Polarisation

Transverse Waves and Polarisation

Shake a horizontal string up and down: a wave runs along x while each point moves along y — the displacement is perpendicular to the propagation, a transverse wave. Because the vibration stays in one plane (x-y), it is a linearly (plane) polarised wave — here y-polarised. Shaking in the x-z plane instead gives a z-polarised wave.

If the plane of vibration changes randomly moment to moment (still always perpendicular to propagation), the wave is unpolarised.

Light is transverse: its electric field $\vec{E}$ oscillates at right angles to the direction of travel. Ordinary light (sodium lamp, sun) is unpolarised — $\vec{E}$ points every which way in the transverse plane.

Key Point: only transverse waves can be polarised. That light can be polarised is direct proof it is transverse (sound, being longitudinal, cannot be). A favourite one-mark gem.

Polaroids, the Half Rule, and Malus' Law

A polaroid is a sheet of aligned long-chain molecules. The $\vec{E}$-components along the chains are absorbed; only the component perpendicular to them passes — that perpendicular direction is the pass-axis.

Unpolarised light through one polaroid P$_1$:

• The transmitted light is linearly polarised along P$_1$'s pass-axis.
• Its intensity is exactly halved ($I_0/2$) — averaging $\cos^2$ over all random orientations gives $\tfrac12$.
• Rotating P$_1$ changes nothing — the input was symmetric.

Polarised light through a second polaroid P$_2$ at angle $\theta$ to the first — Malus' law:

$$\boxed{I = I_0\cos^2\theta}$$

• $\theta = 0$ (axes parallel): full transmission.
• $\theta = 90^{\circ}$ (crossed polaroids): zero — total darkness.
• Rotating P$_2$ from 0 to 90 degrees dims the light from full to nothing.

[NEET Important] Two-step intensity through P$_1$ then P$_2$ for unpolarised input: first the half rule ($I_0 \to I_0/2$), then Malus ($\to \frac{I_0}{2}\cos^2\theta$). Do them in order — skipping the half is the standard error.

The Three-Polaroid Classic

Cross two polaroids (P$_1$, P$_3$ at 90 degrees) so no light gets through. Now slip a third polaroid P$_2$ between them at angle $\theta$ to the first — and light reappears! Tracking the intensity (NCERT's crossed-polaroid example):

• After P$_1$: $I_1$ (the polarised intensity entering the sandwich).
• After P$_2$ (angle $\theta$): $I_1\cos^2\theta$.
• After P$_3$ (angle $90^{\circ} - \theta$ to P$_2$): $I_1\cos^2\theta\cos^2(90^{\circ} - \theta) = I_1\cos^2\theta\sin^2\theta$.

$$\boxed{I_3 = I_1\cos^2\theta\sin^2\theta = \frac{I_1}{4}\sin^2 2\theta}$$

Maximum at $\theta = 45^{\circ}$ ($I_3 = I_1/4$); zero at $\theta = 0$ or $90^{\circ}$. The middle polaroid 're-aims' the polarisation so the last one can pass a component — a striking, deeply exam-favourite result.

[JEE Tip] With unpolarised input intensity $I_0$, $I_1 = I_0/2$, so the three-polaroid output is $\frac{I_0}{8}\sin^2 2\theta$, peaking at $I_0/8$ when $\theta = 45^{\circ}$. Memorise the $\frac{1}{8}$.

Real-world polarisation: Polaroid sunglasses cut glare (reflected light is partly polarised), photographers' filters deepen skies, and LCD screens switch pixels by rotating polarisation.

Solved Examples

Example 1: The half rule [NEET Numerical]

Unpolarised light of intensity 80 W/m$^2$ passes through one polaroid. What emerges?

Solution:

1. One polaroid on unpolarised light transmits half.
2. $I = 80/2 = 40$ W/m$^2$, now linearly polarised along the pass-axis.
3. Rotating this single polaroid would not change the 40 — the input was directionless.

Example 2: Malus at 30 degrees [NEET Numerical]

Polarised light of 40 W/m$^2$ falls on a polaroid whose axis is at 30 degrees to the light's polarisation. Find the transmitted intensity.

Solution:

1. Malus: $I = I_0\cos^2\theta = 40\cos^2 30^{\circ}$.
2. $= 40\times(0.866)^2 = 40\times0.75 = 30$ W/m$^2$.
3. Parallel would give 40; crossed would give 0.

Example 3: Two polaroids from unpolarised [JEE Numerical]

Unpolarised light of 100 W/m$^2$ passes through P$_1$ then P$_2$, with P$_2$ at 60 degrees to P$_1$. Find the final intensity.

Solution:

1. After P$_1$ (half rule): $100/2 = 50$ W/m$^2$.
2. After P$_2$ (Malus): $50\cos^2 60^{\circ} = 50\times0.25 = 12.5$ W/m$^2$.
3. The two-step recipe — half first, then $\cos^2\theta$.

Example 4: Crossed polaroids [NEET Numerical]

What fraction of unpolarised light passes through two polaroids whose axes are perpendicular?

Solution:

1. After P$_1$: half. After P$_2$ at 90 degrees: $\cos^2 90^{\circ} = 0$.
2. Zero transmission — crossed polaroids are dark.
3. The basis of LCD switching and glare elimination.

Example 5: The three-polaroid revival (NCERT's crossed-polaroid example) [JEE Numerical]

Unpolarised light (intensity $I_0$) passes through three polaroids: P$_1$, then P$_2$ at $\theta$, then P$_3$ crossed with P$_1$. Find the output and the angle of maximum.

Solution:

1. After P$_1$: $I_1 = I_0/2$. After P$_2$: $\frac{I_0}{2}\cos^2\theta$.
2. After P$_3$ (at $90^{\circ}-\theta$ to P$_2$): $\frac{I_0}{2}\cos^2\theta\sin^2\theta = \frac{I_0}{8}\sin^2 2\theta$.
3. Maximum at $\theta = 45^{\circ}$: $I = I_0/8$. Without the middle polaroid the crossed pair gives zero — the inserted sheet revives the light.

Example 6: Finding the angle [JEE Numerical]

Through two polaroids, the transmitted intensity is half the maximum (for polarised input). Find the angle between their axes.

Solution:

1. $\cos^2\theta = \tfrac12 \Rightarrow \cos\theta = \frac{1}{\sqrt2}$.
2. $\theta = 45^{\circ}$.
3. At 45 degrees a polaroid passes exactly half the polarised intensity — a handy benchmark.

Example 7: Why polarisation proves transversality

State the deep significance of the polaroid experiments.

Solution:

1. Only transverse waves have a vibration direction perpendicular to travel that can be selectively absorbed.
2. Polaroids do exactly that to light — so light's vibrations are transverse.
3. Sound (longitudinal) cannot be polarised; the fact that light can is direct evidence of its transverse, electromagnetic nature.

Example 8: Rotating the single polaroid

Unpolarised light passes through one polaroid. Why does rotating it leave the transmitted intensity constant?

Solution:

1. Unpolarised light has $\vec{E}$ uniformly distributed over all transverse directions.
2. Any pass-axis orientation samples the same average — half the intensity.
3. So rotation changes nothing (unlike a second polaroid, where Malus' law makes rotation dramatic).

Example 9: Three polaroids at 30 degrees [JEE Numerical]

Unpolarised light of 64 W/m$^2$ passes P$_1$, then P$_2$ at 30 degrees to P$_1$, then P$_3$ at 30 degrees to P$_2$ (i.e. 60 degrees to P$_1$). Find the output.

Solution:

1. After P$_1$: $64/2 = 32$ W/m$^2$.
2. After P$_2$: $32\cos^2 30^{\circ} = 32\times0.75 = 24$ W/m$^2$.
3. After P$_3$: $24\cos^2 30^{\circ} = 24\times0.75 = 18$ W/m$^2$. Apply Malus at each junction in turn.

Example 10: Sunglasses and glare

Why do polaroid sunglasses cut glare from a wet road?

Solution:

1. Light reflected off horizontal surfaces is partially horizontally polarised.
2. Polaroid lenses are mounted with a vertical pass-axis, blocking that horizontal glare.
3. Diffuse (unpolarised) light from the scene still passes at half intensity — so you see the road, minus the dazzle. Malus' law, worn on the face.