17-Phys-A7 Optics · Undated paper

Question 4 of 10: Polarization — sunglasses, Jones matrices

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

Question 4: Polarization — sunglasses, Jones matrices (7 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.

a) Sunglasses and reflected glare. Glare is light that has reflected off a horizontal surface (water, wet road, snow) at a large angle of incidence; near the Brewster angle for that surface, the reflected light is strongly and predominantly linearly polarized parallel to the reflecting surface (horizontal). A polarizing sunglass lens has its transmission axis oriented vertically, so it strongly attenuates this horizontally-polarized reflected glare while passing most of the (unpolarized, or vertically-polarized) directly-viewed light — the effect is destructive absorption of one linear component by the polarizing filter, not interference.

b) Jones matrix of the polarization-converting component.

Given. $S_{in}=\begin{pmatrix}1\\1\end{pmatrix}$, $S_{out}=\begin{pmatrix}1\\i\end{pmatrix}$, unknown $M=\begin{pmatrix}E&F\\G&H\end{pmatrix}$ with exactly two elements equal to zero.

Find. $E,F,G,H$.

Approach. $M\,S_{in}=S_{out}$ gives two equations, $E+F=1$ and $G+H=i$, in four unknowns — underdetermined in general, but requiring two elements to vanish picks out a unique, physically natural solution: the diagonal choice $F=0,\ G=0$ (so the $x$- and $y$-components of the field are handled independently, with no cross-coupling between them — the signature of a wave plate with its fast/slow axes aligned to $x,y$).

1. Diagonal ansatz. With $F=G=0$: $E=1$ (from $E+F=1$) and $H=i$ (from $G+H=i$).
2. Result. $M=\boxed{\begin{pmatrix}1&0\\0&i\end{pmatrix}}.$ Check: $M\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}1(1)+0(1)\\0(1)+i(1)\end{pmatrix}=\begin{pmatrix}1\\i\end{pmatrix}=S_{out}$ ✓. $|\det M|=|i|=1$, confirming $M$ is unitary (lossless), as a real wave-plate should be.

c) Name of the component. $M=\text{diag}(1,i)$ retards the $y$-component of the field by $90^\circ$ ($e^{i\pi/2}$) relative to the $x$-component while leaving each component’s amplitude unchanged — this is exactly the action of a quarter-wave plate with its fast axis along $x$ (slow axis along $y$). A $45^\circ$-oriented input like $S_{in}=(1,1)$ becomes circularly polarized output, matching the input/output pair given.

d) Output for $S_{in}=(1,0)$.

Approach. Apply the same $M$ found in (b) to the new input.

1. Multiply. $S_{out}=M\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}1&0\\0&i\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\boxed{\begin{pmatrix}1\\0\end{pmatrix}}$ — unchanged.

Light polarized purely along the plate’s fast axis carries no component along the slow axis, so the quarter-wave plate has nothing to retard — light launched along either principal axis of a wave plate always emerges with its polarization state unaltered (only its overall phase can shift), which is exactly what the calculation reproduces.