Using a grating to measure wavelength
Shine the light normally on a grating of known d, measure the angle θ of a chosen order n, and calculate λ = d sin θ / n. Measure the same order on both sides of the normal and take half the angle between them to remove any error in locating the normal. Higher orders give a larger θ and therefore a smaller percentage uncertainty, but are dimmer and so harder to locate.
λ = (d sin θ)/n · θₙ = ½(θ_right − θ_left)What this actually means
The method is three steps: know d from the lines-per-mm marking, measure θ for a known order, substitute. That is the whole of learning outcome (m); the structure of a spectrometer is not required.
Measuring both sides and halving is the standard precision improvement, and it is worth a mark. It cancels any systematic error in setting the zero of the angular scale.
Justify the higher-order choice properly. A fixed absolute uncertainty of, say, 0.1° is a much smaller fraction of 50° than of 15°, so the percentage uncertainty in θ and therefore in λ falls.
The counter-argument is intensity. Higher orders sit lower on the single-slit diffraction envelope, so they are fainter and the crosswire is harder to centre. State both sides if the question says discuss.
A grating beats a double slit for this job because its maxima are far sharper and sit at large, easily measured angles, so the position of each maximum is pinned down much more precisely.
Measuring the angle from the grating surface instead of from the normal, or forgetting to divide by the order n.
Prove it — watch it be true
- Switch to the measure-λ grading mode with the wavelength hidden.
- Read the angle of the first-order maximum on both sides and take half the difference.
- Compute λ = d sin θ / n and check your value against the graded answer, then repeat with the second order and compare the uncertainties.