The Diffraction Grating and Measuring Wavelength
For normally incident coherent monochromatic light in the far field, a grating's principal reinforcement directions satisfy d sin(theta) = n lambda, with theta measured from the grating normal. Many illuminated slits make narrow principal maxima useful for wavelength measurement.
What a learner can do afterwards
- Converts lines per millimetre into a slit spacing before using the grating equation
- Finds a wavelength from a measured angle for a stated order
- Uses the sine bound to determine whether positive grating orders exist, finds the largest permitted positive order when one exists, and distinguishes permitted orders from observable maxima
1 · Read
Line density N tells you how many grating lines fit in one millimetre. Centre-to-centre spacing is d = (1/N) mm = 0.001/N metres. A density is not a spacing. One millimetre is 0.001 m, and one nanometre is 10^−9 m.
Assume coherent monochromatic light falls perpendicularly on an evenly spaced transmission grating, with observation in the far field. Let d be centre-to-centre slit spacing, lambda the wavelength, n an integer order and theta the angle from the grating normal. Neighbouring paths differ by d sin(theta). Full reinforcement requires n wavelengths: d sin(theta) = n lambda. Order zero is straight ahead.
Orders require |n| lambda/d ≤ 1. If d/lambda < 1, no positive order exists; zero remains. Otherwise the largest permitted positive order is floor(d/lambda), including grazing at 90 degrees from the normal, which misses an ordinary forward screen. Each finite slit has its own diffraction pattern. This slit envelope varies the grating peaks' brightness; its zeros can hide permitted orders. Allowed orders can be weak, so the sine bound does not guarantee observation.
To calculate sin(theta), rearrange the normal-incidence equation to sin(theta) = n lambda/d. Convert wavelength and spacing to the same units first. For d = 4.0 × 10^−6 m, lambda = 800 nm and n = 1, the ratio is 0.2. A ratio above 1 would mean that order has no physical angle.
For fixed spacing, wavelength and illumination per slit, increasing the number of illuminated slits narrows principal maxima. At their centres all contributions reinforce. Away from them, differing phases produce stronger cancellation and nearby minima. A finite grating still has finite-width principal maxima and weaker secondary maxima, not zero intensity everywhere else. A slit-envelope minimum can suppress a principal order.
Keep the units attached to the reciprocal density. If N is lines per millimetre, 1/N is a spacing in millimetres. Calling that number metres skips a factor of 0.001. Check the operation separately from the unit: dividing by the density is correct, but the resulting millimetres still need conversion.
The unit changes after conversion. A reciprocal density does not automatically give metres.
To recover wavelength, rearrange d sin(theta) = n lambda to lambda = d sin(theta)/n. Use a known nonzero order and theta measured from the grating normal. For normal incidence with d = 2.4 × 10^−6 m, sin(theta) = 0.5 and n = 3, lambda = 4.0 × 10^−7 m.
Convert density to spacing. For normal incidence, use d sin(theta) = n lambda with theta from the normal. The sine bound permits orders; it does not promise they are visible.
Original lesson, questions and flashcards by the dataset authors.
The rewritten readings were written for this dataset. Videos credited to Khan Academy.
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