# Dependence of the angle of rotation on the wavelength of plane polarized light

Wavelength difference is a big deal, I know. It can solely change the whole interaction between the chiral molecule & the light. But I am not sure what's the mechanism by which light of different wavelengths produces different angles of rotation.

So, why and how wavelength matters?

The rotation of plane polarised light by a solution of, say, sucrose depend on the ability of the oscillating magnetic filed of the light to induce an electric dipole moment in the molecule and the ability of the oscillating electric field of the light to induce a magnetic dipole moment.

For these interaction to have any magnitude it is supposed that the electrons in a molecule move in a helical path, or, alternatively that there are two linear dipoles generated by electron motion which are in planes that are at some angle to one another. Although this model is clearly artificial it emphasises the fact that movement of charge must follow somewhat crooked pathways under the influence of the radiation. The theory of this is very complicated but the result from a quantum calculation is relatively straightforwards and is that the molecular rotation M at wavelength $\lambda$ is given by

$$M_\lambda = a\sum_i \frac{\lambda_{0i}^2X}{\lambda^2-\lambda_{0i}^2}$$

where i represents all the electronic states of the molecule and the wavelength $\lambda_{0i} = c/\omega_{0i}$ where $\omega_{0i} = (E_i-E_0)/\hbar$ is the frequency of the $i^{th}$ excited state at energy $E_i$. The parameter $X=\mathrm{Im}(\mu_{m_{0i}}\cdot\mu_{e_{i0}})$ is the value imaginary part of the complex dot product of the induced magnetic and electric dipole moments and a are a set of constants independent of wavelength.

From this formula it is seen that the polarisation rotation depends on the wavelength. (The dependence of rotation angle on wavelength has been called optical rotatory dispersion.) We can also conclude

• that experiments should be performed where the molecule has little absorbance but that the rotation will be larger as an absorption band is approached. (in an absorption band elliptically polarised light can be formed)

• The strength of any optical transition is not important as the dot product X depends on induced moments and the angle between them. This also means that weak optical transitions can be as important as strong one in rotation the polarisation. If the induced dipoles are perpendicular then the rotation vanishes as the dot product is zero.

• That many excited states i can be involved and their effects can cancel to some extent so the value of the rotation is hard to predict.

• The polarisation rotation of two mirror image molecules are equal and opposite in size and if a molecule is identical with its mirror image polarisation rotation must be zero and the term X is zero.

• When the wavelength $\lambda < \lambda_{0i}$ i.e. the wavelength crosses a transition then the signal changes sign and this is called the Cotton effect.

• 'strength of any optical transition' what do you mean by it? "it is supposed that the electrons in a molecule move in a helical path" is this assumption rational and legitimate? – Mockingbird May 1 '17 at 1:02
• 'strength of transition' just means how intense the absorption spectrum is, i.e. how big the extinction coefficient is. The 'helical path' thing is just a model to understand the effect. Linear dipoles at an angle work just as well and is more intuitive. It simply means that the radiation induces some complicated electron response that can be broken down into a helical path among other motions. – porphyrin May 1 '17 at 7:20
• The text book by Kuhn & Forsterling ' Principles of Physical Chemistry' section 8.11 has a brief explanation but an old book W. Kauzmann 'Quantum Chemistry' has the all the gory details of how a chiral molecule can have a spiral path of induced electron motion. – porphyrin May 1 '17 at 7:47
• Any internet link? – Mockingbird May 2 '17 at 2:27

Light travels at different speeds in different materials. The ratio of the speed at which light travels in a vacuum compared to the speed light travels in a given material is given by the index of refraction ($n$) of the material.

$${{n}=\mathrm{\frac{speed~ of~ light ~in~ vacuum}{speed ~of~ light~ in~ material}}}$$

The index of refraction is a basic property of a material. Further, the index of refraction of a material is wavelength dependent.

The beam of plane polarized light that we use in our polarimeter experiments is created by combining a beam of left circularly polarized light with a beam of right circularly polarized light. Those two beams interfere with each other to produce the plane polarized light we use in the polarimeter experiment. It is important to note that both the right and left circularly polarized beams are chiral as they trace out a right and left handed helix.

If the material in the polarimeter is not chiral, then the angle of the plane polarized light will not be rotated as it passes through the sample. However, if a chiral material is placed in the path of a plane polarized beam, then the left circularly polarized component of the beam will be rotated a different amount than the right circularly polarized component (the 2 beams are chiral and interact differently with the chiral sample) and we observe a rotation of our light beam as it passes through the polarimeter. In other words, the index of refraction of the left circularly polarized light ($n_L$) is different from the index of refraction of the right circularly polarized light ($n_R$) in chiral media.

Since $n$ itself is wavelength dependent, both $n_L$ and $n_R$ are also wavelength dependent and the observed rotation from our sample will change as we change the wavelength of our light beam.

This property of $n_L$ and $n_R$ is the basis of the Cotton effect. If you would like to read more about the Cotton effect and its use in chemistry see this earlier answer.

• "combining a beam of left circularly polarized light with a beam of right circularly polarized light" would you elaborate it? – Mockingbird May 1 '17 at 7:30
• @Mockingbird Circularly polarized light can be used to create plane polarized light and vice-versa, see here for example. Therefore, a plane polarized beam of light may be treated as the combination of two circularly polarized in-phase beams of the same frequency and each with half the amplitude of the resultant plane polarized beam. Viewing a plane polarized beam this way makes it easier to understand its interaction with chiral molecules. – ron May 1 '17 at 14:01
• So it means the plane polarized light becomes circularly polarised light as the 2 circularly constituent polarized lights get out of phase. But, the resultant electric wave vector would be rotating, so the measured rotation angle will depend on the length of the polarimeter, right? – Mockingbird May 2 '17 at 0:01
• @Mockingbird Very close. The plane polarized light does not "become" circularly polarized. Rather the plane polarized light can be viewed as the superposition of 2 circularly polarized beams (one right circularly polarized and the other left). Since the 2 circularly polarized beams interact diasterotopically with the chiral media (they have different indices of refraction in chiral media), the plane of polarization rotates as the beam travels through the chiral sample. So yes, the observed rotation is dependent upon polarimeter length. – ron May 2 '17 at 0:12
• They differ only when they pass through chiral media. – ron May 2 '17 at 13:56