Q.What is meant by the polarisation of light? Two polaroids A and B are placed crossed with each other. How a third polaroid is to be placed in between them such that the light transmitted from B have intensity $\dfrac{1}{32}$ that of the intensity of unpolarised light incident on A? OR What is Huygen's theory of secondary wavelets? A monochromatic light of wavelength 6000 $\overset{\circ}{\text{A}}$ is incident on the surface of a glass having refractive index 1.5 from air. Find wavelength and speed for each of following:

Ordinary (unpolarised) light passing through a single polaroid always emerges at exactly half the incident intensity, regardless of the polaroid's orientation.

Sodium light is unpolarised — its electric field vibrates randomly in all directions perpendicular to propagation. When such light of intensity $I_0$ passes through a single polaroid, the polaroid transmits only the component of vibration along its transmission axis. Averaging over all random orientations of the incident field gives a transmitted intensity of exact …

Polaroids work by selectively absorbing one plane of polarisation, which controls the intensity of transmitted light.

A polaroid consists of long chain molecules aligned in a particular direction; it transmits light whose electric vector is parallel to this pass-axis and absorbs the perpendicular component. When two polaroids are used together, rotating one relative to the other varies the transmitted intensity according to Malus's law, $I = I_0\cos^2\theta$. In sunglasses, polaroids cut down glare (reduce the intensity of reflected/scattered light); in 3D movie syst …

Unpolarised light is halved by the first polaroid, then reduced further by $\cos^2\theta$ (Malus' law) through the second polaroid at $30°$, giving $3I/8$.

Working

Step 1 — first polaroid: Unpolarised light of intensity $I$ falling on a polaroid always transmits half its intensity as plane-polarised light:

$I_1 = \dfrac{I}{2}$

Step 2 — second polaroid: This linearly polarised light of intensity $I_1$ now falls on the second polaroid, whose transmission axis is inclined at $\theta = 30°$ to the first. By Malus' law: …

By Malus's law, $I=I_0\cos^2\theta$, transmitted intensity is maximum when the analyzer's pass axis is parallel to the polarizer's.

When unpolarized light passes through a polarizer, it becomes linearly polarized along the polarizer's transmission axis, with intensity $I_0$ (half the incident intensity). If this polarized light then falls on an analyzer whose pass axis makes an angle $\theta$ with the polarizer's axis, Malus's law gives the transmitted intensity as

$I = I_0 \cos^2\theta$ …

Intensity transmitted through a pair of polaroids follows Malus's law, $I=I_0\cos^2\theta$, where $\theta$ is the angle between the two polaroid axes — maximum when the axes are parallel, minimum (zero) when they are perpendicular.

Unpolarized light first becomes polarized (and halved in intensity, $I_0/2$) after passing through the first (upper) polaroid. Passing through the second (lower) polaroid then attenuates it further according to Malus's law:

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

where $\theta$ is the angle between the transmission axes of the upper and lower polaroids.

In each of the four pairs (a)–(d), the lower polaroid's axis is marked at a fixed angle and the upper polaroid's axis is rotated to a different absolute orientation, which changes $\theta$ (and hence the relative angle between the two axes) from pair to pair. To identify the answer:

• The pair where the two marked axis lines are CLOSEST to being parallel to each other (smallest $\theta$, ideally $\theta \to 0°$) transmits the MAXIMUM intensity ($I \to I_0/2$, its largest possible value).
• The pair where the two marked axis lines are CLOSEST to being perpendicular ($\theta \to 90°$) transmits the MINIMUM intensity ($I \to 0$). …

The intensity of polarised light passing through an analyser is I = I₀ cos²θ.

When plane-polarised light of intensity $I_0$ falls on an analyser (a polaroid), only the component of its electric field along the analyser's transmission axis is passed.

If $\theta$ is the angle between the plane of polarisation of the incident light and the analyser's axis, the transmitted amplitude is $E_0\cos\theta$, and since intensity is proportional to the square of amplitude: …