A 200 mW horizontally polarized laser beam

[jlmessick88]

Homework Statement

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

Homework Equations

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

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!

 

[Doc Al]

The Attempt at a Solution

200= E(incident) cos (25)
E(incident) = 2.206*10^2

(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.

 

[jlmessick88]

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

 

[Doc Al]

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

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

 

[jlmessick88]

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

 

[Doc Al]

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

Yes, that's what I would say.

 

[kikko]

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

 

[Redbelly98]

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.)