Question

The average intensity of light emerging from a polarizing sheet is 0.708 W/m2, and that of the horizontally polarized light incident on the sheet is 0.960 W/m2. Determine the angle that the transmission axis of the polarizing sheet makes with the horizontal.

Answer 1

Answer:

Angle θ = 30.82°

Explanation:

From Malus’s law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ

where;

I_o is the intensity of the polarized wave before passing through the filter.

In this question,

I is 0.708 W/m²

While I_o is 0.960 W/m²

Thus, plugging in these values into the equation, we have;

0.708 W/m² = 0.960 W/m² •cos²θ

Thus, cos²θ = 0.708 W/m²/0.960 W/m²

cos²θ = 0.7375

Cos θ = √0.7375

Cos θ = 0.8588

θ = Cos^(-1)0.8588

θ = 30.82°