Question

Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second polarizing filter. It emerges from the second filter with intensity 138 W/m2 .
a. What is the angle from vertical of the axis of the second polarizing filter?b. Express your answer to two significant figures and include the appropriate units

Answer 1

To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.

From the law of Malus intensity can be defined as

$I' = \frac{I_0}{2} cos^2 \theta$

Where

$\theta =$Angle From vertical of the axis of the polarizing filter

$I_0 =$ Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

$I = \frac{I_0}{2}$

Replacing we have that

$I = \frac{370}{2}$

$I = 185W/m^2$

Re-arrange the equation,

$I'= \frac{I_0}{2}cos^2\theta$

Re-arrange to find \theta

$cos^2\theta = \frac{2I'}{I_0}$

$cos^2\theta = \frac{2*138}{370}$

$\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})$

$\theta = 0.5282rad$

$\theta = 30.27\°$

The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°