Consider three polarizer’s P1, P2 and P3 placed along an axis. The pass axis of P1 and P3 are at right angles to each other while the pass axis of P2 makes an angle θ with that of P1. A beam of unpolarized light of intensity I0 is incident on P1 as shown. The intensity of light emerging from P3 is

( a ) 0                                        ( b )   I ₀ / 2           

( c ) ( I ₀ / 8 ) sin ² 2 θ               ( d ) ( I ₀ / 4 ) sin ² 2 θ

Solution :

According to Malus Law , when a polarizer is placed in a polarized beam of light , the intensity , I , of the light that passes through is given by ( I ₀ ) cos ² θ .

where I ₀ is the initial intensity and θ is the angle between the light's initial polarization direction and the axis of the polarizer .

When a beam of unpolarized light {\displaystyle \cos ^{2}\theta }is passes through 1 ^st polarizer , the intensity of polarized light transmitted by polarizer P ₁ is half the value of intensity of unpolarized light that incident on polarizer P _{1 .}

Mathematically it can be given as ,

I ₁ = I ₀ / 2           - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ( 1 )

Now , when the light transmitted from polarizer P _{1 ,} the transmitted plane polarized light of intensity I ₀ / 2 incident on second polarizer P _{2 ,} such that the pass axis of polarizer P ₂  which makes an angle θ with that of P ₁ .

Therefore , according to Malus law , the intensity of light transmitted by polarizer P ₂ can be given as ,

I ₂ = I ₁ cos ² θ

From equation ( 1 )

I ₁ = I ₀ / 2

Therefore , the above equation becomes ,

I ₂ = ( I ₀ / 2 ) cos ² θ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ( 2)

Now , the light transmitted from polarizer P ₂ of intensity I ₀ / 2 cos ² θ incident on third polarizer P _{3 ,} the pass axis of polarizer P ₃ which makes an angle ( 90 – θ ) with that of P ₂ .

The intensity of light transmitted by polarizer P ₃ can be given as ,

I ₃ = I ₂ cos ² θ

From equation ( 2 )

I ₂ = I ₀ / 2 cos ² θ

Therefore , the above equation becomes ,

I ₃ = ( I ₀ / 2 ) cos ² θ . cos ² ( 90 – θ )

I ₃ = ( I ₀ / 8 ) sin ² 2 θ

The intensity of light emerging from P ₃ is ( I ₀ / 8 ) sin ² 2 θ

Answer : ( C ) ( I ₀ / 8 ) sin ² 2 θ