# Unpolarized Light

1. Apr 4, 2009

### hardwork

PART 1: Unpolarized light has an intensity of 100 W/m^2. It passes though a polarizer whose transmission axis is rotated 18 degrees with respect to the vertical. What is the intensity of the light that emerges from the polarizer?

I = (1/2)(Io)(cos angle)^2
I = (1/2)(100 W/m^2)(cos 18)^2
I = 45 W/m^2 -- I am not sure if this is correct.

PART 2: This light passes through a second polarizer and the intensity of the light that emerges is 40 W/m^2. What angel does the transmission axis of the second polarizer make with the vertical?

I don't understand this part. Would I use the same equation above and plug in 40 W/m^2 for I, 100 W/m^2 for Io, and solve for cos?

Thank you!

2. Apr 4, 2009

### rl.bhat

PART 1: From a polarizer 50% of light will be transmitted irrespective of the orientation of the transmission axis.
For part 2, use Malus's law.

3. Apr 5, 2009

### hardwork

Thank you!
For PART 1, would it just be I = (1/2)(100 W/m^2) then? So, I = 50 W/m^2 ?
I still don't understand PART 2. What do I plug into Malus' Law?

4. Apr 5, 2009

### rl.bhat

I = Imax*cos^2(theta)

5. Apr 5, 2009

### hardwork

Part 1 is okay then? The answer is 50?

Part 2: 40 W/m^2 = 100 W/m^2 cos^2(theta)
and then solve for cos?

6. Apr 5, 2009

### rl.bhat

No.
For second polarizer Imax is 50 w/m^2

7. Apr 5, 2009

### hardwork

Oh! Thank you so much for helping me. I'm sorry I'm a little confused.
Would the answer be 27 degrees?

I am not sure if I solved for the angle correctly. I'm forgetting my math skills here. haha

8. Apr 5, 2009

### rl.bhat

Part 2: 40 W/m^2 = 50 W/m^2 cos^2(theta)
cos^2(theta) = 0.8
Find cos theta. Add 18 degrees to this angle to get the result.

9. Apr 5, 2009

### hardwork

cos^2(theta) = 0.8
then take the sqrt of both sides?
cos(theta) = 0.9
cos-1(theta) = 26 degrees

Angle: 26 + 18 = 44 degrees ... I hope.

10. Apr 5, 2009

### rl.bhat

Yes. Correct. You can take 26.6 as well.

11. Apr 5, 2009

### hardwork

That was very helpful. Thanks, again!!