Polarization — Question 1

Section 33-6 · Physics pages 5472–5473

Question 1 A single intensity measurement leaves two analyzer orientations A linearly x-polarized beam of intensity I0 passes through an ideal analyzer whose axis is at angle theta to x. The transmitted intensity is I0/4. Axes are equivalent modulo 180 degrees. x y axis Schematic; labels, not drawing lengths, specify the data. (a) Find all analyzer angles in the interval from zero through less than 180 degrees. (b) For each angle, predict the transmitted fraction after a small positive rotation delta, to first order with delta in radians.
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Question 1 - Solution A single intensity measurement leaves two analyzer orientations x y axis (a) Method and calculation I/I0 = cos2 θ = 1/4 ⇒| cos θ| = 1/2. θ = 60◦or 120◦. (b) Interpretation and further calculation I(θ + δ) I0 = cos2 θ −sin 2θ δ + O(δ2). I(60◦+ δ)/I0 ≃1/4 − √ 3δ/2, I(120◦+ δ)/I0 ≃1/4 + √ 3δ/2 . Physical check The intensity alone determines an unsigned projection; the response to a known rotation resolves the two choices.

Question 1

A single intensity measurement leaves two analyzer orientations

A linearly x-polarized beam of intensity I0 passes through an ideal analyzer whose axis is at angle theta to x. The transmitted intensity is I0/4. Axes are equivalent modulo 180 degrees.

Diagram shown in the original SVG worksheet.

  1. Find all analyzer angles in the interval from zero through less than 180 degrees.

  2. For each angle, predict the transmitted fraction after a small positive rotation delta, to first order with delta in radians.

Question 1 - Solution

A single intensity measurement leaves two analyzer orientations

Diagram shown in the original SVG worksheet.

(a) Method and calculation

I/I0=cos⁡2θ=1/4⇒|cos⁡θ|=1/2.I/I_0=\cos^2\theta=1/4\Rightarrow|\cos\theta|=1/2.θ=60∘or120∘.\boxed{\theta=60^\circ\ \text{or}\ 120^\circ}.

(b) Interpretation and further calculation

I(θ+δ)I0=cos⁡2θ−sin⁡2θδ+O(δ2).\frac{I(\theta+\delta)}{I_0}=\cos^2\theta-\sin2\theta\,\delta+O(\delta^2).I(60∘+δ)/I0≃1/4−3δ/2,I(120∘+δ)/I0≃1/4+3δ/2.\boxed{I(60^\circ+\delta)/I_0\simeq1/4-\sqrt3\delta/2,\quad I(120^\circ+\delta)/I_0\simeq1/4+\sqrt3\delta/2}.

Physical check

The intensity alone determines an unsigned projection; the response to a known rotation resolves the two choices.