The grating equation d sin θ = nλ
● VERBATIM — examined word-for-wordFor a diffraction grating with slit separation d, the principal maxima occur at angles given by d sin θ = nλ, where n = 0, 1, 2, … is the order. If the grating is specified as N lines per metre, then d = 1/N. Every pair of adjacent slits contributes a path difference of d sin θ, so at these angles all the slits are in phase.
d sin θ = nλ · d = 1/NWhat this actually means
The condition is between adjacent slits, and that is enough. If neighbouring slits differ by exactly nλ, then slits two apart differ by 2nλ, three apart by 3nλ, and so on, so every one of the thousands of beams arrives in phase.
Get d right first. A grating marked 600 lines per mm has N = 6.00 × 10⁵ lines per metre, so d = 1.67 × 10⁻⁶ m. Converting per-mm to per-metre is where most grating marks are lost.
Angles are large here, often tens of degrees, so you must use sin θ. Never approximate sin θ ≈ tan θ ≈ θ on a grating problem.
This gives maxima, not minima. Compare with sin θ = λ/b for a single slit, which gives minima. Different letters (d versus b) for exactly that reason.
The angular gaps between orders are not equal. From the RI worked example with λ = 656 nm and d = 2.50 µm, the orders sit at 15.2°, 31.7° and 51.9°, so the gaps grow: 15.2°, then 16.5°, then 20.2°.
Leaving d in lines per millimetre, or using tan θ instead of sin θ because the small-angle habit carried over from double slits.
Prove it — watch it be true
- Set the grating to 600 lines per mm and the wavelength to 656 nm.
- Read the true ray angles for n = 1, 2 and 3 and check each against d sin θ = nλ by hand.
- Confirm from the order table that the angular gap between consecutive orders increases with n.