Polarisation

Scope note, stated honestly up front: under the CBSE (NCERT) 2025-26 rationalised syllabus, polarisation of light has been removed from the Wave Optics chapter. Malus's law, Brewster's angle, and polaroids are not part of the current theory list, and none of the material here should be revised as exam content this year — the examinable core of Wave Optics is Huygens' principle, reflection/refraction of wavefronts, Young's double-slit interference (final fringe-width formula $\beta=\lambda D/d$, no derivation required), and qualitative single-slit diffraction. Because older NCERT prints, many reference books, and older question banks still carry polarisation as a full topic with numericals, it is easy to mistakenly over-prepare it; this article exists precisely so that a curious student can still learn the physics correctly, understand why polaroid sunglasses and 3-D glasses work, and recognise the vocabulary if it appears in a non-CBSE context (competitive exams, older textbooks, general reading) — without confusing any of it for board-exam material.

Picture light as a wave in which the electric field vector traces out a line, or oscillates back and forth, in a plane perpendicular to the direction the wave is travelling — call that direction the $z$-axis. For an unpolarised beam, if you could freeze a snapshot at successive instants, the oscillation direction in the $xy$-plane would point in a new random direction almost every instant, because it is really the sum of light from countless independently-vibrating atoms. A polaroid is built from long-chain molecules aligned in one direction; it absorbs the component of the electric field along that molecular direction and transmits the perpendicular component (or, in the simplified picture used at this level, it is described as transmitting field oscillations along one transmission axis and blocking the perpendicular one). After the light has passed through, only oscillation along the transmission axis survives — the beam is now linearly polarised, and its transmitted intensity is lower than the original, since only one direction's worth of random contributions got through.

There are two everyday routes to polarised light besides a manufactured polaroid sheet. First, reflection: light reflected off a non-metallic surface such as water, glass, or a road becomes partially polarised, with the reflected field favouring oscillation parallel to the reflecting surface; at one special angle of incidence, the Brewster angle $i_B$, the reflected ray becomes completely polarised. Second, scattering: sunlight scattered by air molecules (the blue sky) is also partially polarised, which is why polarising filters on cameras can deepen a blue sky in a photograph. The workhorse quantitative relation of this topic links the intensities on either side of a second polaroid. If linearly polarised light of intensity $I_0$ (already polarised along some axis by a first polaroid, often called the polariser) meets a second polaroid (the analyser) whose transmission axis is turned through an angle $\theta$ from the first, the transmitted intensity is

where: $I_0$ is the intensity of the light incident on the analyser (already linearly polarised), $\theta$ is the angle between the transmission axes of the two polaroids, and $I$ is the intensity that emerges. This is Malus's law. The physical reading: only the component of the electric field along the analyser's axis gets through, and since intensity goes as the square of the field amplitude, a field component $E_0\cos\theta$ gives an intensity $I_0\cos^2\theta$. When the axes are aligned ($\theta=0°$), all the light that reached the analyser passes through, $I=I_0$. When the axes are perpendicular ($\theta=90°$), $\cos^2 90°=0$ and essentially no light gets through — the crossed-polaroids condition, the classic dark-field demonstration. At $\theta=45°$, exactly half the intensity survives, since $\cos^2 45°=\tfrac12$. The companion relation for reflection-induced polarisation is Brewster's law:

where: $i_B$ is the Brewster angle of incidence measured from the normal to the reflecting surface, and $n$ is the refractive index of the reflecting medium (relative to the medium the light travels in before striking it, usually air, so $n$ is simply the material's refractive index). At this special angle, the reflected and refracted rays turn out to be exactly perpendicular to each other, and the reflected ray is purely linearly polarised, vibrating perpendicular to the plane of incidence. For ordinary window glass, $n\approx1.5$, giving $i_B=\tan^{-1}(1.5)\approx56.3°$.

Board level - checking Malus's law at standard angles

Unpolarised light of intensity $I_u$ first passes through a polaroid $P_1$ (which, on average, halves its intensity, a standard starting assumption for unpolarised light through a single polaroid), and then through a second polaroid $P_2$ held at $\theta=60°$ to $P_1$. If $I_u=200$ units, find the intensity emerging from $P_2$.

Step 1. Find the intensity after $P_1$. Passing unpolarised light through a single polaroid transmits, on average, half the incident intensity (since the random oscillation direction is equally likely to have any component along the transmission axis, and the average of $\cos^2\theta$ over all angles is $\tfrac12$). So $I_1=\tfrac12 I_u=\tfrac12(200)=100$ units. The light emerging from $P_1$ is now linearly polarised along $P_1$'s axis.

Step 2. Apply Malus's law across $P_2$. With $\theta=60°$ between the two transmission axes, $I_2=I_1\cos^2\theta=100\times\cos^2 60°=100\times(0.5)^2=100\times0.25$.

Step 3. Compute the final answer. $I_2=\mathbf{25\ units}$ — only one-eighth of the original unpolarised intensity survives both polaroids, since $\tfrac12\times\tfrac14=\tfrac18$ and $\tfrac18\times200=25$.

Competitive level - two-polaroid extinction check with Brewster's angle

A beam of unpolarised light of intensity $I_0=400$ units passes through polaroid $P_1$, then through polaroid $P_2$ held at $\theta=30°$ to $P_1$, and would go on to strike a glass surface of refractive index $n=1.5$. Find (a) the intensity emerging from $P_2$, and (b) the Brewster angle for the glass surface, to confirm whether the light could in principle be additionally polarised on reflection at that angle.

Step 1. Intensity after $P_1$, as in the board example: unpolarised light through a single polaroid gives half the incident intensity, $I_1=\tfrac12(400)=200$ units, now linearly polarised along $P_1$'s axis.

Step 2. Intensity after $P_2$ via Malus's law. With $\theta=30°$, $\cos 30°=\dfrac{\sqrt3}{2}\approx0.866$, so $\cos^2 30°=0.75$. Then $I_2=I_1\cos^2\theta=200\times0.75=150$ units.

Step 3. Find the Brewster angle for $n=1.5$ using $\tan i_B=n$. So $i_B=\tan^{-1}(1.5)$. Since $\tan 56°\approx1.483$ and $\tan 57°\approx1.540$, the angle lies just above $56°$; a calculator gives $i_B\approx56.3°$.

Step 4. State both final answers together. The intensity emerging from $P_2$ is 150 units, and the Brewster angle for this glass is approximately 56.3° — meaning that if this beam (already reduced to 150 units by the two polaroids) were instead directed at the glass surface at $56.3°$ to the normal, the reflected component at that specific angle would itself come out completely linearly polarised, perpendicular to the plane of incidence. Timing note: in a competitive-exam setting, budget under a minute for each of Steps 1–2 (they are routine substitution into Malus's law) and reserve the remaining time for Step 3, where $\tan^{-1}(1.5)$ is easy to mis-estimate — memorise that $\tan 45°=1$ and $\tan 60°\approx1.73$ as anchor points so you can sanity-check that $56°$–$57°$ is the right neighbourhood without needing a lookup table.

• Forgetting the crucial factor of $\tfrac12$ when unpolarised light passes through the first polaroid — Malus's law with $\cos^2\theta$ applies only from the second polaroid onward, once the light is already linearly polarised; applying $\cos^2\theta$ directly to unpolarised light is a common numerical slip.
• Confusing the angle $\theta$ in Malus's law (between the two transmission axes) with the angle of incidence $i_B$ in Brewster's law (measured from the normal to a reflecting surface) — the two angles describe different physical setups and are not interchangeable.
• Assuming crossed polaroids ($\theta=90°$) block light because they are somehow "opposite" filters — the real reason is purely geometric: the field surviving the first polaroid has zero component along a perpendicular axis, so $\cos^2 90°=0$ exactly.
• Thinking sound waves can be polarised the way light can — sound is a longitudinal wave, oscillating along its own direction of travel, with no perpendicular direction to restrict, so "polarised sound" is not a meaningful concept in this sense.
• On the current CBSE 2025-26 syllabus specifically: treating any of this — Malus's law, Brewster's angle, polaroids — as examinable board content. It is enrichment only this year; do not spend revision time deriving or memorising these formulae for your exam.

• State clearly that polarisation is not part of the CBSE 2025-26 Class 12 Wave Optics theory syllabus, and correctly prioritise board-exam revision toward Huygens' principle, Young's double-slit interference, and single-slit diffraction instead.
• Explain, conceptually, the difference between unpolarised and linearly polarised light in terms of the direction of electric-field oscillation.
• State and apply Malus's law, $I=I_0\cos^2\theta$, to find transmitted intensity through a second polaroid, including the extra factor of $\tfrac12$ when the very first polaroid acts on unpolarised light.
• State Brewster's law, $\tan i_B=n$, and explain what makes the reflected ray at that angle completely polarised.
• Explain why polarisation is possible only for transverse waves, using sound (longitudinal) as the contrasting example.
• Give at least two everyday applications of polarisation (polaroid sunglasses cutting glare, camera filters, LCD screens, or 3-D glasses).

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