Fine structure of sodium: D-line separation with a Michelson interferometer

Theory

The sodium yellow light is a doublet: D₂ (3p ²P₃/₂ → 3s ²S₁/₂) at 588.995 nm and D₁ (3p ²P₁/₂ → 3s) at 589.592 nm, with intensity ratio 2 : 1 (the ratio of the statistical weights 2J+1). Moving mirror M₂ by \(d\) changes the path difference by \(2d\). At the centre of the pattern, each line gives

$$ I_i \propto 1+\cos\!\left(\frac{4\pi d}{\lambda_i}\right) .$$

The two fringe systems drift in and out of step. Their sum has high contrast when the bright fringes coincide and minimum contrast when bright falls on dark. Successive visibility minima are separated by

$$ \Delta d = \frac{\lambda_1\lambda_2}{2(\lambda_2-\lambda_1)} \approx \frac{\bar\lambda^2}{2\Delta\lambda}\;\;\Rightarrow\;\; \Delta\lambda = \frac{\bar\lambda^2}{2\,\Delta d} \approx 0.597\ \text{nm},\quad \Delta d \approx 0.291\ \text{mm}.$$

Because D₂ is twice as strong as D₁, the visibility never reaches zero: \(V_{min} = (2-1)/(2+1) = 1/3\).

Counting \(N\) fringes that pass the centre while the mirror moves \(\Delta x\) gives the mean wavelength \(\bar\lambda = 2\Delta x/N\).

Fine-structure splitting

$$ \Delta E = hc\,\frac{\Delta\lambda}{\bar\lambda^2} = hc\,\Delta\tilde\nu,\qquad \Delta\tilde\nu \approx 17.2\ \text{cm}^{-1}\ (2.13\ \text{meV}).$$

This is the spin–orbit interaction \(\xi\,\mathbf L\cdot\mathbf S\) in the 3p level, split as \(\Delta E = \tfrac32\xi\) between J = 3/2 and J = 1/2.

Procedure

1. Adjust the interferometer for circular fringes (M₁ ⟂ M₂′). The compensating plate equalises the glass path of the two arms.
2. λ̄: record the micrometer reading and reset the counter. Run the motor (or turn the fine drum) until about 100–500 fringes have passed the centre, then record the reading and count again.
3. Δλ: turn the coarse wheel until the fringes almost disappear, and refine with the fine drum to the position of poorest contrast. Record it. Move on to the next minimum, about 0.29 mm further, and repeat for 5–8 successive minima.
4. Plot the minimum positions against their order k. The slope is Δd.

Micrometer

Main scale in mm, coarse drum 0.01 mm per division, fine drum 0.0001 mm (0.1 µm) per division. In this simulation the coarse wheel and fine buttons move M₂ directly.

Precautions & errors

• Judging the visibility minimum by eye is the main error, typically ±10–20 µm. Use many minima and a straight-line fit.
• Remove backlash by always turning in the same direction.
• Count fringes at the centre; miscounting by one in 500 gives 0.2% error in λ̄.

Viva questions

Why does the visibility never become zero?

Because the D₂ and D₁ intensities are unequal (2 : 1). At the minima, the bright fringes of D₂ fall on the dark fringes of D₁, but D₂ still dominates.

Why are the fringes circular?

The path difference 2d cos θ depends only on the angle θ to the axis, so the fringes are of equal inclination (Haidinger fringes) and are seen at infinity.

What would happen with white light?

Fringes appear only within a few micrometres of zero path difference, because the coherence length of white light is about 1 µm. This is used to find the position d = 0.