回折エンベロープおよび欠失干渉順
導入済み| English | 意味 |
|---|---|
| missing order/ˈmɪsɪŋ ˈɔːdə/ | 欠級 |
| diffraction envelope/dɪˈfrækʃn ˈenvələʊp/ | 回折包絡線 |
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?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
このケースに当てはまる答えはどれか？
振幅の重ね合わせを用いて遠方領域単スリット強度を導出・評価する
The single-slit centre is dark because sinβ/β becomes 0/0 there.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.