Young's Interference Experiment — Question 10

Section 35-3 · Physics pages 5770–5771

Question 10 A uniform medium and a changed slit spacing can preserve a fringe pattern An experiment in air uses wavelength λ0 = 600 nm, slit spacing d1 = 0.40 mm, and screen distance L1 = 2.0 m. It is rebuilt entirely in a uniform medium of index n = 1.50 with L2 = 3.0 m. Neglect dispersion. P(y) d L uniform medium n Schematic; labels, not drawing lengths, specify the data. (a) Find the new slit spacing needed to preserve the paraxial fringe spacing. (b) If instead the old slit spacing is retained, find the spacing ratio and determine whether the full angular bright-fringe pattern is also preserved.
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Question 10 - Solution A uniform medium and a changed slit spacing can preserve a fringe pattern P(y) d L uniform medium n (a) Method and calculation ∆y1 = λ0L1/d1, ∆y2 = λ0L2/(nd2). ∆y2 = ∆y1 =⇒d2 = d1L2/(nL1) = 0.40 mm . (b) Interpretation and further calculation d2 = d1 : ∆y2/∆y1 = L2/(nL1) = 1 ; ∆y1 = 3.00 mm. sin θm,2 = mλ0 nd1 = sin θm,1 n ⇒angular pattern contracts . Physical check Equal paraxial screen spacing can conceal unequal angular spacing; exact large-angle screen positions would not remain identical.

Question 10

A uniform medium and a changed slit spacing can preserve a fringe pattern

An experiment in air uses wavelength λ0=600nm\lambda_0=600\,\mathrm{nm}, slit spacing d1=0.40mmd_1=0.40\,\mathrm{mm}, and screen distance L1=2.0mL_1=2.0\,\mathrm m. It is rebuilt entirely in a uniform medium of index n=1.50n=1.50 with L2=3.0mL_2=3.0\,\mathrm m. Neglect dispersion.

Diagram shown in the original SVG worksheet.

  1. Find the new slit spacing needed to preserve the paraxial fringe spacing.

  2. If instead the old slit spacing is retained, find the spacing ratio and determine whether the full angular bright-fringe pattern is also preserved.

Question 10 - Solution

A uniform medium and a changed slit spacing can preserve a fringe pattern

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Δy1=λ0L1/d1,Δy2=λ0L2/(nd2).\Delta y_1=\lambda_0L_1/d_1,\qquad\Delta y_2=\lambda_0L_2/(nd_2).Δy2=Δy1⇒d2=d1L2/(nL1)=0.40mm.\Delta y_2=\Delta y_1\Longrightarrow\boxed{d_2=d_1L_2/(nL_1)=0.40\,\mathrm{mm}}.

(b) Interpretation and further calculation

d2=d1:Δy2/Δy1=L2/(nL1)=1;Δy1=3.00mm.d_2=d_1:\quad\boxed{\Delta y_2/\Delta y_1=L_2/(nL_1)=1};\quad\Delta y_1=3.00\,\mathrm{mm}.sin⁡θm,2=mλ0nd1=sin⁡θm,1n⇒angular pattern contracts.\sin\theta_{m,2}=\frac{m\lambda_0}{nd_1}=\frac{\sin\theta_{m,1}}{n}\Rightarrow\boxed{\text{angular pattern contracts}}.

Physical check

Equal paraxial screen spacing can conceal unequal angular spacing; exact large-angle screen positions would not remain identical.