Polarization — Question 4

Section 33-6 · Physics pages 5478–5479

Question 4 Analyzer extrema measure the linearly polarized part of a mixture A beam is an incoherent sum of x-polarized intensity Ip and unpolarized intensity Iu. A rotating ideal analyzer records a maximum of 7.0 W/m squared and a minimum of 2.0 W/m squared. x y axis incoherent polarized + unpolarized input Schematic; labels, not drawing lengths, specify the data. (a) Derive the analyzer intensity and infer Ip and Iu. (b) Find total incident intensity and the degree of linear polarization, defined as Ip divided by the total.
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Question 4 - Solution Analyzer extrema measure the linearly polarized part of a mixture x y axis incoherent polarized + unpolarized input (a) Method and calculation I(θ) = Ip cos2 θ + Iu/2, Imax = Ip + Iu/2, Imin = Iu/2. Ip = Imax −Imin = 5.0 W/m2, Iu = 2Imin = 4.0 W/m2 . (b) Interpretation and further calculation Iin = Imax + Imin = 9.0 W/m2 . PL = Imax −Imin Imax + Imin = 5/9 . Physical check The minimum is half the unpolarized component, not the entire unpolarized incident intensity.

Question 4

Analyzer extrema measure the linearly polarized part of a mixture

A beam is an incoherent sum of x-polarized intensity Ip and unpolarized intensity Iu. A rotating ideal analyzer records a maximum of 7.0 W/m squared and a minimum of 2.0 W/m squared.

Diagram shown in the original SVG worksheet.

  1. Derive the analyzer intensity and infer Ip and Iu.

  2. Find total incident intensity and the degree of linear polarization, defined as Ip divided by the total.

Question 4 - Solution

Analyzer extrema measure the linearly polarized part of a mixture

Diagram shown in the original SVG worksheet.

(a) Method and calculation

I(θ)=Ipcos⁡2θ+Iu/2,Imax=Ip+Iu/2,Imin=Iu/2.I(\theta)=I_p\cos^2\theta+I_u/2,\quad I_{\max}=I_p+I_u/2,\ I_{\min}=I_u/2.Ip=Imax−Imin=5.0W/m2,Iu=2Imin=4.0W/m2.\boxed{I_p=I_{\max}-I_{\min}=5.0\,\mathrm{W/m^2},\quad I_u=2I_{\min}=4.0\,\mathrm{W/m^2}}.

(b) Interpretation and further calculation

Iin=Imax+Imin=9.0W/m2.\boxed{I_{\mathrm{in}}=I_{\max}+I_{\min}=9.0\,\mathrm{W/m^2}}.𝒫L=Imax−IminImax+Imin=5/9.\boxed{\mathcal P_L=\frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}}=5/9}.

Physical check

The minimum is half the unpolarized component, not the entire unpolarized incident intensity.