Wavelength and Frequency — Question 4

Question 4

A phase difference determines a family of possible wavelengths

Two sensors separated by $d=.60\,\mathrm{m}$ see a positive spatial phase advance of $\pi/2$ modulo $2\pi$. The wave number is positive. An independent bound states $.30<\lambda<1.0\,\mathrm{m}$.

1. Find all candidate wavelengths before applying the bound.

2. Use the bound to choose the wavelength and explain why phase alone was ambiguous.

Question 4 - Solution

A phase difference determines a family of possible wavelengths

Method and mathematical working

Keep the unknown integer cycle count instead of immediately equating the measured phase to the full phase advance. Use the wavelength bound only after forming all positive candidates.$$\begin{aligned}\frac d\lambda&=n+\frac14,\qquad n=0,1,2,\ldots.\\[4pt]\lambda_n&=\frac{0.60}{n+1/4}\,\mathrm m,\\\lambda_0&=2.40,\quad\lambda_1=0.480,\quad\lambda_2=0.2667\quad(\mathrm m).\end{aligned}$$

(a) Calculation and conclusion

$kd=\pi/2+2\pi n$, with $n=0,1,2,\ldots$. Since $k=2\pi/\lambda$, $\lambda=.60/(n+1/4)\,\mathrm{m}$. The first candidates are $2.40,.480,.2667,\ldots\,\mathrm{m}$.

(b) Interpretation and limits

Only $.480\,\mathrm{m}$ lies in the stated range, so $n=1$. Phase measurements wrap after each complete cycle and cannot reveal the number of intervening full cycles without extra spatial or physical information.

Physical check

The candidate sequence decreases with $n$. Testing the neighboring candidates on both sides of the admitted value verifies that the interval selects exactly one wavelength.