# A 200 mW horizontally polarized laser beam

1. Jul 2, 2009

### jlmessick88

1. The problem statement, all variables and given/known data
A 200 mW horizontally polarized laser beam passes through a polarizing filter whose axis is 25 degrees from vertical.

What is the power of the laser beam as it emerges from the filter?
P=mW

2. Relevant equations

E (transmitted) = E (incident)cos()

3. The attempt at a solution
200= E(incident) cos (25)
E(incident) = 2.206*10^2

Not sure if this is correct..this is my last attempt at the question, need to know if I'm approaching this correctly at all?!?!
Thanks!

2. Jul 2, 2009

### Staff: Mentor

(1) Don't mix up the amplitude (E) with the intensity (what you want), which is governed by the Law of Malus. See: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polcross.html#c3"
(2) The incident power is given; that's what the 200 mW refers to.
(3) Note that the incident wave is horizontally polarized, while the angle given is with respect to the vertical.

Last edited by a moderator: Apr 24, 2017
3. Jul 2, 2009

### jlmessick88

so instead, if i understand you...it should be 200*cos(25) = 1.81*10^2???

4. Jul 2, 2009

### Staff: Mentor

No, but you're getting closer. Read all three of my points, not just the second one.

5. Jul 2, 2009

### jlmessick88

so in regards to #3...instead of being 25 degrees...it would be 65 degrees?? if i'm picturing this correctly

6. Jul 2, 2009

### Staff: Mentor

Yes, that's what I would say.

7. Feb 28, 2011

### kikko

200*cos(65) = 84.5 =/= An answer Mastering Physics will mark as right.

8. Mar 1, 2011

### Redbelly98

Staff Emeritus
See Doc Al's link to the Law of Malus in Post #2. What does the link say about the transmitted intensity? (Don't confuse it with the electric field.)