Diffraction by a Double Slit — Question 3

Section 36-6 · Physics pages 5956–5957

Question 3 Exactly missing orders require a rational separation-to-width ratio A double slit has d/a = √ 2. Nominal bright orders satisfy d sin θ = mλ; envelope zeros satisfy a sin θ = pλ for nonzero integer p. Assume any orders discussed are propagating. d P slit width a d/a = √ 2 or 7/5 Schematic; labels, not drawing lengths, specify the data. (a) Prove that no nonzero nominal bright order is exactly missing. (b) If fabrication changes the ratio to d/a = 7/5, identify the first positive missing order and the corresponding envelope zero.
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Question 3 - Solution Exactly missing orders require a rational separation-to-width ratio d P slit width a d/a = √ 2 or 7/5 (a) Method and calculation Coincidence =⇒d a = m p . √ 2 /∈Q =⇒no nonzero exact missing order . (b) Interpretation and further calculation d/a = 7/5 : 5m = 7p =⇒m = 7j, p = 5j. m = 7, p = 5 for the first positive coincidence . Physical check An irrational ratio can still produce very weak orders close to an envelope zero; absence of exact cancellation does not guarantee visibility.

Question 3

Exactly missing orders require a rational separation-to-width ratio

A double slit has d/a=2d/a=\sqrt2. Nominal bright orders satisfy dsin⁡θ=mλd\sin\theta=m\lambda; envelope zeros satisfy asin⁡θ=pλa\sin\theta=p\lambda for nonzero integer p. Assume any orders discussed are propagating.

Diagram shown in the original SVG worksheet.

  1. Prove that no nonzero nominal bright order is exactly missing.

  2. If fabrication changes the ratio to d/a=7/5d/a=7/5, identify the first positive missing order and the corresponding envelope zero.

Question 3 - Solution

Exactly missing orders require a rational separation-to-width ratio

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Coincidence⇒da=mp.\text{Coincidence}\Longrightarrow\frac da=\frac mp.2∉ℚ⇒no nonzero exact missing order.\sqrt2\notin\mathbb Q\Longrightarrow\boxed{\text{no nonzero exact missing order}}.

(b) Interpretation and further calculation

d/a=7/5:5m=7p⇒m=7j,p=5j.d/a=7/5:\quad5m=7p\Longrightarrow m=7j,\ p=5j.m=7,p=5for the first positive coincidence.\boxed{m=7,\quad p=5\quad\text{for the first positive coincidence}}.

Physical check

An irrational ratio can still produce very weak orders close to an envelope zero; absence of exact cancellation does not guarantee visibility.