# Intensity of light with given power through a polarizer

1. ### Linus Pauling

190
1. A 200 mW horizontally polarized laser beam passes through a polarizing filter whose axis is 25 degrees from vertical.

2. Malus's Law

3. Ok, so I will use theta = 65 degrees in the I = I0cos(theta) equation, right? But how do I calculate I0 knowing only power and not area?

2. ### ehild

12,941
How is the intensity of light defined? And what does Malus's Law say? The equation you use is wrong.

ehild

3. ### Linus Pauling

190
I = P/A = 0.5c*ep0*E2

And Malus's law states that the intensity of the light transmitted is the product of the intensity of the incident light and the square of the cosine of the angle theta with respect to the polarizer's axis. I don't get what you're getting at....

In terms of the theta I use, this is how I am thinking about it: I would use 25 degrees if the incident light was vertically polarized. Since it is horizontally polarized, i.e. is at 90 degrees, I should use 90-25 = 65 degrees...

4. ### ehild

12,941
It is all right if you use the square of the cosine, but you wrote I=Io*cos(theta) in the previous post.
As for intensity, it is power/area. The incident power of the laser beam is given and the area is the same before and after crossing the polarizer. You did not write what is the question, but presumably it is the power of the laser beam after the polarizer.

ehild