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
The setup explicitly says that the full 10 mm width is illuminated, so a beam restricted to fewer than about 491 lines cannot explain the observation unless that supplied premise is wrong. With all 5000 lines illuminated, the calculated grating resolution is adequate. The entrance slit and telescope can still limit the observed image. The doublet separation in second order is approximately 7.4 times 10^-4 rad, or 0.043 degrees. If the slit image subtends an angular width comparable to or larger than that, the two images can overlap despite adequate grating resolving power. Investigate slit-image width, focus, alignment and telescope resolution, rather than declaring the grating defective. Narrowing the slit can improve its image width while reducing intensity. A third-order calculation gives a larger angular separation, approximately 0.11 degrees, but its practical usefulness also depends on intensity and instrument response. This is an optical model comparison, not evidence that the stated full-width beam was actually narrow.