17-Phys-A7 Optics · December 2013

Question 4 of 6: Polarization, a quarter-wave plate, and Fresnel reflection at a water surface

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4: Polarization, a quarter-wave plate, and Fresnel reflection at a water surface (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — polarization states and their transformation. The unique states are unpolarized, linear (at some azimuth), circular (left/right), elliptical (left/right, the general case), and partially polarized (a polarized state plus an unpolarized remainder).

Part (b).

1. QWP thickness. A quarter-wave plate needs a birefringent path-length difference of $\lambda/4$: $$(n_e-n_o)\,d=\frac{\lambda}{4}\Rightarrow d=\frac{\lambda}{4(n_e-n_o)}=\frac{600\text{ nm}}{4(1.5005-1.5)}=\boxed{300\ \mu\text{m}}$$

Parts (c)/(d) — Fresnel reflectance, internal and external. Given. Water $n_w=1.33$, air $n_a=1.00$.

1. Part (c) — internal reflection (water → air). Normal incidence: $R=\left(\dfrac{n_w-n_a}{n_w+n_a}\right)^2=\left(\dfrac{0.33}{2.33}\right)^2=\boxed{2.0\%}$ (both polarizations, T=98.0%). Brewster angle (R∥=0): $\theta_B=\arctan(n_a/n_w)=\boxed{36.9^\circ}$. Critical angle: $\sin\theta_c=n_a/n_w\Rightarrow\boxed{\theta_c=48.75^\circ}$, beyond which $R=1$ for both polarizations (total internal reflection, T=0).
2. Part (d) — external reflection (air → water). Normal incidence: same magnitude, $R=\boxed{2.0\%}$. Brewster angle: $\theta_B=\arctan(n_w/n_a)=\boxed{53.1^\circ}$ (R∥=0). No critical angle exists going into the denser medium: R rises smoothly to 100% only as $\theta\to90^\circ$ (grazing incidence).

Part (e). Glare off a horizontal surface like water or a wet road is reflected at close to the EXTERNAL Brewster angle ($53.1^\circ$ from the vertical here), where the reflected light is nearly 100% polarized perpendicular to the plane of incidence — i.e. horizontally, since the plane of incidence for glare off a horizontal surface is vertical. Polarizing sunglasses are most effective at rejecting this glare when their transmission axis is oriented vertically, blocking the horizontally polarized reflected component while passing the (unpolarized) light scattered from the scene itself.