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Wave Optics - Polarisation

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Polarisation is the phenomenon of restricting the vibrations of light (electric field vector) to a particular direction in a plane perpendicular to the direction of propagation.

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Unpolarised light consists of electric field vibrations in all possible directions perpendicular to the direction of travel. When passed through a polaroid, the intensity of unpolarised light I0I_0 becomes I=I02I = \frac{I_0}{2}.

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The plane containing the direction of vibration and the direction of propagation is called the Plane of Vibration.

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The plane passing through the direction of propagation and perpendicular to the plane of vibration is called the Plane of Polarisation.

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Malus' Law states that when completely plane-polarised light is incident on an analyser, the intensity II of light transmitted through the analyser is proportional to the square of the cosine of the angle θ\theta between the transmission axes of the polariser and the analyser.

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Brewster's Law states that when light is incident at the polarising angle θp\theta_p, the reflected light is completely plane-polarised and is perpendicular to the refracted ray. The refractive index of the medium is given by μ=tan⁡θp\mu = \tan \theta_p.

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Polarisation by scattering occurs when sunlight is scattered by molecules in the atmosphere. The scattered light seen in a direction perpendicular to the incident ray is plane-polarised.

📐Formulae

I=I0cos⁡2θI = I_0 \cos^2 \theta

μ=tan⁡θp\mu = \tan \theta_p

θp+r=90∘\theta_p + r = 90^\circ

Itransmitted=Iincident2 (for unpolarised light falling on a polaroid)I_{transmitted} = \frac{I_{incident}}{2} \text{ (for unpolarised light falling on a polaroid)}

💡Examples

Problem 1:

Unpolarised light of intensity I0I_0 is incident on two polaroids kept such that the angle between their pass axes is 60∘60^\circ. What is the intensity of the light transmitted through the second polaroid?

Solution:

  1. Intensity after the first polaroid: I1=I02I_1 = \frac{I_0}{2}
  2. Using Malus' Law for the second polaroid: I2=I1cos⁡2θI_2 = I_1 \cos^2 \theta
  3. Given θ=60∘\theta = 60^\circ, we have: I2=I02cos⁡2(60∘)I_2 = \frac{I_0}{2} \cos^2(60^\circ) I2=I02(12)2I_2 = \frac{I_0}{2} \left(\frac{1}{2}\right)^2 I2=I02×14=I08I_2 = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}

Explanation:

The first polaroid reduces the intensity of unpolarised light by half. The second polaroid further reduces the intensity based on the angle between the transmission axes following the squared cosine relationship.

Problem 2:

Light is incident on a transparent glass plate of refractive index μ=1.732\mu = 1.732 such that the reflected ray is completely plane-polarised. Calculate the angle of refraction.

Solution:

  1. According to Brewster's Law: μ=tan⁡θp\mu = \tan \theta_p
  2. 1.732=3=tan⁡θp⇒θp=60∘1.732 = \sqrt{3} = \tan \theta_p \Rightarrow \theta_p = 60^\circ
  3. We know that at Brewster's angle, θp+r=90∘\theta_p + r = 90^\circ
  4. Finding rr: 90−6030\begin{array}{r} 90 \\ -60 \\ \hline 30 \end{array} So, r=30∘r = 30^\circ.

Explanation:

The Brewster's angle is found using the tangent of the refractive index. Since the reflected and refracted rays are perpendicular at this angle, their sum equals 90∘90^\circ.

Polarisation Class 12 Notes & Examples | CBSE Physics