assume $0<D\leq4.0\,\mu\mathrm m$ and no additional boundary phases.

1. Find all allowed values of D in the stated interval.

2. At the allowed D, calculate the phase at $550\,\mathrm{nm}$ reduced to $[0,2\pi)$.

Question 6 - Solution

Two wavelengths resolve an optical-path ambiguity

(a) Method and calculation

$$D=m(500\,\mathrm{nm})=q(600\,\mathrm{nm})\Longrightarrow5m=6q.$$$$m=6j,\ q=5j\Longrightarrow D=3.00j\,\mu\mathrm m;\quad\boxed{D=3.00\,\mu\mathrm m}.$$

(b) Interpretation and further calculation

$$\frac{D}{550\,\mathrm{nm}}=\frac{60}{11}=5+\frac{5}{11}.$$$$\boxed{\phi_{550}=\frac{10\pi}{11}\ \mathrm{rad}\simeq2.856\,\mathrm{rad}}.$$

Physical check

The finite prior interval removes higher common multiples; one wavelength alone would leave several possible path differences.