Length — Question 2

Question 2

An optical ruler in air

The SI metre is defined through the fixed vacuum light speed $c=299\,792\,458\ \mathrm{m/s}$ and the second. For this exercise, light of frequency $f$ has wavelength $$\lambda_0=\frac{c}{f}\quad\text{in vacuum},\qquad \lambda_{\mathrm{air}}=\frac{c}{nf}\quad\text{in air},$$ where the refractive index $n$ describes the reduction of the wave’s phase speed. Take $f=5.000\times10^{14}\ \mathrm{s^{-1}}$ and $n=1.00030$ as exact model inputs.

An ideal measurement determines that a one-way air path contains exactly $N=1\,000\,000$ spatial wavelengths. Thus its length is $L=N\lambda_{\mathrm{air}}$. Here $N$ counts wavelengths along the path; it is not a round-trip interference-fringe count.

1. Find the vacuum wavelength and the wavelength in air, in nanometres. Which is shorter, and why?

2. Find the air-path length in metres. A program instead multiplies $N$ by $\lambda_0$. Find its length error in millimetres and its fractional error relative to the correct length.

3. Does using air require defining a different metre? Explain the distinction between changing a unit definition and correcting a measurement model for the medium.

Keep sufficient digits until the final comparison. No optical theory beyond the supplied wavelength model is required.

Question 2 - Solution

An optical ruler in air

(a) The frequency is shared; the phase speed changes.

$$\frac{\lambda_{\mathrm{air}}}{\lambda_0}=\frac1n<1,\qquad \frac{L_{\mathrm{naive}}}{L_{\mathrm{true}}}=n.$$

$$\lambda_0=\frac{299\,792\,458}{5.000\times10^{14}}\ \mathrm{m} =\boxed{599.584916\ \mathrm{nm}},$$ $$\lambda_{\mathrm{air}}=\frac{599.584916}{1.00030}\ \mathrm{nm} \approx\boxed{599.405094\ \mathrm{nm}}.$$ At the same frequency, the smaller phase speed $c/n$ gives a smaller distance per cycle. The wavelength in air is therefore shorter.

(b) Use the wavelength actually present in the path.

$$L=\frac{0.599584916}{1.00030}\ \mathrm{m} \approx\boxed{0.599405094\ \mathrm{m}}.$$ The faulty program gives $L_{\mathrm{wrong}}=0.599584916\ \mathrm{m}$. Its error is positive: $$L_{\mathrm{wrong}}-L\approx\boxed{0.179822\ \mathrm{mm}}.$$ Since $L_{\mathrm{wrong}}=nL$, its fractional error relative to the correct length is $$\frac{L_{\mathrm{wrong}}-L}{L}=n-1=0.00030=\boxed{0.030\%}.$$ Using vacuum wavelengths assigns too much length to each wavelength counted in air.

(c) Correct the realization, not the definition.

The metre remains the same SI unit. The model links the measured wave spacing in air to that unit through $c$, $f$, and $n$. A vacuum definition does not require every practical measurement to occur in vacuum; it requires the relevant medium effects to be accounted for.

The long decimals here result from exact stipulated inputs. In a real realization, frequency, refractive index, and path determination have uncertainties, so these computed digits alone would not justify an equally precise measured length.