Question

Unpolarized light with intensity 300 W/m^2 passes first through a polarizing filter with its axis vertical, then through a second polarizing filter. It emerges from the second filter with intensity 121 W/m^2. What is the angle from vertical of the axis of the second polarizing filter?

Answer 1

Answer:

The angle from vertical of the axis of the second polarizing filter is 50.57⁰.

Explanation:

Given;

intensity of the unpolarized light, I₀ = 300 W/m²

intensity of emergent polarized light, I = 121 W/m²

let the angle from vertical of the axis of the second polarizing filter = θ

Apply Malus's law, intensity of emergent polarized light is given as;

I = I₀Cos²θ

$Cos^2 \theta = \frac{I}{I_o} \\\\Cos^2 \theta =\frac{121}{300} \\\\Cos^2 \theta =0.4033\\\\Cos \theta = \sqrt{0.4033} \\\\Cos \theta = 0.6351\\\\\theta = Cos^{-1} (0.6351)\\\\\theta = 50.57 ^0$

Therefore, the angle from vertical of the axis of the second polarizing filter is 50.57⁰.