# Homework Help: Polarizer Easy Q Help

1. Nov 26, 2007

### weathercheck

1. The problem statement, all variables and given/known data
Three polarizing filters are stacked, with the polarizing axis of the second and third filters at angles of 17.4^\circ and 59.8^\circ, respectively, to that of the first. If unpolarized light is incident on the stack, the light has an intensity of 74.0 W/cm^2 after it passes through the stack.

If the incident intensity is kept constant, what is the intensity of the light after it has passed through the stack if the second polarizer is removed?

Im getting the original Intensity to be 421.

Anyone help me with this?

regards

2. Nov 26, 2007

### Staff: Mentor

3. Nov 26, 2007

### weathercheck

I said

74 = Io/4 (.7030)

4. Nov 26, 2007

### Staff: Mentor

I'm expecting to see something relating to the given angles.

5. Nov 26, 2007

### weathercheck

I = IoCos^2(phi)

6. Nov 26, 2007

### weathercheck

So go through second filter, Io/2 and third, Io/4

7. Nov 26, 2007

### weathercheck

well?

8. Nov 26, 2007

### Staff: Mentor

Good. This will be used to find the transmitted intensity passing through the second and third filters.

I don't understand this. If the incident intensity is Io, what is the intensity after passing through the first filter? Second filter? Third filter?

9. Nov 26, 2007

### weathercheck

Io/2, Io/4, Io/8 ??

10. Nov 26, 2007

### Staff: Mentor

Nope. What happened to the equation (Malus's law) you quoted in your earlier post? The angles make a difference.

The Io/2 thing only applies to unpolarized light going through a polarizer. But after the light goes through the first filter, it becomes polarized light.

11. Nov 26, 2007

### weathercheck

Can you explain it a wee bit for me, like when the light goes through the first screen there is no angle so surely it does not change the intensity when it hits the second screen. When it hits the second screen it I = Io Cos^2(17.4), third I2=I1Cos^2(55.2)

??

12. Nov 26, 2007

### Staff: Mentor

When unpolarized light passes through a polarizer, the intensity is changed to Io/2.
Not exactly. Read up on the http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polcross.html#c3".

For clarity, label the intensity at each point:
Initial intensity: I_0
After passing through filter 1: I_1
After passing through filter 2: I_2
After passing through filter 3: I_3

Figure out how the intensity changes at each point.

Last edited by a moderator: Apr 23, 2017
13. Nov 26, 2007

### weathercheck

So im getting say.

When passing through first filter..

I_1=Io
I_2=.91Io
I_3=(.91)(.545)Io

??

14. Nov 26, 2007

### Joza

Use Malus's Law to work backward to the original intensity falling on the first polarizer.

Intensity=(Intensity falling on polarizer)(cos angle)^2

Then just use the formula for the angle that the 3rd polarizer makes with the 1st.

Understand???

Are you in Ireland, if you don't mind me asking???

15. Nov 26, 2007

### Staff: Mentor

No. (Review my previous posts.)
Good: $$I_2 = I_1 \cos^2(17.4)$$
Excellent. Of course that's really I_3 = (.91)(.545)I_1.

Just fix that value for I_1 and you're good to go.

16. Nov 26, 2007

### weathercheck

So I_1 = 1o/2

=> I_3=(.91)(.545).Io/2

Correct?

thanks alot

17. Nov 26, 2007

### Staff: Mentor

That looks good. You are given I_3, now solve for I_0.

18. Nov 26, 2007

### weathercheck

I got Io=298.41

19. Nov 26, 2007

### weathercheck

Then for the q said, I2 =(1/2Io)(.25302)

and found it to be 37.751

20. Nov 26, 2007

### Staff: Mentor

Sounds good to me.

21. Nov 26, 2007

### weathercheck

Wahey

Tis right,

Thanks alot ;)