Diffraction Grating and a Misread Laser Label
A student shines a red laser through a diffraction grating labelled "600 lines per mm" and observes bright spots on a screen 1.20 m away. The first-order maximum appears 0.52 m from the central maximum. The student computes a wavelength and gets 722 nm, which seems too high for a red laser (typical 630–680 nm). Suspicious, the student checks the grating and finds the label actually reads 500 lines per mm - the "600" was a misprint on an old sticker. The screen distance and spot position are unchanged.
1. Recompute the wavelength using the correct grating spacing and explain the direction of the error.
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Grating equation: $d\sin\theta = m\lambda$.
Spacing for 500 lines/mm: $d = \dfrac{1}{500\ \text{mm}^{-1}} = 2.0\times10^{-6}$ m.
The angle from geometry: $\tan\theta = \dfrac{0.52}{1.20} = 0.433$, so $\theta \approx 23.4°$, $\sin\theta \approx 0.397$.
$\lambda = d\sin\theta = 2.0\times10^{-6} \times 0.397 \approx 7.9\times10^{-7}$ m $\approx 794$ nm - even higher, so something else is off.
Recheck: the student's earlier 722 nm used the wrong $d = 1/600$ mm $= 1.67\times10^{-6}$ m, giving $\lambda = 1.67\times10^{-6}\times0.397 \approx 662$ nm - which is actually reasonable for red! So the misprint was the correct value; the corrected 500 lines/mm gives an implausible 794 nm. The conclusion: the grating really is 600 lines/mm and the "correction" is the error. The lesson is to check consistency against a known physical range (red ≈ 630–680 nm).
2. Using the self-consistent value (600 lines/mm, λ ≈ 662 nm), predict where the second-order maximum would appear on the screen.
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For $m = 2$: $\sin\theta_2 = \dfrac{2\lambda}{d} = \dfrac{2 \times 6.62\times10^{-7}}{1.67\times10^{-6}} = \dfrac{1.324\times10^{-6}}{1.67\times10^{-6}} \approx 0.794$.
$\theta_2 \approx 52.6°$.
Screen position $y = L\tan\theta_2 = 1.20 \times \tan 52.6° = 1.20 \times 1.307 \approx 1.57$ m from the centre.
Note the second-order spot is far out and the small-angle approximation $y \approx mL\lambda/d$ would badly underestimate it here (it would give ~0.95 m), because $\theta$ is large. This case shows why the exact $\tan\theta$ relation must be used for high orders.