• 1. What is "Polarisation" and the "State of Polarisation" (SOP) ?
• 2. Why do wa need to know, and define the SOP of light?
• 3. How to obtain polarised light?
• 4. Polarised light by Reflection (at the Brewster angle)
• 5. Propagation of plane polarised light through one or more polariser (the Malus' law)

• 'Polarisation' is a property of 'Light'

• 'Light' is an electromagnetic wave, comprising of rapidly varying electric and magnetic fields.

• The electric and magnetic fields are perpendicular to each other, and perpendiculer to the direction of propagation.

• 'Polarisation of Light' refers to the direction of oscillation of the electric field (of light)

• Light is an electromagnetic wave

• eg. an e.m wave propagating in the x-direction

• Polarisation refers to the direction of oscillation of the electric field

• y-polarised wave

• Projection of the Electric field on a plane perpendicular to the direction of propagation is a line

• State of polarisation of light is given by the projection of the locus of the tip of the electric field vector on a plane perpendicular to the direction of propagation.

• E = $E_o sin (kx-\omega t)$

• $\omega = 2 \pi v$

• $v = \frac {c}{\lambda}$

• Z - polarised waves

• E - field oscillations are confined to the x-z plane

• Also called Plane polarised light

• Light from common sources such as sun, electric bulb, fluorescent lamp, etc.

• eg. Battery torch

• The beam of light comprises of large number of "component waves"

• The component waves are emitted by different atomic oscillators, from different parts of the Source.

• They may have different planes of oscillations

• The combination forms a randomly polarised beam or "unpolarisod light"

• Randomly Polarised light

• Electric field vectors of the randomly oriented polarisations are resolved into their components along 'y' and 'z' directions.

• In the equivalent representation, unpolarised light comprises of two equal components of the electric field of light in 'y' and 'z' directions.

• $\vec E = \hat j E_y + \hat z E_z$

• Assuming the following cartesian coordinats system with x-axis to be the direction of propagation.

• Coordinats system $,\quad$ unpolarised wave

• Y- polarised wave $,\quad$ z- polarised wave

• How to obtain polarised light?

• Ans: By passing unpolarised light through a polariser.

• There are different types of polarises, based on different working principle. The simplest, least expensive, and must widely used are "sheet polarisers" or a "polaroid sheet".

• The Polaroid comprises of a shut of certain long-chain polymeric molecules, which are almost aligned like a wire-grid.

• The polarization component which is parallel to the long chain suffers loss (or attenuation).

• $\therefore$ The vertical component (in figure) is attenuated; the horrigontal component parses through the sheet with very little loss. Therefore, the horrigontal axis is called the "pass axis" of the polariser.

• Input Intensity $I_0$

• $\frac {I_0}{2} \rightarrow$ is the output intensity (neglecting absorption ) of the vertical component )

• Q. What if we rotate the polariser?

• Ans. No change in the Intensity of light at the output.

• Polarisation by Reflection of light (at the Brewster Angle)

• Recall: Reflection of light at a plane inteface.

• Snell's Law: $\frac{\sin i}{\sin r}=\frac{n_2}{n_1}=n_{21}$

• $r_B=\left(90-i_B\right) \longrightarrow$ from fig.

• $\therefore \frac{\sin i_B}{\sin r_B}=\frac{\sin i_s}{\cos i_B}=\tan i_B$

• ie. $\tan i_B = n_{21}$ - called Brewster's Law

• $L_B$ - Brewster Angle.

• What is special about the Brewster Angle ?

• Example : For air glass interface, $n_1 = 1 , n_2 = 1.5$ (seq)

• $tan i_B = r_{21} = \frac {1.5}{1} \quad gives \quad L_B = 56.3$

• The reflected light does not have the inplane component , why?

• $\vec E = \hat y E_0 sin (kx -\omega t)$

• I = $\vec |E| = |E_0|^2 |sin (kr-\omega t)|^2$

• = $|E_0|^2 . \frac {1}{2}$

• $\vec E = \hat z E_0 sin (kx -\omega t)$

• I = $|E_0|^2 . \frac {1}{2}$

• $\vec E = (\hat y E_0 cos \theta + \hat z E_0 sin \theta) \times sin (kx -\omega t)$

• I = $|E|^2 = E^2_0 \frac {1}{2}$

• $\vec E_2 = \hat y E_0 cos \theta$

• $I_2 = \vec {E_0^2} = E_0^2 cos^2 \theta$

• 0r $I_2 = I_1 cos^2 \theta$

• Malus law

• $\vec E_1 = \hat y E_0 cos \theta + \hat z E_0 sin\theta$

• $I_1 = \vec {|E_1|}^2 = E_0^2 cos^2 \theta + E_0^2 sin^2 \theta$

• or $I_1 = E_0^2$

)

• Unpolarised Light- passing through two "crossed polarisers"

• $I_3 = 0$

• $I_2 = \frac {I_0}{2}$

• $I_1 = I_0$

• What if we rotate any one of the polarisers?

• $I_4$ = ?

• $I_2 = \frac {I_0}{2} cos^2 \theta$

• $I_2 = \frac {I_0}{2} (\vec E_2 \hat y \frac{E_0}{\sqrt 2})$

• $I_4 = I_0 (input) = \vec{|E_0|}^2$

• $I_4=\left[\frac{I_3}{2} \cos ^2 \theta\right] \cos ^2(90-\theta)$

• $=\frac{I_0}{2}(\cos \theta \cdot \sin \theta)^2$

• $=\frac{I_ 1}{2}\left(\frac{\sin 2 \theta}{2}\right)^2 =\frac{I_0}{8} \sin ^2 2 \theta$

• $\theta$ is the angle between the pass axis and the plans of polarisation.

• Note- $I_4$ is maximum when $\theta=45^{\circ}$

• $I_4$ is zero when $\theta=0^{\circ}$ or $\theta = 90^{\circ}$

• when $2^{\text {nd }}$ and $3^{\text {rd }}$ polarisers are "crossed".