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Physics, 05.06.2020 15:59 dianamunoz580

When discussing the dispersion in gases, we model the gas molecule as a damped harmonic oscillator, and obtain the induced dipole moment by an incident EM wave. Since we know that an oscillating dipole will radiate, the gas molecule will scatter the incident wave by such re-radiation. (a) Find the general expression for the scattering cross section of the gas molecule.

(b) When considering the sunlight scattered by molecules in the atmosphere (known as the Rayleight scattering), we know that the sunlight frequency (optical range) is much smaller than the resonance frequency which is typically in the ultraviolet range, and we can further ignore the damping. Derive the approximation of the scattering cross section (textbook Eq. 10.41). The strong frequency dependence explains why the sky looks blue. Next let’s analyze the polarization of the scattered light. Consider the incident light traveling in the z direction.

(c) First, assume that the incident light is linearly polarized along the x axis. What is the polarization direction of the light scattered into the z direction and the y direction, respectively?

(d) Next, consider unpolarized incident light, i. e., its E^-> field direction remains in the x-y plane, but the exact orientation fluctuates randomly and uniformly along all directions in the x-y plane. Argue that the scattered light along the z-direction is also unpolarized, but the scattered light along the x and y directions is fully polarized. The scattered light along other directions is partially polarized.

This effect helps photographers to obtain a stronger contrast of the blue sky. Since the blue light scattered by the atmosphere is typically partially polarized while the scattered light by clouds is unpolarized (because such scattered light experiences multiple reflections and scatterings by the dense clouds thus loses its polarization), photographers use a polarizer to block the polarized blue light thus reduce the brightness of the blue (so the sky looks more blueish) which increases the contrast of the picture. Such effect is strongest when the camera is pointed at 90 degrees from the sunlight.

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