## Malus Law:

** Statement:** It states that the intensity of polarized light transmitted through the analyzer is directly proportional to the square of Cos θ where θ is the angle between the plane of the polarizer and the plane of the analyzer.

i.e. I ∝ cos^{2} θ |

** Proof:** Let ‘

**‘ be the amplitude of light transmitted through the polarizer and θ be the angle between the plane of the polarizer and the plane of the analyzer.**

*a*Resolve ‘** a**‘ into two components i.e.,

**along OA and**

*a cos θ***along OB. Clearly,**

*a sin θ***component will pass through the analyzer. We know,**

*a cos θ*Intensity ∝ (Amplitude)^{2}⇒ I ∝ (a cos θ) ^{2}⇒ I ∝ a ^{2} cos^{2} θ⇒ I = k a ^{2} cos^{2} θPut ka ^{2} = I_{o} = constant⇒ I = I _{o} cos^{2} θ⇒ I ∝ cos ^{2} θ |

Two polaroids are crossed to each other. Now one of them is rotated through 60°. What percentage of incident unpolarized light will pass through the system?Sample Problem: Initially the polaroids are at 90°. When one is rotated through 60°, the angle between their transmission axes is 30° i.e., θ = 30°.Solution:Let I _{o} be the intensity of the incident unpolarized beam. The intensity of the beam transmitted by the polarizer will be I_{o}/2 as the other I_{o}/2 is absorbed by the polaroid.The intensity of light transmitted by the analyzer will, therefore, be given by I _{o} = I_{m} cos^{2} θwhere I _{m} = I_{o}/2, θ = 30°, henceI _{θ} = (I_{o}/2) cos^{2} 30° = (I_{o}/2) (√3/2)^{2}∴ (I _{θ}/I_{o}) x 100 = (3/8) x 100 = 37.5% |