Question

The polarization of a helium-neon laser can change with time. The light from a 1.5 mW laser is initially horizontally polarized; as the laser warms up, the light changes to be vertically polarized. Suppose the laser beam passes through a polarizer whose axis is 21 ∘ from horizontal. By what percent does the light intensity transmitted through the polarizer decrease as the laser warms up?

(Iwarm-Icold)/Icold= ? %

Answer 1

Let’s say I_o is the initial intensity of light before going through the polarizer

I_o = 1.5 mW/m²

When the laser is cold it is horizontally polarized. The angle between the polarization of laser and the polarizer is θ = 21^o

The transmitted intensity is obtained by Law of Malus

I_cold = I_ocos²θ

I_cold = 1.5 x (cos21)²

I_cold = 1.307 mW/m²

When the laser is warm it is horizontally polarized. The angle between the polarization of laser and the polarizer is

φ = 90 - θ

φ = 90- 21^o

φ = 69^o

The transmitted intensity is obtained by Law of Malus

I_warm = I_ocos²φ

I_warm = 1.5 x (cos69)²

I_warm = 0.193 mW/m²

(I_warm - I_cold)/I_cold = (0.193 – 1.307 )/ 1.307 = -0.85

The mimes sign indicated that the intensity got reduced with warm up.

So intensity decreases by 85%