= Magnetic reversal measurement of Faraday rotation
{c}
{title2=$\theta_{\rm odd}(B)=[\theta_{\rm obs}(B)-\theta_{\rm obs}(-B)]/2$}
Consider an original model at fixed <vacuum> <wavelength> $600\,\mathrm{nm}$ and <temperature> $20^\circ\mathrm C$. Stipulate a transparent sample with <Verdet constant> $V=20\,\mathrm{rad\,T^{-1}\,m^{-1}}$, length $L=0.050\,\mathrm m$, uniform adjustable $B_z$, and a field-independent analyzer-zero offset $\theta_0=0.020\,\mathrm{rad}$. These are chosen values, not measured glass properties. On a continuously tracked small-<angle> branch,
$$
\theta_{\rm obs}(B)=\theta_0+VLB,\qquad
\theta_{\rm obs}(+0.10\,\mathrm T)=0.120\,\mathrm{rad},\qquad
\theta_{\rm obs}(-0.10\,\mathrm T)=-0.080\,\mathrm{rad}.
$$
Taking half the difference gives $\theta_{\rm odd}=0.100\,\mathrm{rad}$; half the sum recovers $\theta_0=0.020\,\mathrm{rad}$. Hence $V=\theta_{\rm odd}/(LB)=20\,\mathrm{rad\,T^{-1}\,m^{-1}}$. A field-reversal difference of $0.200\,\mathrm{rad}$ is twice the single-field rotation, not that rotation itself. More generally this subtraction removes every contribution that is even under $B\mapsto-B$, provided the sample and apparatus remain otherwise matched.
\Image[/faraday-magnetic-reversal-inference.png]
{title=Chosen <Faraday rotation> model: reversing the <magnetic field> removes a constant angular offset}
{description=Lines and marked values are exact synthetic predictions for $V=20\,\mathrm{rad\,T^{-1}\,m^{-1}}$, $L=0.050\,\mathrm m$ and offset $0.020\,\mathrm{rad}$. Left: observed orientation and its odd part. Right: normalized <Malus's law> curves at $B_z=-0.10,0,+0.10\,\mathrm T$; shifted maxima recover the three orientations. The $45^\circ$ analyzer line illustrates a signed intensity response. No points are experimental observations.}
An ideal <polarimeter analyzer> at laboratory <angle> $a$ reads normalized <light intensity> $I(a,B)/I_0=\cos^2[a-\theta_{\rm obs}(B)]$. Here $I_0$ is the <light intensity> immediately before the analyzer, separately calibrated for each condition. A scan locates the maximum at $a=\theta_{\rm obs}$ modulo $\pi$. At $a=\pi/4$, expansion about zero gives $I/I_0=1/2+\theta_{\rm obs}+O(\theta_{\rm obs}^3)$, with <angles> in <radians>. At crossed extinction, the response instead begins as $\theta_{\rm obs}^2$ and loses the sign near zero. The favorable $45^\circ$ slope follows directly by differentiating <Malus's law>; its sign reverses if the angular convention is reversed.
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