Coherence — Question 5

Section 35-4 · Physics pages 5780–5781

Question 5 An extended uniform source can erase otherwise stable fringes A uniform incoherent line source of width W lies distance S before two slits of spacing d. A source element at coordinate h contributes phase offset 2πdh/(λS). Its fringe intensity adds incoherently to those of the other elements. Define γ = (1/W) ∫W/2 −W/2 ei2πdh/(λS)dh and V = |γ|. P(y) d L W Schematic; labels, not drawing lengths, specify the data. (a) Evaluate the integral and derive the first source width producing zero contrast. (b) For λ = 600 nm, S = 0.50 m, and d = 0.30 mm, calculate that width and the visibility at half that width.
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Question 5 - Solution An extended uniform source can erase otherwise stable fringes P(y) d L W (a) Method and calculation γ = sin[πdW/(λS)] πdW/(λS) . πdW0/(λS) = π =⇒W0 = λS/d . (b) Interpretation and further calculation W0 = 1.00 mm . W = W0/2 : V = 2/π = 0.63662 . Physical check Each source element can produce a clear pattern, while their differently shifted patterns cancel the total modulation.

Question 5

An extended uniform source can erase otherwise stable fringes

A uniform incoherent line source of width W lies distance S before two slits of spacing d. A source element at coordinate h contributes phase offset 2πdh/(λS)2\pi dh/(\lambda S). Its fringe intensity adds incoherently to those of the other elements. Define γ=(1/W)∫−W/2W/2ei2πdh/(λS)dh\gamma=(1/W)\int_{-W/2}^{W/2}e^{i2\pi dh/(\lambda S)}dh and V=|γ|V=|\gamma|.

Diagram shown in the original SVG worksheet.

  1. Evaluate the integral and derive the first source width producing zero contrast.

  2. For λ=600nm\lambda=600\,\mathrm{nm}, S=0.50mS=0.50\,\mathrm m, and d=0.30mmd=0.30\,\mathrm{mm}, calculate that width and the visibility at half that width.

Question 5 - Solution

An extended uniform source can erase otherwise stable fringes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

γ=sin⁡[πdW/(λS)]πdW/(λS).\boxed{\gamma=\frac{\sin[\pi dW/(\lambda S)]}{\pi dW/(\lambda S)}}.πdW0/(λS)=π⇒W0=λS/d.\pi dW_0/(\lambda S)=\pi\Longrightarrow\boxed{W_0=\lambda S/d}.

(b) Interpretation and further calculation

W0=1.00mm.\boxed{W_0=1.00\,\mathrm{mm}}.W=W0/2:V=2/π=0.63662.W=W_0/2:\quad\boxed{V=2/\pi=0.63662}.

Physical check

Each source element can produce a clear pattern, while their differently shifted patterns cancel the total modulation.