Excitation and Polarization of Light

[scarecrow]

Let's say I have a molecule with a carbonyl (C=O) group on it. Let's assume that the carbonyl group is positioned in a such a way that when we look at it the carbon is below the oxygen (so it's vertical). Therefore, it is stretching up and down.

What would happen if p-polarized light (E vector is vertical) is directed at the carbonyl group? S-polarized light (E vector is horizontal)?

I believe that the carbonyl stretch will be excited by the p-polarized light since both vectors are aligned. Nothing will happen to the C=O stretch if s-polarized light is directed at it since the vectors are perpendicular.

Am I on the right path?



Thanks

 

[Loren Booda]

I think in terms of diffraction gratings, with many parallel slits, or hydrocarbon chains. Let randomly polarized light pass through a first grating - yielding vertically polarized light - and then through a similar, second grating. The light remains vertically polarized. If, however, the second polarizer is oriented perpendicular (now horizontal) to the first, the light will effectively be blocked.

 

[scarecrow]

I think in terms of diffraction gratings, with many parallel slits, or hydrocarbon chains. Let randomly polarized light pass through a first grating - yielding vertically polarized light - and then through a similar, second grating. The light remains vertically polarized. If, however, the second polarizer is oriented perpendicular (now horizontal) to the first, the light will effectively be blocked.

I think you misunderstood me or I wasn't clear enough. I wanted to know what happens to the carbonyl stretch.

Anyways, I think I know the answer now. If the electric dipole moment vector (u) is perpendicular to the Electric Field vector (in this case it would mean the light is s-polarized), then the dot product is zero (H' = u*E) and then H' = 0 (perturbed Hamiltonian).

The other case would be that H' is nonzero if the two vectors are not perpendicular. Therefore the carbonyl stretch would be excited.

 

[Loren Booda]

Thanks for pointing that out. I tried to reason that if the incident polarization were perpendicular to the structure of the polarizer in question, the wave would be absorbed. In this logic of mine, which appears to be incorrect, the absorption would then indicate excitation. Your formula is more likely sound.