Interference Pattern from a Diffraction Grating
A student uses a diffraction grating with 600 lines per millimetre and a red laser of wavelength 632.8 nm. The screen is 1.50 m from the grating. The student measures the distance from the central maximum to the first-order bright fringe as 0.62 m and to the second-order fringe as 1.40 m. The student's partner says the second-order measurement must be wrong because it is more than twice the first-order distance, and the fringes should be evenly spaced. The student insists the fringes are not evenly spaced.
1. Calculate the expected positions of the first- and second-order maxima and judge which partner is correct.
Read the worked answer
The grating spacing is d = 1/(600 lines/mm) = 1/600 mm = 1.67e-6 m.
For the first order: sin θ1 = λ/d = 632.8e-9/1.67e-6 = 0.379, so θ1 = 22.3 degrees. The screen distance is y1 = L tan θ1 = 1.50 × tan(22.3°) = 1.50 × 0.410 = 0.615 m. This matches the measured 0.62 m.
For the second order: sin θ2 = 2λ/d = 0.758, so θ2 = 49.3 degrees. Then y2 = 1.50 × tan(49.3°) = 1.50 × 1.163 = 1.74 m. The measured 1.40 m does not match; it is too small.
The student is correct that fringes are not evenly spaced, because the condition is d sin θ = nλ, which is linear in sin θ, not in θ or in y. The partner's 'evenly spaced' assumption is wrong. However, the second-order measurement is also inconsistent with theory, so the partner is right to distrust that number, just for the wrong reason. The discrepancy suggests a measurement error or that the screen was not perpendicular to the beam.
2. Explain why the third-order maximum cannot be observed with this setup, and what would change if the grating had 300 lines per millimetre.
Read the worked answer
For the third order, sin θ3 = 3λ/d = 3 × 0.379 = 1.14, which exceeds 1. No real angle satisfies this, so the third-order maximum does not exist for this wavelength and grating. The maximum observable order is the largest integer n with nλ/d ≤ 1, which is n = 2 here.
With 300 lines per millimetre, d doubles to 3.33e-6 m, so λ/d halves to 0.190. Then sin θ3 = 0.570, giving θ3 = 34.7 degrees and y3 = 1.50 tan(34.7°) = 1.04 m, which is observable. Higher orders become possible because a coarser grating produces smaller diffraction angles.
The judgement being tested is that the grating equation has a hard cutoff at sin θ = 1, and that reducing line density widens the available angular range. Students who assume all orders always exist will predict a third fringe that cannot appear.