Exercises — Polarization — Malus's law, Brewster's angle derivation

WHAT kind of input? Unpolarized — $\vec{E}$ points in every transverse direction randomly. WHY not Malus directly? Malus needs a single angle $\theta$. Unpolarized light has no single angle; it is an even mix of all of them. So we average ${\cos }^2\theta$ over a full turn, which gives $\frac{1}{2}$. $I=\frac{1}{2}{I}_0.$ The output is now polarized along this polarizer's axis (important for the next problems).

Brewster's law is $\tan {\theta }_B=n_2/n_1$, where $n_2$ is the medium light goes into and $n_1$ the medium it comes from. $\tan {\theta }_B=\frac{2.42}{1}=\mathrm{2.42.}$ That's the equation. (Solving it: ${\theta }_B=\arctan 2.42\approx 67.5^{\circ }$ — done fully in L2.)

$I=I_0{\cos }^{2}90^{\circ }=I_0\cdot 0^2=0.$ WHY zero? $\cos 90^{\circ }=0$: the wiggle is entirely perpendicular to the slot, so no component lies along the axis. This is a crossed polarizer.

Step 1: $\tan {\theta }_B=2.42\Rightarrow {\theta }_B=\arctan (2.42)=67.54^{\circ }.$ Step 2 (WHY we can get ${\theta }_r$ for free): at Brewster's angle the reflected and refracted rays are perpendicular, so ${\theta }_r=90^{\circ }-{\theta }_B=22.46^{\circ }.$ Check with Snell: $\frac{\sin 67.54^{\circ }}{\sin 22.46^{\circ }}=\frac{0.924}{0.382}=2.42$ ✓.

Through A: angle $0^{\circ }$, so $I_A=I_0{\cos }^2{0}^{\circ }=I_0$ (A is aligned — nothing lost). Through B: light leaving A is polarized along A's axis, so it meets B at $60^{\circ }$: $I_B=I_A{\cos }^{2}60^{\circ }=I_0\cdot {\left(\frac{1}{2}\right)}^2=\frac{1}{4}{I}_0=0.25I_0.$

Step 1 (A): input unpolarized → halve: $I_A=\frac{1}{2}{I}_0$, now polarized along A. Step 2 (B): polarized light meets B at $30^{\circ }$: $I_B=I_A{\cos }^{2}30^{\circ }=\frac{1}{2}{I}_0\cdot {\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{1}{2}{I}_0\cdot \frac{3}{4}=\frac{3}{8}{I}_0=0.375I_0.$

Angles between consecutive polarizers are what matter — this is the key idea in the figure above. After A: unpolarized halved → $I_A=\frac{1}{2}{I}_0$, polarized at $0^{\circ }$. After B: angle from A to B is $30^{\circ }$: $I_B=I_A{\cos }^{2}30^{\circ }=\frac{1}{2}{I}_0\cdot \frac{3}{4}=\frac{3}{8}{I}_0.$ Light is now polarized along B (at $30^{\circ }$). After C: angle from B ($30^{\circ }$) to C ($75^{\circ }$) is $75^{\circ }-30^{\circ }=45^{\circ }$: $I_C=I_B{\cos }^{2}45^{\circ }=\frac{3}{8}{I}_0\cdot \frac{1}{2}=\frac{3}{16}{I}_0=0.1875I_0.$

A → B: angle $\varphi$, so $I_B=I_0{\cos }^2\varphi$ (A aligned, no loss at A). B → C: C is at $90^{\circ }$, B at $\varphi$, so the angle between them is $90^{\circ }-\varphi$: $I_C=I_B{\cos }^2(90^{\circ }-\varphi )=I_0{\cos }^2\varphi {\sin }^2\varphi .$ WHY light reappears: with no middle filter the field along C's axis is zero. The middle filter re-projects the field onto a slanted axis, giving it a nonzero component along C. Maximize: use $\cos \varphi \sin \varphi =\frac{1}{2}\sin 2\varphi$, so $I_C=\frac{1}{4}{I}_0{\sin }^{2}2\varphi$. This peaks when $\sin 2\varphi =1$, i.e. $2\varphi =90^{\circ }\Rightarrow \varphi =45^{\circ }$, giving $I_C^{\max }=\frac{1}{4}{I}_0=0.25I_0.$

(a) Air → water: $\tan {\theta }_B=1.33\Rightarrow {\theta }_B=\arctan (1.33)=53.06^{\circ }.$ (b) The glare is polarized horizontally; the glasses pass vertical. Angle between them is $90^{\circ }$: $I=I_{\text{glare}}{\cos }^{2}90^{\circ }=0.$ They block all of the Brewster-reflected glare. (c) At ${\theta }_B$ the reflected ray is 100% polarized perpendicular to the plane of incidence — that plane is vertical for a horizontal lake, so the polarization is horizontal. A vertical filter is exactly crossed with it, so glare is killed while ordinary (partly vertical) scenery light survives.

After A: $\frac{1}{2}{I}_0$. After B: $I=\frac{1}{2}{I}_0{\cos }^2\theta$. Set equal to threshold: $\frac{1}{2}{I}_0{\cos }^2\theta =0.30I_0\Rightarrow {\cos }^2\theta =0.60\Rightarrow \cos \theta =\mathrm{0.7746.}$ $\theta =\arccos (0.7746)=39.23^{\circ }.$ Any larger $\theta$ dims the output below threshold, so ${\theta }_{\max }=39.23^{\circ }$.

(a) From glass to air: $n_2=1$ (into air), $n_1=1.5$ (from glass): $\tan {\theta }_B^{\prime }=\frac{n_{\text{air}}}{n_{\text{glass}}}=\frac{1}{1.5}=0.6667\Rightarrow {\theta }_B^{\prime }=\arctan (0.6667)=33.69^{\circ }.$ This equals the earlier refraction angle ${\theta }_r$ — the picture above shows why the internal Brewster ray retraces the external refracted ray. (b) ${\theta }_B+{\theta }_B^{\prime }=56.31^{\circ }+33.69^{\circ }=90.00^{\circ }.$ WHY exactly $90^{\circ }$: $\tan {\theta }_B=n_2/n_1$ and $\tan {\theta }_B^{\prime }=n_1/n_2$ are reciprocals. Since $\tan (90^{\circ }-x)=1/\tan x$, the two angles are complementary. (c) The external refracted ray and the internal Brewster ray are the same line by ray reversibility: a ray refracted into the glass at Brewster incidence is, run backwards, a ray leaving the glass at its Brewster angle.

The reflected beam is already polarized (horizontal), so we do not halve it — use Malus with real angles. Reflected beam: $I_1=0.08I_0$, polarized at $0^{\circ }$ (horizontal). Through P1 (at $40^{\circ }$): angle from beam to P1 is $40^{\circ }$: $I_2=I_1{\cos }^{2}40^{\circ }=0.08I_0\cdot (0.766{)}^2=0.08I_0\cdot 0.5868=0.04695I_0.$ Light is now polarized along P1 (at $40^{\circ }$). Through P2 (at $70^{\circ }$): angle from P1 ($40^{\circ }$) to P2 ($70^{\circ }$) is $30^{\circ }$: $I_3=I_2{\cos }^{2}30^{\circ }=0.04695I_0\cdot 0.75=0.03521I_0.$ Final: $I_3\approx 0.0352I_0.$

Unpolarized through one polarizer? ::: $\frac{1}{2}{I}_0$, output polarized along the axis. Malus angle between consecutive filters, not from the first? ::: True — each filter re-polarizes the light. Brewster angle air→glass ($n=1.5$)? ::: $\arctan 1.5=56.3^{\circ }$. Why do vertical sunglasses kill lake glare? ::: Brewster glare is horizontal, crossed at $90^{\circ }$ with vertical. ${\theta }_B$ and the reverse-direction ${\theta }_B^{\prime }$ sum to? ::: $90^{\circ }$ (reciprocal tangents). Max output of crossed polarizers with a middle one? ::: $\frac{1}{4}{I}_0$ at the middle angle $45^{\circ }$.