10.7 Polarisation

Consider holding a long string that is held horizontally, the other end of which is assumed to be fixed. If we move the end of the string up and down in a periodic manner, we will generate a wave propagating in the \(+x\) direction (Fig. 10.17). Such a wave could be described by the following equation
\(
y(x, t)=a \sin (k x-\omega t) \dots(10.15)
\)
where \(a[latex] and [latex]\omega(=2 \pi \nu)\) represent the amplitude and the angular frequency of the wave, respectively; further,
\(
\lambda=\frac{2 \pi}{k} \dots(10.16)
\)
represents the wavelength associated with the wave. We had discussed propagation of such waves in Chapter 14 of Class XI textbook. Since the displacement (which is along the \(y\) direction) is at right angles to the direction of propagation of the wave, we have what is known as a transverse wave. Also, since the displacement is in the \(y\) direction, it is often referred to as a \(y\)-polarised wave. Since each point on the string moves on a straight line, the wave is also referred to as a linearly polarised wave. Further, the string always remains confined to the \(x-y\) plane and therefore it is also referred to as a plane polarised wave.

In a similar manner we can consider the vibration of the string in the \(x\) – \(z\) plane generating a \(z\)–polarised wave whose displacement will be given by
\(
z(x, t)=a \sin (k x-\omega t) \dots(10.17)
\)
It should be mentioned that the linearly polarised waves [described by Eqs. (10.15) and (10.17)] are all transverse waves; i.e., the displacement of each point of the string is always at right angles to the direction of propagation of the wave. Finally, if the plane of vibration of the string is changed randomly in very short intervals of time, then we have what is known as an unpolarised wave. Thus, for an unpolarised wave the displacement will be randomly changing with time though it will always be perpendicular to the direction of propagation.

Light waves are transverse in nature; i.e., the electric field associated with a propagating light wave is always at right angles to the direction of propagation of the wave. This can be easily demonstrated using a simple polaroid. You must have seen thin plastic like sheets, which are called polaroids. A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarised light wave is incident on such a polaroid then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules; this direction is known as the pass-axis of the polaroid.

Thus, if the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet \(P_{1,}\) it is observed that its intensity is reduced by half. Rotating \(P_1\) has no effect on the transmitted beam and transmitted intensity remains constant. Now, let an identical piece of polaroid \(P_2\) be placed before \(P_1\). As expected, the light from the lamp is reduced in intensity on passing through \(P_2\) alone. But now rotating \(P_1\) has a dramatic effect on the light coming from \(P_2\). In one position, the intensity transmitted by \(P_2\) followed by \(P_1\) is nearly zero. When turned by \(90^{\circ}\) from this position, \(P_1\) transmits nearly the full intensity emerging from \(P_2\) (Fig. 10.18).

The experiment at figure 10.18 can be easily understood by assuming that light passing through the polaroid \(P_2\) gets polarised along the passaxis of \(P_2\). If the pass-axis of \(P_2\) makes an angle \(\theta\) with the pass-axis of \(P_1\), then when the polarised beam passes through the polaroid \(P_2\), the component \(E \cos \theta\) (along the pass-axis of \(P_2\) ) will pass through \(P_2\). Thus, as we rotate the polaroid \(P_1\) (or \(P_2\) ), the intensity will vary as:
\(
I=I_0 \cos ^2 \theta \dots(10.18)
\)
where \(I_0\) is the intensity of the polarized light after passing through \(P_1\). This is known as Malus’ law. The above discussion shows that the intensity coming out of a single polaroid is half of the incident intensity. By putting a second polaroid, the intensity can be further controlled from \(50 \%\) to zero of the incident intensity by adjusting the angle between the pass-axes of two polaroids.

Polaroids can be used to control the intensity, in sunglasses, windowpanes, etc. Polaroids are also used in photographic cameras and 3D movie cameras.

Example 10.2: Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids?

Solution: Let \(I_0\) be the intensity of polarised light after passing through the first polariser \(P_1\). Then the intensity of light after passing through second polariser \(P_2\) will be
\(
I=I_0 \cos ^2 \theta
\)
where \(\theta\) is the angle between pass axes of \(P_1\) and \(P_2\). Since \(P_1\) and \(P_3\) are crossed the angle between the pass axes of \(P_2\) and \(P_3\) will be \((\pi / 2-\theta)\). Hence the intensity of light emerging from \(P_3\) will be
\(
\begin{aligned}
I & =I_0 \cos ^2 \theta \cos ^2\left(\frac{\pi}{2}-\theta\right) \\
& =I_0 \cos ^2 \theta \sin ^2 \theta=\left(I_0 / 4\right) \sin ^2 2 \theta
\end{aligned}
\)
Therefore, the transmitted intensity will be maximum when \(\theta=\pi / 4\).