Diffraction envelopes and missing interference orders
| English | Français |
|---|---|
| missing order | missing order |
| diffraction envelope | diffraction envelope |
A decision before an answer
- Two equal slits can have the right path difference for constructive interference and still deliver no intensity at that angle.
- Your goal: Derive and evaluate the far-field single-slit intensity using amplitude superposition.
Add aperture amplitudes
- For a uniformly illuminated slit of width a, far-field contributions across its opening arrive with a phase gradient k sinθ. Add their complex amplitudes before squaring: the normalised integral over x from −a/2 to a/2 is sinβ/β, where β=πa sinθ/λ. The intensity ratio is (sinβ/β)².
- At θ=0 take the limit sinβ/β→1, rather than calling the centre undefined or dark. Zeros occur at a sinθ=mλ with nonzero integer m. The central peak lies between the first zeros and is twice as wide as one adjacent zero-to-zero interval in sinθ. Side-peak maxima are not exactly halfway between their zeros.
Two equal finite slits have d=4a. Which nominal interference orders are cancelled by the first single-slit minima?
d sinθ=nλ and a sinθ=±λ coincide when n=±d/a=±4.
Convert angles to the screen
- At small angles on a distant screen L away, y≈Lθ and sinθ≈θ give first zeros y≈±Lλ/a. Thus the central width is 2Lλ/a. Use metres consistently: a millimetre slit and a nanometre wavelength differ by six powers of ten.
- For a wider angle, use θ=asin(mλ/a) and y=L tanθ; the small-angle y expression then becomes inaccurate. Far-field conditions need nearly parallel rays from different parts of the aperture, for example L much greater than a²/λ, or the equivalent focal-plane arrangement. Near-field Fresnel patterns cannot be assigned this intensity law blindly.
Double slit width a is doubled while d, λ and L stay fixed in the small-angle far field. What changes?
Spacing λL/d is unchanged; central envelope width 2λL/a halves.
Multiply interference and envelope
- For two coherent identical uniformly illuminated slits of width a and centre separation d, the normalised pattern is (sinβ/β)² cos²α, with α=πd sinθ/λ and central intensity as the normalisation. The cos² factor produces interference orders d sinθ=nλ; the single-slit factor gives the broader envelope zeros.
- Small-angle neighbouring interference spacing is λL/d. Increasing d narrows fringe spacing; increasing a narrows the envelope. These are separate changes. The equal-height narrow-slit interference formula alone does not predict the diminishing brightness or missing orders of finite apertures. Incoherent sources do not maintain the same phase-dependent cross term.
For λ=500 nm, a=0.10 mm, d=0.50 mm and L=2 m, first envelope zeros are approximately ±10 mm, central width 20 mm and interference spacing 2 mm. Since d/a=5, nominal orders ±5 are missing. Orders −4 through +4 give nine interference centres in the central envelope. At sinθ=λ/(2a), β=π/2, so the single-slit intensity is 4/π²=0.4053 of its central value.
With λ=600 nm, L=1 m and a=0.20 mm, small-angle central single-slit width is ____ mm.
2Lλ/a=0.006 m=6 mm.
Cancel coincident orders
- A missing order occurs when nλ/d=mλ/a simultaneously, giving n=m d/a. If d/a is an integer r, orders ±r, ±2r and so on are cancelled by aperture zeros. Do not count an order at an envelope boundary as a visible bright fringe.
- In the central envelope, the nominal interference-order centres satisfy |n|<d/a; when r is an integer there are 2r−1 such centres. For finite slit width, exact local maxima are shifted slightly by the changing envelope, so the interference-order locations are an approximation to observed peak centres. Identify what quantity is being requested before treating every cos² maximum as an exact maximum of the product.
Add field amplitudes before squaring, distinguish a from d, and exclude dark boundary orders. An interference maximum alone does not guarantee a maximum of the full pattern.
Which answer fits this case?
Derive and evaluate the far-field single-slit intensity using amplitude superposition
The single-slit centre is dark because sinβ/β becomes 0/0 there.
The limit is one; the apparent 0/0 is removable, and the centre is the intensity maximum.
Keep the distinctions
- diffraction envelope 衍射包络 — The aperture-dependent intensity factor that modulates the interference pattern.
- missing order 缺级 — An interference order cancelled because it coincides with an aperture intensity zero.
- Derive and evaluate the far-field single-slit intensity using amplitude superposition.
- Separate double-slit interference spacing from finite-aperture envelope width.
- Identify missing orders and state the limits of far-field and small-angle formulas.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.