Find Intensity of Light After Removing Second Polarizer | Polarizer Easy Q"

[weathercheck]

Homework Statement

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

 

[Doc Al]

Show how you arrived at your answer.

 

[weathercheck]

I said74 = Io/4 (.7030)

 

[Doc Al]

What physics principles did you use to get your answer?

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

 

[weathercheck]

I = IoCos^2(phi)

 

[weathercheck]

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

 

[weathercheck]

well?

 

[Doc Al]

I = IoCos^2(phi)

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

What about the first?

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

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?

 

[weathercheck]

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

 

[Doc Al]

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

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.

 

[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)

??

 

[Doc Al]

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 unpolarized light passes through a polarizer, the intensity is changed to Io/2.

When it hits the second screen it I = Io Cos^2(17.4), third I2=I1Cos^2(55.2)

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.

 

[weathercheck]

So I am getting say.

When passing through first filter..

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

??

 

[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

Sorry about the format there^^^^.

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?

 

[Doc Al]

So I am getting say.

When passing through first filter..

I_1=Io

No. (Review my previous posts.)

I_2=.91Io

Good: $$I_2 = I_1 \cos^2(17.4)$$

I_3=(.91)(.545)Io

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.

 

[weathercheck]

So I_1 = 1o/2

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

Correct?

thanks alot

 

[Doc Al]

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

 

[weathercheck]

I got Io=298.41

 

[weathercheck]

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

and found it to be 37.751

 

[Doc Al]

Sounds good to me.

 

[weathercheck]

Wahey

Tis right,

Thanks a lot ;)