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.

Solved Problems

1. Successive polarizer and analyzer

Unpolarized light of intensity $12.0\,\mathrm{W\,m^{-2}}$ passes through an ideal polarizer and then an analyzer whose axis is $30^\circ$ from the polarizer axis. Find the final intensity.

Step 1: Average over the unpolarized input. An ideal polarizer transmits half:

\[I_P=\frac{I_0}{2}=6.00\,\mathrm{W\,m^{-2}}.\]

Step 2: Apply Malus’s law to the analyzer.

\[\begin{aligned} I&=I_P\cos^230^\circ\\ &=6.00\left(\frac{\sqrt3}{2}\right)^2\\ &=4.50\,\mathrm{W\,m^{-2}}. \end{aligned}\]

The total transmission fraction is $(1/2)(3/4)=3/8$, and $(3/8)(12.0)=4.50$, which independently checks the result.

2. Specific rotation and a second solution

A solution rotates plane-polarized light through $+13.2^\circ$ in a tube of length $2.00\,\mathrm{dm}$ at concentration $0.200\,\mathrm{g\,mL^{-1}}$. Find its specific rotation and predict the rotation for a $3.00\,\mathrm{dm}$ tube at concentration $0.100\,\mathrm{g\,mL^{-1}}$ at the same temperature and wavelength.

Step 1: Calculate the specific rotation.

\[[\alpha]^T_\lambda =\frac{\alpha}{lc} =\frac{13.2}{(2.00)(0.200)} =33.0\,\mathrm{degree\,dm^{-1}(g\,mL^{-1})^{-1}}.\]

Step 2: Apply it to the second tube.

\[\alpha^{\prime}=[\alpha]^T_\lambda l^{\prime}c^{\prime} =(33.0)(3.00)(0.100)=9.90^\circ.\]

The positive sign is retained because neither the substance nor the viewing convention has changed.

Descriptive Questions

1. How does a rotating analyzer distinguish plane-polarized light from circular and unpolarized light?
2. How are circularly polarized light and its handedness produced and detected with a quarter-wave plate?
3. Why does elliptically polarized light show unequal analyzer maxima and minima but no complete extinction?
4. How does circular birefringence rotate a plane-polarized vibration without making it elliptical in an ideal medium?

Numerical Problems

1. Without an active sample, an analyzer gives extinction at $15^\circ$. After inserting the sample, the nearest recorded extinction is at $167^\circ$. Since an analyzer axis repeats after $180^\circ$, find the signed rotation of smallest magnitude.

   Answer: $\alpha=(167^\circ-15^\circ)-180^\circ=-28.0^\circ$, a negative rotation under the stated convention.

2. Circularly polarized light has intensity $8.00\,\mathrm{W\,m^{-2}}$. Find the intensity after an ideal linear analyzer at any azimuth.

   Answer: $I=4.00\,\mathrm{W\,m^{-2}}$, independent of analyzer azimuth.

3. Principal electric-field amplitudes of elliptically polarized light are in the ratio $3:1$. Find the ratio of maximum to minimum analyzer intensities and state whether complete extinction occurs.

   Answer: $I_{\max}/I_{\min}=9$; no complete extinction occurs.

4. An optically active medium has $n_L-n_R=2.00\times10^{-6}$ at $\lambda=500\,\mathrm{nm}$. Find the rotation after $l=0.100\,\mathrm{m}$.

   Answer: $\alpha=0.400\pi\,\mathrm{rad}=72.0^\circ$.

The solved results and all numerical answers are verified by exact residuals in the Unit III polarization Maxima worksheet.

References

1. Optical rotation - Wikipedia
2. F. A. Jenkins and H. E. White, Fundamentals of Optics, McGraw-Hill, sections on polarization analysis and optical activity.
3. Max Born and Emil Wolf, Principles of Optics, Cambridge University Press, sections on polarization and circular birefringence.
4. Ajoy Ghatak, Optics, McGraw Hill Education, chapters on production and detection of polarized light.