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 33 ∘ from horizontal.

By what percent does the light intensity transmitted through the polarizer decrease as the laser warms up?

Express your answer using two significant figures.

Answer 1

As we know that the intensity of the light varies as per relation
I = I₁Cos²()
Where I is the intensity after the polarisation , I₁ is the intenstiy after the initial polarisation (i.e. horizontal polarisation)
is the angle between the polariser and the direction of the polarised light.
Here the value of = 33 , therefore
I = I₁Cos²(33) = 0.7033 I₁
But when the laser get warms up then the light will become vertically polarised
then the will become = 90-33 = 57 degree
hence
I = I₁Cos²(57) = 0.2966 I₁
Hence we calculate the decrease in the intenstiy due to laser warm up
percentage decrease = ((Final intensity - Intiial intensity) / Intiial intensity)*100
= ((0.2966I₁ - 0.7033I₁)  / 0.7033I₁) *100 = 57.823 %
Hence this amount of percentage will decrease.