Question

Unpolarized light is passed through an optical filter that is oriented in the vertical direction.
If the incident intensity of the light is 46 W/m2 , what is the intensity of the light that emerges from the filter? (Express your answer to two significant figures.)

Answer 1

In order to solve this problem it is necessary to apply the concepts related to intensity and specifically described in Malus's law.

Malus's law warns that

$I = I_0 cos^2\theta$

Where,

$\theta=$ Angle between the analyzer axis and the polarization axis

$I_0 =$Intensity of the light before passing through the polarizer

The intensity of the beam from the first polarizer is equal to the half of the initial intensity

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

Replacing with our the numerical values we get

$I = \frac{46}{2}$

$I = 23W/m^2$

Therefore the  intensity of the light that emerges from the filter is $23W/m^2$