衍射包络与缺失的干涉级次
GRE Subject Test · GRE 物理 知识点 32 6:28 英语旁白 · 英中内嵌字幕
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Two equal slits can have exactly the right path difference for a bright fringe and still send no light in that direction.
两条相同的狭缝可能恰好具有形成亮纹的光程差，却仍然没有光射向那个方向。
This topic builds the single-slit pattern by adding amplitudes, separates the double-slit fringe spacing from the envelope width, and finds the orders that go missing.
本主题通过叠加振幅得到单缝图样，把双缝条纹间距与包络宽度区分开，并找出缺失的级次。
For a uniformly illuminated slit of width a, far-field contributions across its opening arrive with a phase gradient k sin theta.
对于宽度为 a 的均匀照明狭缝，其开口处的远场贡献具有相位梯度 k sin θ。
Add their complex amplitudes before squaring.
在平方之前先叠加复振幅。
The normalised integral over x from negative a slash 2 to positive a slash 2 is sin beta divided by beta, where beta equals pi a sin theta divided by lambda.
从负 a/2 到正 a/2 对 x 进行归一化积分的结果为 sin β 除以 β，其中 β 等于 πa sin θ 除以 λ。
The intensity ratio is sin beta divided by beta, all squared.
强度比值为 (sin β / β)²。
At theta equals zero, take the limit of sin beta divided by beta approaching one, rather than calling the centre undefined or dark.
在 θ = 0 处，取极限 sin β / β 趋近于 1，而不是称中心未定义或黑暗。
Zeros occur at a sin theta equals m lambda, with nonzero integer m.
零点出现在 a sin θ = mλ 处，其中 m 为非零整数。
The central peak lies between the first zeros and is twice as wide as one adjacent zero-to-zero interval in sin theta.
中央主极大位于第一对零点之间，其在 sin θ 方向上的宽度是相邻一对零点间距的两倍。
Side-peak maxima are not exactly halfway between their zeros.
次级峰极大值并不严格位于其对应零点的正中间。
At small angles on a distant screen L away, y is approximately equal to L times theta and sine of theta is approximately equal to theta.
在距离为L的远处屏幕上，当角度很小时，y约等于L乘以theta，且sin theta约等于theta。
This gives first zeros at y is approximately equal to plus or minus L lambda over a.
由此可得第一极小值位置为y约等于正负L lambda除以a。
Thus the central width is 2 L lambda over a.
因此中央主极大的宽度为2L lambda除以a。
Use metres consistently: a millimetre slit and a nanometre wavelength differ by six powers of ten.
请统一使用米作为单位：一毫米狭缝与一纳米波长的量级相差六个数量级。
For a wider angle, use theta equals arcsine of m lambda over a and y equals L times tangent of theta. The small-angle y expression then becomes inaccurate.
对于较大角度，应使用theta等于m lambda除以a的反三角函数，以及y等于L乘以tan theta；此时小角度近似下的y表达式将不再准确。
Far-field conditions need nearly parallel rays from different parts of the aperture, for example L much greater than a squared over lambda, or the equivalent focal-plane arrangement.
远场条件要求来自孔径不同部分的射线近乎平行，例如L远大于a平方除以lambda，或采用等效的焦平面装置。
Near-field Fresnel patterns cannot be assigned this intensity law blindly.
近场的菲涅耳衍射图样不能盲目套用该光强分布公式。
For two coherent, identical, uniformly illuminated slits of width a and centre separation d, the normalised pattern is sine beta over beta, squared, times cosine squared alpha, with alpha equal to pi d sine theta over lambda, and the central intensity as the normalisation.
对于两条相干、相同、均匀照明的狭缝，缝宽为 a、中心间距为 d，归一化图样等于 sin β 除以 β 的平方，再乘以 cos α 的平方，其中 α 等于 π d sin θ 除以 λ，并以中央强度归一化。
The cosine squared factor produces interference orders where d sine theta equals n lambda; the single-slit factor gives the broader envelope zeros.
余弦平方因子在 d sin θ 等于 n λ 处产生干涉级次；单缝因子给出较宽的包络零点。
Small-angle neighbouring interference spacing is lambda L over d.
小角度下相邻干涉条纹间距为 λ L 除以 d。
Increasing d narrows the fringe spacing; increasing a narrows the envelope.
增大 d 使条纹间距变小；增大 a 使包络变窄。
These are separate changes.
这是两种独立的变化。
The equal-height narrow-slit interference formula alone does not predict the diminishing brightness or missing orders of finite apertures.
仅用等高窄缝干涉公式无法预测有限缝宽造成的亮度递减或缺级。
Incoherent sources do not maintain the same phase-dependent cross term.
非相干光源不能保持同样的相位相关交叉项。
A missing order occurs when n lambda over d equals m lambda over a simultaneously, giving n equals m times d over a.
缺级发生在 nλ/d 与 mλ/a 同时成立时，此时 n 等于 m 乘以 d 除以 a。
If d over a is an integer r, orders plus or minus r, plus or minus 2r, and so on are cancelled by aperture zeros.
若 d 除以 a 为整数 r，则 ±r、±2r 等级次会被孔径零点抵消。
Do not count an order at an envelope boundary as a visible bright fringe.
不要将包络边界处的级次计为可见亮纹。
In the central envelope, the nominal interference-order centres satisfy absolute value of n is less than d over a. When r is an integer, there are 2r minus 1 such centres.
在中央包络内，名义干涉级中心满足 |n| < d/a；当 r 为整数时，共有 2r−1 个这样的中心。
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.
对于有限狭缝宽度，精确局部极大值会因包络变化而略微偏移，因此干涉级位置仅为观测峰中心的近似值。
Identify what quantity is being requested before treating every cos squared maximum as an exact maximum of the product.
在将每个 cos² 极大值视为乘积的精确极大值之前，请先明确题目所求的物理量。
For two finite slits, keep the fringe spacing and the envelope apart.
对于两条有限宽度的狭缝，要把条纹间距和包络分开。
The pattern is sine beta over beta, squared, times cos squared alpha: beta uses the slit width a, alpha the separation d.
图样等于 sin β 除以 β 的平方，乘以 cos² α：β 用缝宽 a，α 用缝间距 d。
If d equals 5 a, the first envelope zero falls exactly on order 5, so orders plus and minus 5 vanish.
如果 d 等于 5a，包络的第一个零点恰好落在第 5 级上，所以正负第 5 级消失。
The central envelope holds the nine orders from minus 4 to plus 4.
中央包络内有从 −4 到 +4 的九个级次。
These are interference centres; the envelope shifts the exact peaks slightly.
这些是干涉中心；包络会使实际峰值略微偏移。
A double slit has d equals three a over two.
双缝间距 d 等于 3a/2。
Which positive interference orders are missing?
哪些正级干涉条纹会缺失？
How many nominal interference centres lie strictly between the first envelope zeros?
在第一级包络零点之间，严格来说有多少个名义干涉极大中心？
Explain why rounding d over a to an integer is invalid.
请解释为什么将 d/a 四舍五入为整数是无效的。
Four marks.
【4分】
Coincidence requires n equals m times d over a, which is 3m over 2, with both m and n being nonzero integers.
重合条件要求 n 等于 m 乘以 d 除以 a，即 3m 除以 2，且 m 和 n 均为非零整数。
Thus m must be even, and the missing orders are n equals 3, 6, 9, and so on.
因此 m 必须为偶数，缺级顺序为 n 等于 3、6、9 等。
Central maxima obey the absolute value of n less than 3 over 2, so n equals negative 1, 0, and 1, giving three orders.
中央主极大满足 |n| 小于 3/2，故 n 取 -1、0、1，共三个级次。
The fractional ratio d over a describes actual geometry; rounding changes which zeros coincide.
d/a 的分数形式描述实际几何结构；四舍五入会改变哪些极小值发生重合。
Award integer condition, missing sequence, count, and explanation.
评分要点包括：整数条件、缺级序列、计数及解释。
Add field amplitudes before squaring, distinguish a from d, and exclude dark boundary orders.
在平方之前先叠加场振幅。 区分a与d，并排除暗边界级次。
An interference maximum alone does not guarantee a maximum of the full pattern.
仅凭干涉极大值并不能保证整个图样中一定存在极大值。
diffraction envelope: The aperture-dependent intensity factor that modulates the interference pattern.
衍射包络。 调制干涉图样的孔径相关强度因子。
missing order: An interference order cancelled because it coincides with an aperture intensity zero.
缺级。 因与孔径强度零点重合而被抵消的干涉级次。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不阅读定义的情况下回忆其含义，然后更改原始传递中的一个参数并检验极限情况。