Young's Interference Experiment — Question 7

Section 35-3 · Physics pages 5764–5765

Question 7 A controlled source phase creates a fractional fringe translation Two narrow slits have equal amplitudes and an imposed relative phase. At screen coordinate y the total phase is δ(y) = 2πdy/(λL) + π/2. Let d = 0.25 mm, λ = 500 nm, and L = 1.00 m. P(y) d L ϕ0 = π/2 Schematic; labels, not drawing lengths, specify the data. (a) Find the bright fringe closest to the origin and the two neighboring dark fringes. (b) Find the phase change needed to move that bright fringe upward by 0.20 mm without relabeling its order.
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Question 7 - Solution A controlled source phase creates a fractional fringe translation P(y) d L ϕ0 = π/2 (a) Method and calculation ∆y = λL/d = 2.00 mm; ym = (m −1/4)∆y. ybright = −0.500 mm ; ydark = (m + 1/4)∆y ⇒−1.500, +0.500 mm . (b) Interpretation and further calculation δϕ0 + 2π ∆yδy = 0. δϕ0 = −2π(0.20/2.00) = −0.20π rad . Physical check The direction of translation depends on the explicitly defined phase convention; holding the same order fixes the sign.

Question 7

A controlled source phase creates a fractional fringe translation

Two narrow slits have equal amplitudes and an imposed relative phase. At screen coordinate y the total phase is δ(y)=2πdy/(λL)+π/2\delta(y)=2\pi dy/(\lambda L)+\pi/2. Let d=0.25mmd=0.25\,\mathrm{mm}, λ=500nm\lambda=500\,\mathrm{nm}, and L=1.00mL=1.00\,\mathrm m.

Diagram shown in the original SVG worksheet.

  1. Find the bright fringe closest to the origin and the two neighboring dark fringes.

  2. Find the phase change needed to move that bright fringe upward by 0.20mm0.20\,\mathrm{mm} without relabeling its order.

Question 7 - Solution

A controlled source phase creates a fractional fringe translation

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Δy=λL/d=2.00mm;ym=(m−1/4)Δy.\Delta y=\lambda L/d=2.00\,\mathrm{mm};\quad y_m=(m-1/4)\Delta y.ybright=−0.500mm;ydark=(m+1/4)Δy⇒−1.500,+0.500mm.\boxed{y_{\mathrm{bright}}=-0.500\,\mathrm{mm}};\quad y_{\mathrm{dark}}=(m+1/4)\Delta y\Rightarrow\boxed{-1.500,\ +0.500\,\mathrm{mm}}.

(b) Interpretation and further calculation

δϕ0+2πΔyδy=0.\delta\phi_0+\frac{2\pi}{\Delta y}\delta y=0.δϕ0=−2π(0.20/2.00)=−0.20πrad.\boxed{\delta\phi_0=-2\pi(0.20/2.00)=-0.20\pi\ \mathrm{rad}}.

Physical check

The direction of translation depends on the explicitly defined phase convention; holding the same order fixes the sign.