physics-class-12-unit-10-chapter-09-optics-polarisation-of-light-l-9-9-7iagvl8mjqk

### <Success style="font-size:25px"> Optics Polarisation Of Light L-9 </Success>
### <primary style="font-size:24px"> Index </primary>
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- 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)

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<footer style = "font-size: 18px"> $\rightarrow$ Optics Polarisation of Light $\rightarrow$ <span style = "color:blue"> Index </span> $\rightarrow$ Polaristation of Light $\rightarrow$ Light Waves </footer>

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### <primary style="font-size:24px"> Polaristation of Light </primary>
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* '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)

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<footer style = "font-size: 18px">Optics Polarisation of Light $\rightarrow$ Index $\rightarrow$ <span style = "color:blue"> Polaristation of Light </span> $\rightarrow$ Light Waves $\rightarrow$ State of Polarisation (SOP) </footer>

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- 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

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<footer style = "font-size: 18px">Index $\rightarrow$ Polaristation of Light $\rightarrow$ <span style = "color:blue"> Light Waves </span> $\rightarrow$ State of Polarisation (SOP) $\rightarrow$ State of Polarisation (SOP) </footer>

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### <primary style="font-size:24px"> State of Polarisation (SOP) </primary>
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- 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.

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<footer style = "font-size: 18px">Polaristation of Light $\rightarrow$ Light Waves $\rightarrow$ <span style = "color:blue"> State of Polarisation (SOP) </span> $\rightarrow$ State of Polarisation (SOP) $\rightarrow$ Plane Polarised Light </footer>

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### <primary style="font-size:24px"> Plane Polarised Light </primary>
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- Z - polarised waves

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

- Also called Plane polarised light

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<footer style = "font-size: 18px">State of Polarisation (SOP) $\rightarrow$ State of Polarisation (SOP) $\rightarrow$ <span style = "color:blue"> Plane Polarised Light </span> $\rightarrow$ Unpolarised Light $\rightarrow$ Unpolarised Light </footer>

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### <primary style="font-size:24px"> Unpolarised Light </primary>
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- 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"

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<footer style = "font-size: 18px">State of Polarisation (SOP) $\rightarrow$ Plane Polarised Light $\rightarrow$ <span style = "color:blue"> Unpolarised Light </span> $\rightarrow$ Unpolarised Light $\rightarrow$ Unpolarised Light </footer>

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### <primary style="font-size:24px"> Unpolarised Light </primary>
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- 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.

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<footer style = "font-size: 18px">Plane Polarised Light $\rightarrow$ Unpolarised Light $\rightarrow$ <span style = "color:blue"> Unpolarised Light </span> $\rightarrow$ Unpolarised Light $\rightarrow$ Representation of Polarised Light </footer>

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### <primary style="font-size:24px"> Representation of Polarised Light </primary>
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- 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

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<footer style = "font-size: 18px">Unpolarised Light $\rightarrow$ Unpolarised Light $\rightarrow$ <span style = "color:blue"> Representation of Polarised Light </span> $\rightarrow$ Polarised Light $\rightarrow$ Polaroid Sheet </footer>

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### <primary style="font-size:24px"> Polarised Light </primary>
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- 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".

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<footer style = "font-size: 18px">Unpolarised Light $\rightarrow$ Representation of Polarised Light $\rightarrow$ <span style = "color:blue"> Polarised Light </span> $\rightarrow$ Polaroid Sheet $\rightarrow$ Unpolarised Light- Passing through a Polariser </footer>

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### <primary style="font-size:24px"> Polaroid Sheet </primary>
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- 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.

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<footer style = "font-size: 18px">Representation of Polarised Light $\rightarrow$ Polarised Light $\rightarrow$ <span style = "color:blue"> Polaroid Sheet </span> $\rightarrow$ Unpolarised Light- Passing through a Polariser $\rightarrow$ Polarisation by Reflection of Light </footer>

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### <primary style="font-size:24px"> Polarisation by Reflection of Light </primary>
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- 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 ?

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<footer style = "font-size: 18px">Polaroid Sheet $\rightarrow$ Unpolarised Light- Passing through a Polariser $\rightarrow$ <span style = "color:blue"> Polarisation by Reflection of Light </span> $\rightarrow$ Brewster Angle $\rightarrow$ Electromagnetic Wave </footer>

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### <primary style="font-size:24px"> Brewster Angle </primary>
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- 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?

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<footer style = "font-size: 18px">Unpolarised Light- Passing through a Polariser $\rightarrow$ Polarisation by Reflection of Light $\rightarrow$ <span style = "color:blue"> Brewster Angle </span> $\rightarrow$ Electromagnetic Wave $\rightarrow$ Electric Field and Intensity of Light </footer>

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### <primary style="font-size:24px"> Electromagnetic Wave </primary>
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<footer style = "font-size: 18px">Polarisation by Reflection of Light $\rightarrow$ Brewster Angle $\rightarrow$ <span style = "color:blue"> Electromagnetic Wave </span> $\rightarrow$ Electric Field and Intensity of Light $\rightarrow$ Electric Field and Intensity of Fight </footer>

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### <primary style="font-size:24px"> Electric Field and Intensity of Light </primary>
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- $\vec E = \hat y E_0 sin (kx -\omega t)$

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

- = $|E_0|^2 . \frac {1}{2} $

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<footer style = "font-size: 18px">Brewster Angle $\rightarrow$ Electromagnetic Wave $\rightarrow$ <span style = "color:blue"> Electric Field and Intensity of Light </span> $\rightarrow$ Electric Field and Intensity of Fight $\rightarrow$ Linearly Polarised Light through a Polariser </footer>

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### <primary style="font-size:24px"> Electric Field and Intensity of Fight </primary>
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- $\vec E = \hat z E_0 sin (kx -\omega t)$

- 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 = $|E|^2 = E^2_0 \frac {1}{2}$

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<footer style = "font-size: 18px">Electromagnetic Wave $\rightarrow$ Electric Field and Intensity of Light $\rightarrow$ <span style = "color:blue"> Electric Field and Intensity of Fight </span> $\rightarrow$ Linearly Polarised Light through a Polariser $\rightarrow$ Crossed Polariser </footer>

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### <primary style="font-size:24px"> Linearly Polarised Light through a Polariser </primary>
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- $\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 $I_1 = E_0^2$

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<footer style = "font-size: 18px">Electric Field and Intensity of Light $\rightarrow$ Electric Field and Intensity of Fight $\rightarrow$ <span style = "color:blue"> Linearly Polarised Light through a Polariser </span> $\rightarrow$ Crossed Polariser $\rightarrow$ Crossed Polariser </footer>

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### <primary style="font-size:24px"> Crossed Polariser </primary>
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- 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?

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<footer style = "font-size: 18px">Electric Field and Intensity of Fight $\rightarrow$ Linearly Polarised Light through a Polariser $\rightarrow$ <span style = "color:blue"> Crossed Polariser </span> $\rightarrow$ Crossed Polariser $\rightarrow$ Crossed Polariser </footer>

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### <primary style="font-size:24px"> Crossed Polariser </primary>
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- $ 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".

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<footer style = "font-size: 18px">Crossed Polariser $\rightarrow$ Crossed Polariser $\rightarrow$ <span style = "color:blue"> Crossed Polariser </span> $\rightarrow$ Thank You $\rightarrow$  </footer>