Question bank — Polarization — Malus's law, Brewster's angle derivation

False — the $\frac{1}{2}$ holds only for unpolarized light. Already-polarized light follows $I=I_0{\cos }^2\theta$ (with $\theta$ the angle between $\vec{E}$ and the axis), which ranges from $0$ (crossed) to $I_0$ (aligned).

False — Malus needs a defined polarization direction to measure $\theta$ from. Unpolarized light has no single direction, so you average ${\cos }^2\theta$ over all angles and get $\frac{1}{2}$ instead.

True — ${\cos }^{2}90^{\circ }=0$, so no surviving component. But inserting a third polarizer between them can revive some light because it rotates the polarization in steps.

False — only the polarization component in the plane of incidence vanishes. The perpendicular component still reflects, so the reflected ray is dim but 100% polarized, not gone.

False — the law is $\tan {\theta }_B=n_2/n_1$, a ratio of entering-index over starting-index. Only when the first medium is air ($n_1=1$) does it collapse to $\tan {\theta }_B=n_2$.

True — with $n_1=1$ fixed, ${\theta }_B=\arctan (n_2)$ and $\arctan$ is increasing, so a larger $n_2$ gives a larger ${\theta }_B$ (though it can never reach $90^{\circ }$). If $n_1$ also changed, only the ratio $n_2/n_1$ would matter.

False — at $\theta =0$, ${\cos }^{2}0=1$, so $I=I_0$ at most. A passive filter can never amplify; it can only pass or absorb.

False — it is polarized perpendicular to the plane of incidence (parallel to the surface). That's why horizontal glare off water/roads dominates.

The error is forgetting the square. Detectors read intensity, and $I\propto E^2$, so the projection factor squares: $I=I_0{\cos }^2\theta$.

Wrong function — the derivation gives tangent, $\tan {\theta }_B=n_2$ (with $n_1=1$). It comes from Snell plus ${\theta }_r=90^{\circ }-{\theta }_B$, where $\sin (90^{\circ }-{\theta }_B)=\cos {\theta }_B$ turns the ratio into $\tan$.

The $\frac{1}{2}$ applies only at the first polarizer, because after it the light is polarized. Every later polarizer uses ${\cos }^2\theta$ relative to the previous axis.

Inverted ratio. The derivation gives $\tan {\theta }_B=n_2/n_1=n_{\text{into}}/n_{\text{from}}$. Going air→glass, that's $n_{\text{glass}}$, not its reciprocal.

They are perpendicular, ${\theta }_B+{\theta }_r=90^{\circ }$. That right angle is the whole physical reason the in-plane reflection cancels.

No change — ${\cos }^2\theta$ has period $180^{\circ }$ since $\cos (\theta +180^{\circ })=-\cos \theta$ and squaring kills the sign. The axis is a line, not an arrow.

No energy is added; the middle filter realigns the polarization so the last filter is no longer fully crossed with the surviving field. Each stage still only absorbs, never amplifies.

$\theta$ is measured to the transmission axis, and a vector's component along an axis is magnitude times the cosine of the angle from it. Using $\sin$ would measure the component being absorbed.

The energy density of an EM wave scales with the square of the field amplitude, and intensity is energy flow per area — so doubling $E$ quadruples $I$.

${\cos }^2\theta$ oscillates symmetrically between $0$ and $1$ about the value $\frac{1}{2}$ (since ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$ and $\cos 2\theta$ averages to zero). So half the intensity survives.

The two polarization components reflect with different strengths at an interface, so the mix is unbalanced — partially polarized. At Brewster's angle one component drops to zero, making it total.

The refracted ray drives the glass electrons to oscillate along its transverse field. A charge radiates nothing along its own oscillation axis, and that forbidden direction lines up with the reflected ray precisely when the two rays are perpendicular (see the third figure).

Glare from horizontal surfaces (water, roads) is polarized horizontally. A vertical axis is crossed with that glare, so ${\cos }^{2}90^{\circ }=0$ blocks it while letting useful vertical light through.

The geometric condition ${\theta }_r=90^{\circ }-{\theta }_B$ alone relates two angles; Snell links those angles to the indices $n_1,n_2$. Substituting one into the other converts the geometry into $\tan {\theta }_B=n_2/n_1$.

Zero — ${\cos }^{2}90^{\circ }=0$. The field is entirely perpendicular to the axis, so the polarizer absorbs all of it (crossed polarizers).

The formula still formally holds: $\tan {\theta }_B=n_2/n_1=1$, so ${\theta }_B=45^{\circ }$. But equal indices mean there is no genuine optical interface, so no reflection occurs to polarize — the number is mathematically valid yet has nothing to act on.

Yes — $\tan {\theta }_B=n_2/n_1<1$, giving a smaller Brewster angle. It exists as long as total internal reflection hasn't taken over first; check the critical angle separately.

No — ${\theta }_B=\arctan (n_2/n_1)$ and $\arctan$ approaches but never reaches $90^{\circ }$ for any finite index ratio. Grazing-incidence polarization by reflection is impossible.

Malus can't be applied directly — there is no reference direction. You average over all $\theta$, recovering the $\frac{1}{2}$ rule instead of a ${\cos }^2\theta$ value.

Each step multiplies intensity by ${\cos }^2(\frac{90^{\circ }}{N})\approx 1$, so the product $[{\cos }^2(\frac{90^{\circ }}{N}){]}^N\to 1$: the light is gradually rotated by a full $90^{\circ }$ with near-zero loss — the limiting behaviour behind the crossed-pair revival trick (see the second figure).

Full transmission, $I=I_0$, since ${\cos }^{2}0=1$. No half-factor applies because the light is polarized, not unpolarized.