Unpolarised vs Polarised Light
A light wave is transverse: the electric field oscillates at right angles to the direction of propagation. For a wave travelling along the $x$-axis, the electric vector can point anywhere in the $y$-$z$ plane. NCERT develops this with the analogy of a string. If the string is shaken only in the $y$-direction it carries a $y$-polarised wave; shaken only in the $z$-direction it carries a $z$-polarised wave. In both cases every point moves on a straight line, so the disturbance is linearly polarised (also called plane polarised, since the vibration stays in one plane).
If, instead, the plane of vibration of the string is changed randomly in very short intervals of time, the wave is unpolarised. The displacement is always perpendicular to the direction of propagation, but its orientation in the transverse plane keeps changing at random. Natural light from the Sun or a sodium lamp is unpolarised: the electric vector takes all transverse directions, rapidly and randomly, during any measurement.
Polaroids and the Pass-axis
A polaroid is a thin sheet built from long-chain molecules aligned in one direction. The electric-field component along the aligned molecules is absorbed; the component perpendicular to them is transmitted. The transmitted direction is called the pass-axis of the polaroid. When unpolarised light passes through a polaroid, the emerging beam is linearly polarised along the pass-axis.
NCERT describes the standard two-polaroid experiment. Light from a sodium lamp passing through one polaroid $P_1$ loses half its intensity, and rotating $P_1$ alone produces no further change. Place a second polaroid $P_2$ in the beam: now rotating one relative to the other has a dramatic effect. When their pass-axes are parallel, transmission is maximum; when they are crossed at $90^\circ$, transmission falls nearly to zero. The polaroid the light reaches first acts as the polariser; the second acts as the analyser.
| Quantity | Incident light | Output / rule |
|---|---|---|
| One ideal polaroid | Unpolarised, intensity $I_0$ | $I_0/2$, now polarised along pass-axis |
| Rotating a single polaroid | Unpolarised | Output unchanged at $I_0/2$ |
| Second polaroid at angle $\theta$ | Polarised, intensity $I_0$ | $I = I_0\cos^2\theta$ (Malus) |
| Crossed polaroids | Polarised | $\theta = 90^\circ \Rightarrow I = 0$ |
Malus's Law
Suppose light is already linearly polarised with intensity $I_0$, and the pass-axis of the next polaroid makes an angle $\theta$ with the plane of polarisation. Only the component of the electric vector along the pass-axis, $E\cos\theta$, is transmitted. Since intensity is proportional to the square of the amplitude, the transmitted intensity is
$$I = I_0\cos^2\theta$$
This is Malus's law. It holds only for light that is already polarised; it does not describe unpolarised light striking the first polaroid, which always loses exactly half its intensity regardless of orientation. At $\theta = 0$, $I = I_0$ (axes parallel); at $\theta = 90^\circ$, $I = 0$ (axes crossed).
The "half" rule versus the "cos²θ" rule
The single most common error in this topic is applying $I_0\cos^2\theta$ to unpolarised light striking the first polaroid. Unpolarised light has no fixed plane, so there is no angle $\theta$ to use. The first polaroid always cuts the intensity to exactly $I_0/2$ and polarises the beam. Malus's law applies only from the second polaroid onward.
First polaroid (unpolarised in): $I \to I_0/2$. After that (already polarised): $I = I_0\cos^2\theta$.
A polaroid sheet is placed between two crossed polaroids $P_1$ and $P_3$. The middle polaroid $P_2$ makes an angle $\theta$ with $P_1$. Discuss the transmitted intensity as $P_2$ is rotated.
Let $I_0$ be the intensity of polarised light after $P_1$. After $P_2$, the intensity is $I_0\cos^2\theta$. Since $P_1$ and $P_3$ are crossed, the angle between $P_2$ and $P_3$ is $(\tfrac{\pi}{2}-\theta)$. Hence the intensity emerging from $P_3$ is
$$I = I_0\cos^2\theta\,\cos^2\!\left(\tfrac{\pi}{2}-\theta\right) = I_0\cos^2\theta\,\sin^2\theta = \tfrac{I_0}{4}\sin^2 2\theta$$
The transmitted intensity is maximum when $\theta = \pi/4$, i.e. when $P_2$ bisects the crossed axes. With no middle polaroid, crossed $P_1$ and $P_3$ transmit zero; inserting $P_2$ revives the beam.
Polarisation is the last pillar of wave optics. Trace the wave model from the start with Huygens' Principle.
Polarisation by Reflection & Brewster's Angle
Polarisation also occurs on reflection. When unpolarised light strikes a transparent surface such as glass or water, the reflected light is, in general, only partially polarised. There is one special angle of incidence, the polarising angle or Brewster's angle $i_B$, at which the reflected light is completely linearly polarised, with its electric vector perpendicular to the plane of incidence.
At Brewster's angle the reflected and refracted rays are mutually perpendicular. Using Snell's law $n = \dfrac{\sin i_B}{\sin r}$ together with the geometric condition $i_B + r = 90^\circ$ gives $\sin r = \sin(90^\circ - i_B) = \cos i_B$, so
$$n = \frac{\sin i_B}{\cos i_B} = \tan i_B$$
This is Brewster's law. The polarising angle depends only on the refractive index of the medium. NIOS notes that for an air–water interface $i_B \approx 53^\circ$, so sunlight reflected from a calm pond is fully polarised when the Sun is about $37^\circ$ above the horizon — the principle behind polaroid sunglasses, which cut glare by blocking this reflected component.
What is polarised at Brewster's angle?
At $i_B$, only the reflected ray is completely (perfectly) polarised, with its electric vector perpendicular to the plane of incidence. The refracted (transmitted) ray remains only partially polarised. The two emerging rays are at $90^\circ$ to each other. A frequent slip is to call both rays fully polarised, or to use $\sin i_B = n$ instead of $\tan i_B = n$.
$\tan i_B = n$ · reflected ray fully polarised (⊥ to plane of incidence) · reflected ⊥ refracted.
Unpolarised light is incident from air on a medium of refractive index $1.73$ at Brewster's angle. Find $i_B$ and describe the reflected and refracted beams.
By Brewster's law $\tan i_B = n = 1.73 = \sqrt{3}$, so $i_B = 60^\circ$. The reflected ray (at $60^\circ$ to the normal) is completely polarised; the refracted ray, with $r = 90^\circ - 60^\circ = 30^\circ$, is partially polarised. The two are at right angles.
Why Polarisation Proves Light is Transverse
Interference and diffraction can occur for any wave, longitudinal or transverse, so they establish only that light is a wave. Polarisation goes further. A vibration can be selectively passed or blocked by a polaroid only if it is confined to a plane perpendicular to the direction of travel; a longitudinal vibration, oscillating along the line of propagation, offers no orientation to select. The string-and-slit demonstration in NIOS makes this concrete: a vertical slit passes a transverse wave only when the rope vibrates vertically. Because light can be polarised, its vibrations must be transverse. NCERT states the conclusion directly: polarisation phenomena are special to transverse waves like light, and they do not arise for longitudinal waves such as sound in air.
| Phenomenon | What it establishes | Wave type required |
|---|---|---|
| Interference | Light is a wave (superposition) | Any wave |
| Diffraction | Light is a wave (bending at edges) | Any wave |
| Polarisation | Light is a transverse wave | Transverse only |
Polarisation in one screen
- Unpolarised light: electric vector in all transverse directions, randomly. Linearly (plane) polarised: vibration confined to one plane.
- One ideal polaroid on unpolarised light gives $I_0/2$ and polarises it; rotating that single polaroid changes nothing.
- Malus's law $I = I_0\cos^2\theta$ applies only to already-polarised light meeting an analyser at angle $\theta$.
- Crossed polaroids transmit zero; a third polaroid between them at $\theta$ gives $\tfrac{I_0}{4}\sin^2 2\theta$, maximum at $45^\circ$.
- Brewster's law $\tan i_B = n$: reflected ray fully polarised (⊥ plane of incidence), refracted ray partial, the two at $90^\circ$.
- Polarisation is the proof that light is a transverse wave; sound (longitudinal) cannot be polarised.