Grating Order Misidentified
A student uses a diffraction grating marked 300 lines/mm with a green laser labelled 532 nm. The grating is perpendicular to the laser beam, and a screen is 1.20 m away. The first-order spot is measured 0.194 m from the central maximum. A brighter spot is measured 0.650 m from the central maximum, and the student labels it second order. Using that assignment, the student computes a wavelength near 0.79 µm, which contradicts the laser label. The student cannot find a spot at the position predicted for second order and concludes that the grating is mislabelled. The laboratory instructor suspects that the 0.650 m spot is a higher order and that the true second-order spot is present but dimmer or overlooked. The student must decide whether the grating spacing is consistent with 300 lines/mm, identify the order of the 0.650 m spot, and judge whether a 1% wavelength measurement is realistic from the second-order position.
1. Use the first-order spot at 0.194 m to calculate the grating spacing in lines per millimetre. Is it consistent with the marked 300 lines/mm?
Read the worked answer
For a screen distance L = 1.20 m and first-order position y = 0.194 m, tanθ = y/L = 0.194/1.20 = 0.1617, so θ = 9.19°. Then sinθ = 0.1597. For first order, d sinθ = λ, so d = 532 × 10⁻⁹/0.1597 = 3.33 × 10⁻⁶ m.
The line density is 1/d = 1/(3.33 × 10⁻⁶ m) = 3.00 × 10⁵ lines/m, which is 300 lines/mm. The first-order measurement is therefore consistent with the marked grating. The grating spacing itself is not the problem; the order assignment for the 0.650 m spot is the likely error.
2. Identify the order of the spot at 0.650 m and calculate where the true second-order spot should be. What wavelength would the student infer if the spot were second order?
Read the worked answer
For y = 0.650 m, tanθ = 0.650/1.20 = 0.5417, so θ = 28.45° and sinθ = 0.4765. The order is m = d sinθ/λ = (3.33 × 10⁻⁶ × 0.4765)/(532 × 10⁻⁹) = 2.98, which is third order, not second.
The true second-order angle satisfies sinθ2 = 2λ/d = 2 × 532 × 10⁻⁹/(3.33 × 10⁻⁶) = 0.3192, so θ2 = 18.61°. Its position is y2 = L tanθ2 = 1.20 × 0.3366 = 0.404 m. The student should look at 0.404 m, not 0.650 m.
If the 0.650 m spot were forced to be second order, the inferred wavelength would be λ = d sinθ/m = (3.33 × 10⁻⁶ × 0.4765)/2 = 7.94 × 10⁻⁷ m, or 0.79 µm. That is visibly inconsistent with a 532 nm laser, which is the clue that the order assignment is wrong.
3. The student wants to use the 0.404 m spot to determine wavelength to 1%. Estimate the uncertainty from a ±2 mm position reading and a ±5 mm screen-distance measurement. Is the target realistic?
Read the worked answer
Use λ = (d/m)(y/sqrt(L² + y²)) with m = 2, y = 0.404 m, L = 1.20 m, and d = 3.33 × 10⁻⁶ m. The position sensitivity is dλ/dy = (d/2)L²/(L² + y²)^(3/2). The denominator is (1.44 + 0.163)^(3/2) = 1.603^(3/2) = 2.03. Thus dλ/dy = (1.667 × 10⁻⁶)(1.44)/2.03 = 1.18 × 10⁻⁶ per metre. A ±0.002 m position uncertainty gives about ±2.4 nm.
The screen-distance sensitivity has magnitude (d/2)yL/(L² + y²)^(3/2) = (1.667 × 10⁻⁶)(0.404)(1.20)/2.03 = 3.98 × 10⁻⁷ per metre. A ±0.005 m distance uncertainty gives about ±2.0 nm. Combining these in quadrature gives about 3.1 nm, or 0.6% of 532 nm.
From geometry alone, a 1% measurement is realistic. However, the grating line spacing and the spot-centring procedure also contribute. If the grating is only known to 1%, that alone consumes the entire budget. The student should calibrate d from a known wavelength or use a grating with a certified spacing before claiming 1%.