Michelson's Interferometer — Question 4

Section 35-7 · Physics pages 5838–5839

Question 4 A slight mirror tilt creates straight fringes across the output beam A Michelson mirror tilts by a small angle α, so the two recombined plane waves differ in direction by approximately 2α. At transverse coordinate y use δ(y) = δ0 + 2kαy, where k = 2π/λ. At λ = 600 nm the measured fringe spacing is p = 3.00 mm. two output directions fringes Schematic; labels, not drawing lengths, specify the data. (a) Infer the mirror tilt in radians and degrees. (b) Calculate the number of full fringe intervals across a 12.0 mm beam and the spacing if the mirror tilt doubles.
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Question 4 - Solution A slight mirror tilt creates straight fringes across the output beam two output directions fringes (a) Method and calculation δ(y + p) −δ(y) = 2kαp = 2π =⇒p = λ/(2α). α = 1.00 × 10−4 rad = 0.0057296◦. (b) Interpretation and further calculation N = W/p = 12.0/3.00 = 4 intervals . α →2α =⇒p →p/2 = 1.50 mm . Physical check The reflected ray turns through twice the mirror angle. Omitting that factor would double the inferred tilt.

Question 4

A slight mirror tilt creates straight fringes across the output beam

A Michelson mirror tilts by a small angle α\alpha, so the two recombined plane waves differ in direction by approximately 2α2\alpha. At transverse coordinate y use δ(y)=δ0+2kαy\delta(y)=\delta_0+2k\alpha y, where k=2π/λk=2\pi/\lambda. At λ=600nm\lambda=600\,\mathrm{nm} the measured fringe spacing is p=3.00mmp=3.00\,\mathrm{mm}.

Diagram shown in the original SVG worksheet.

  1. Infer the mirror tilt in radians and degrees.

  2. Calculate the number of full fringe intervals across a 12.0mm12.0\,\mathrm{mm} beam and the spacing if the mirror tilt doubles.

Question 4 - Solution

A slight mirror tilt creates straight fringes across the output beam

Diagram shown in the original SVG worksheet.

(a) Method and calculation

δ(y+p)−δ(y)=2kαp=2π⇒p=λ/(2α).\delta(y+p)-\delta(y)=2k\alpha p=2\pi\Longrightarrow p=\lambda/(2\alpha).α=1.00×10−4rad=0.0057296∘.\boxed{\alpha=1.00\times10^{-4}\,\mathrm{rad}=0.0057296^\circ}.

(b) Interpretation and further calculation

N=W/p=12.0/3.00=4intervals.\boxed{N=W/p=12.0/3.00=4\ \text{intervals}}.α→2α⇒p→p/2=1.50mm.\alpha\to2\alpha\Longrightarrow\boxed{p\to p/2=1.50\,\mathrm{mm}}.

Physical check

The reflected ray turns through twice the mirror angle. Omitting that factor would double the inferred tilt.