Resolving the Sodium Doublet: Which Order, and Is the Grating at Fault?
A student uses a plane transmission grating marked '500 lines per mm' and 10 mm wide to try to resolve the sodium doublet at 589.0 nm and 589.6 nm. A collimator produces a parallel beam that falls on the full width of the grating, and the diffracted light is observed through a telescope. In the second order the student records a single line at θ = 36.1°, sees no splitting anywhere in the pattern, and writes in the report: 'the doublet needs very high resolving power, so only the highest possible order can resolve it - the fourth order, m = 4, would give the best result. Since no split is seen, the grating must be defective.' The laboratory technician insists that the grating and the lamp are both satisfactory and that the fault lies in the optical arrangement. The grating equation is d sinθ = mλ, and d follows from the ruling density.
1. Determine which orders are observable and comment on the student's claim about the fourth order.
Read the worked answer
d = 1/(500 mm⁻¹) = 1/(5.00 × 10⁵ m⁻¹) = 2.00 × 10⁻⁶ m.
From d sinθ = mλ with λ = 589.3 nm:
sinθ = m × 589.3 × 10⁻⁹/2.00 × 10⁻⁶ = 0.2947m
- m = 1: sinθ = 0.295 → θ = 17.1°
- m = 2: sinθ = 0.589 → θ = 36.1° ✓ (agrees with the student's reading)
- m = 3: sinθ = 0.884 → θ = 62.1°
- m = 4: sinθ = 1.18, which is greater than 1 - impossible
Orders 1 to 3 only. The highest available order is m = 3, so the student's 'fourth order gives the best resolution' is not merely suboptimal, it describes a maximum that does not exist. A visible maximum always has sinθ ≤ 1, and the number of observable orders is the whole number part of d/λ, here 2.00 × 10⁻⁶/5.893 × 10⁻⁷ = 3.39, i.e. 3.
Marking point: the student has confused 'highest order' with 'highest possible order'. Recognition that the missing fourth order is absent by geometry, not by defect, is what the examiner is looking for.
2. Calculate the resolving power required and the resolving power available in second order, then compare the angular separation of the doublet with the Rayleigh limit.
Read the worked answer
Required: R = λ/Δλ = 589.3/0.6 = 982
Available in second order: N = 500 lines mm⁻¹ × 10 mm = 5000 lines R = mN = 2 × 5000 = 10 000
So the grating has about ten times the resolving power needed.
Angular separation: Δθ = mΔλ/(d cosθ). With cos 36.1° = 0.808: Δθ = (2 × 0.6 × 10⁻⁹)/(2.00 × 10⁻⁶ × 0.808) = 7.4 × 10⁻⁴ rad = 0.043°
Rayleigh limit for this grating: Δθ_R = λ/(Nd cosθ) = 589.3 × 10⁻⁹/(5000 × 2.00 × 10⁻⁶ × 0.808) = 7.3 × 10⁻⁵ rad
Since 7.4 × 10⁻⁴/7.3 × 10⁻⁵ ≈ 10, the doublet should be resolved by a comfortable factor of ten. The grating is not the limiting element, which confirms the technician's view that the fault is elsewhere - the calculation, not the component, has to be examined next.
3. Give a quantitative reason why the student might see only one line, and state what should be changed.
Read the worked answer
Resolution reaches the Rayleigh limit only if roughly N lines are illuminated coherently. For a just-resolved doublet in second order:
N ≥ λ/(mΔλ) = 982/2 = 491 lines
At 500 lines mm⁻¹ that is 491/500 ≈ 1 mm of grating width. If the collimated beam is stopped down, or the source is so far away that only a sliver of the grating is lit, N falls below 500 and the two images merge into one. This is the most likely fault, and it is invisible in the final pattern.
A second, equally common cause is the entrance slit. The line seen in the telescope is an image of the source slit, so its angular width is the slit width divided by the collimator focal length. A slit subtending more than 0.043° produces two overlapping images whatever the resolving power of the grating.
Changes to make: open the beam so that it fills the full 10 mm (remove stops, move the lamp closer or use a diverging lens and a second collimator); narrow the source slit and compensate with a longer exposure or a dark-adapted eye; check alignment so the telescope is truly on the second-order maximum.
If intensity is then too low, the third order is preferable: Δθ = 3Δλ/(d cos 62.1°) = 1.9 × 10⁻³ rad = 0.11°, about 2.6 times the separation in second order, at the cost of a fainter image. Third order also gives R = 15 000.