31 Jul 2025

Production and Detection of Polarized Light and Optical Activity

Polarizer-analyzer tests for linear, circular, and elliptical light, Malus's law, and optical rotation from circular birefringence.

waves-and-optics polarized-light malus-law polarization-detection optical-activity

An ideal polarizer transmits the electric-field component along its transmission axis. The same element used to test an unknown state is called an analyzer.

Malus’s law and the test for plane polarization

Let plane-polarized light of amplitude $E_0$ meet an analyzer whose axis $\hat{\mathbf a}$ makes angle $\theta$ with the incident vibration direction $\hat{\mathbf p}$. Projection on the analyzer axis gives

\[\mathbf E_{\mathrm{out}} =(E_0\hat{\mathbf p}\cdot\hat{\mathbf a})\hat{\mathbf a} =E_0\cos\theta\,\hat{\mathbf a}.\]

Since time-averaged intensity is proportional to the squared amplitude,

\[\boxed{I=I_P\cos^2\theta}.\]

This is Malus’s law. A rotating analyzer gives two maxima and two complete extinctions in one revolution, identifying plane-polarized light. Plane polarization can be produced by a Nicol prism or by reflection at the Brewster angle.

An unpolarized beam has all transverse azimuths with equal probability. Averaging Malus’s factor gives

\[\left\langle\cos^2\theta\right\rangle =\frac{1}{2\pi}\int_0^{2\pi}\cos^2\theta\,\mathrm d\theta =\frac12,\]

so an ideal polarizer transmits half its intensity.

Circular polarization

To produce circular polarization:

1. use a polarizer to obtain a linear field;
2. set its vibration at $45^\circ$ to the fast and slow axes of a quarter-wave plate;
3. let the plate introduce a relative phase $\pm\pi/2$ between the equal components.

The resulting field can be written

\[E_x=E_0\cos\psi, \qquad E_y=\pm E_0\sin\psi.\]

For an analyzer at azimuth $\beta$,

\[E_a=E_0(\cos\psi\cos\beta \pm\sin\psi\sin\beta).\]

Time averaging gives

\[\left\langle E_a^2\right\rangle =\frac{E_0^2}{2}(\cos^2\beta+\sin^2\beta) =\frac{E_0^2}{2},\]

independent of $\beta$. A rotating analyzer alone therefore cannot distinguish circular light from unpolarized light. Insert a quarter-wave plate first: it adds or removes a quarter-wave retardation, turning circular light into linear light. The following analyzer then gives complete extinction.

Elliptical polarization

Linearly polarized light incident on a quarter-wave plate at an angle other than $0^\circ$, $45^\circ$, or $90^\circ$ has two unequal non-zero components in quadrature and becomes elliptically polarized. In axes along the ellipse,

\[E_x=a\cos\psi, \qquad E_y=b\sin\psi, \qquad a\ne b.\]

A rotating analyzer transmits average intensity proportional to

\[\left\langle(E_x\cos\beta+E_y\sin\beta)^2\right\rangle =\frac12\left(a^2\cos^2\beta+b^2\sin^2\beta\right).\]

It varies between values proportional to $a^2$ and $b^2$ but never vanishes when both axes are non-zero. To confirm elliptical polarization, align a quarter-wave plate with the ellipse axes. It cancels their $\pi/2$ phase difference, producing linear light; a following analyzer then gives extinction.

A rotating analyzer detects linear light directly; a quarter-wave plate before the analyzer distinguishes circular and elliptical light from unpolarized light.

Optical activity

An optically active medium rotates the azimuth of plane-polarized light without changing it into an ellipse in the ideal lossless case. The mechanism is circular birefringence: left- and right-circular components propagate with different refractive indices $n_L$ and $n_R$.

Choose circular unit vectors

\[\hat{\mathbf e}_L=\frac{\hat{\mathbf x}-i\hat{\mathbf y}}{\sqrt2}, \qquad \hat{\mathbf e}_R=\frac{\hat{\mathbf x}+i\hat{\mathbf y}}{\sqrt2}.\]

A field initially along $x$ is their equal superposition:

\[\hat{\mathbf x}=\frac{\hat{\mathbf e}_L+\hat{\mathbf e}_R}{\sqrt2}.\]

After travelling distance $l$ through the medium, the two phase advances are

\[\phi_L=\frac{2\pi n_Ll}{\lambda}, \qquad \phi_R=\frac{2\pi n_Rl}{\lambda}.\]

With $\bar\phi=(\phi_L+\phi_R)/2$ and $\Delta\phi=\phi_L-\phi_R$, the output field is

\[\begin{aligned} \mathbf E_{\mathrm{out}} &\propto \frac{e^{i\phi_L}\hat{\mathbf e}_L +e^{i\phi_R}\hat{\mathbf e}_R}{\sqrt2}\\ &=e^{i\bar\phi} \left[ \hat{\mathbf x}\cos\left(\frac{\Delta\phi}{2}\right) +\hat{\mathbf y}\sin\left(\frac{\Delta\phi}{2}\right) \right]. \end{aligned}\]

The common phase $e^{i\bar\phi}$ does not affect the vibration direction. The plane has rotated through

\[\boxed{\alpha=\frac{\Delta\phi}{2} =\frac{\pi l}{\lambda}(n_L-n_R)},\]

with the sign fixed by the circular-basis and viewing convention. A rotating analyzer measures this rotation because its extinction position shifts by $\alpha$.

For a solution of concentration $c$, the specific rotation at stated temperature $T$ and wavelength $\lambda$ is defined by

\[\boxed{[\alpha]^T_\lambda=\frac{\alpha}{lc}}.\]

When $l$ is measured in decimetres and $c$ in $\mathrm{g\,mL^{-1}}$, its conventional unit is $\mathrm{degree\,dm^{-1}(g\,mL^{-1})^{-1}}$. Positive and negative rotations are called dextrorotatory and levorotatory, respectively, after the observation convention has been fixed.

Circular birefringence rotates the linear vibration by half the relative circular phase; the analyzer minimum shifts by the same angle.

The analyzer averages and circular-basis recombination are checked with zero residuals in the Unit III polarization Maxima worksheet.