Light through 3 linear polarizers

[Levi Tate]

Homework Statement

What is the most light that can be transmitted through 3 linear polarizers if you start with light polarized in the (i unit vector) direction and end with light polarized in the (j unit vector) direction for theoretically perfect polarizers

2. Homework Equations and attempt at solution.

Well I reason since the light is already polarized, I need to use the cosine squared law

I=IoCos^2∅ Where I is the intensity after going through a polarizer and Io is the initial intensity.

So the light in already polarized through the i direction, here in what I am thinking, If I send this light through 2 polarizers who's filters are parallel with the light in the i direction, then then angle is zero and all the light is retained.

But when I send it through a third filter in the j direction, as required in the problem, that is perpendicular to i of course, then no light emerges.

I know these are theoretically perfect polarizers but my result seems counterintuitive.

Is this right?

 

[opaka]

You can also put the second polarizer at a 45° angle to the first polarizer, and then some light will emerge.

 

[TSny]

Once you decide on what the orientation of the first and last filters must be, then there is only one variable to adjust. Write the final intensity as a function of that variable and see if you can maximize the function.

 

[Levi Tate]

Yes that was the one thing I was wondering, if I put the second polarizer at 45 degrees.

Could you please help me with maximizing the function? There is definitely only 1 variable, which is the second polarizer, but I am not sure how to maximize it.

I am also having trouble visualizing if it at the maximum at 45 degrees, I would like to see it mathematically but I'm not sure now.

 

[opaka]

To find the maximum of a function, take the derivative and set it to zero. So, write an equation for the light passing through the filters, leaving the angle variable only for the middle filter, take the derivative of that function, and solve for the unknown. If you're stuck, post the derivation as far as you can go.

 

[TSny]

Let θ be the angle between the transmission directions of the first and second filters. If I_o is the original intensity before the first filter, then you know that all the light gets through the first filter so the intensity is still I_o before entering the 2nd filter.

How would you write the intensity just after the 2nd filter?

What would be the direction of the polarization of the light after the 2nd filter?

Just before the light strikes the 3rd filter, what is the angle between the polarization direction of the light and the transmission axis of the 3rd filter?

 

[Levi Tate]

See if I use the equation I=IoCos(∅), If I set the equation equal to zero (IoCos∅) and take the inverse cosine of both sides, it's saying that the angle is at a maximum at zero. That is where I get to in that derivation.

I'm not sure how I would write the intensity after the second filter, I know that the incident light will be reduced, so I could write it as Io=Io2Cos(∅2), and similarly for the third filter after the light passes through.

I understand this, it just seems that 45 degrees is the middle ground for the i and j directions of that middle filter the more I think about it, but I am having trouble representing it mathematically.

Then if it is 45 degrees between the third filter and and the incident light, then the light is again reduced in intensity, but I am unsure of that direction as well, maybe (∅ - Pi/4).

I think I understand the problem intuitively but I am obviously having trouble representing it mathematically.

 

[vela]

See if I use the equation I=IoCos(∅), If I set the equation equal to zero (IoCos∅) and take the inverse cosine of both sides, it's saying that the angle is at a maximum at zero. That is where I get to in that derivation.

This is what I get from what you wrote:
$$I=I_0\cos\theta = 0$$
Then you take the inverse cosine of both sides:
$$\arccos(I_0\cos\theta) = \arccos 0$$
Therefore, $\theta=0$. It doesn't make sense. What did you actually do?

By the way, your expression for the intensity is wrong. You had it right in your original post: $I = I_0 \cos^2 \theta$.

 

[Levi Tate]

Yes, that's right, I forgot the square on the cosine,

So that is exactly what I did, I'm not sure how to find the maxium of the cosine, I just, okay, you're right, what I did does make no sense. This is what I meant to write

As TSny and opaka had said, I should have taken the derivative of the equation,

I=IoCos^2(θ)

So then setting IoCos^2(θ)=0

And taking the derivative, I get -2Iocosθsinθ=0

I think I can write this as -Iosin2θ=0That is about as far as I can get here.

 

[Levi Tate]

yeah thanks everybody for all the help so far, I think i understand the problem, I just need to see the math.

 

[vela]

Yes, that's right, I forgot the square on the cosine,

So that is exactly what I did, I'm not sure how to find the maxium of the cosine, I just, okay, you're right, what I did does make no sense. This is what I meant to write

As TSny and opaka had said, I should have taken the derivative of the equation,

I=IoCos^2(θ)

So then setting IoCos^2(θ)=0

When finding the extrema, you don't set I=0. You set the derivative I' equal to 0. There's no reason to write that last line.

And taking the derivative, I get -2Iocosθsinθ=0

I think I can write this as -Iosin2θ=0

That's right, and sine vanishes at 0 and 180 degree, right? That means the extrema occur at θ=0° and θ=90°. If you plug these values back into the original expression for I, you get I=I₀ for θ=0°, which makes sense. If the angle between the polarization of the light and the axis of the polarizer is 0, all the light gets through. For θ=90°, you find I=0. When the two are perpendicular to each other, no light gets through.

So you've derived what you already knew about how to maximize or minimize the light coming through one polarizer, but in this problem, you have two polarizers. (The first one doesn't matter because it let's all the light through.) So reread TSny's last post and figure out the answers to the questions asked. You want to find a function I(θ) that tells you how much light gets through to the end. Once you have that, maximize it.

 

[TSny]

Look at the attached figure which assumes the light is originally polarized horizontally. Go step by step. The green lines show the transmission axis direction for each filter. You have already figured out how the first and last filters must be oriented. So, you just need to find the angle θ for the middle filter such that you get maximum transmission overall.

How would you express I₁ in terms of I_o?

How would you express I₂ in terms of I₁ and θ?

How would you express I₃ in terms of I₂ and θ? Here you will need to think about how to express the angle shown with the question mark in terms of θ.

 

[Levi Tate]

God thank you guys so much. I still am really unsure of how to formulate this mathematically, but I am going to read over polarizers and see if I can bring any new information to the discussion.

 

[Levi Tate]

I don't understand why this was moved here, this is a pretty hard problem in a 500 level optics course at my school, which is a 3rd year optics course. Perhaps because we are doing the easier stuff from the intro books first, although really fast, before we get into physical optics, eh.

So anyways I thought about this for awhile and this is what I've got

I = Io1Cos(0) = Io1

Then the second polarizer must be positioned at a 45 degree angle relative to the first and the third, which is in the j direction, so

Io2 = Io1Cos(45) = [(2)^1/2]/(2)

But then the next angle is 90, so I think I have two parts of the problem represented properly, but of course the cosine of 90 is zero, so the math there says no light comes through, which doesn't make sense.

Help?

 

[TSny]

I = Io1Cos(0) = Io1

Then the second polarizer must be positioned at a 45 degree angle relative to the first and the third, which is in the j direction, so

Io2 = Io1Cos(45) = [(2)^1/2]/(2)

But then the next angle is 90, so I think I have two parts of the problem represented properly, but of course the cosine of 90 is zero, so the math there says no light comes through, which doesn't make sense.

Don't forget to square the cosine functions.

The last angle is not 90^o. What is the direction of polarization of the light just before it reaches the last filter? What is the angle between that polarization direction and the transmission axis direction of the last filter?

 

[Levi Tate]

edited..

 

[Levi Tate]

The angle between the polarization filter and the polarized light is 60 degrees. I'm not sure what my problem is.

 

[TSny]

The angle between the polarization filter and the polarized light is 60 degrees. I'm not sure what my problem is.

Not sure how you got 60^o. Remember, when light passes through a polarizing filter, the light comes out with its polarization direction the same as the direction of the transmission axis of the filter. So, if you set the middle filter at 45^o from the horizontal, then after the light passes through the filter the light will be polarized at 45^o to the horizontal. It then proceeds to the last filter (which has its transmission axis vertical). So, what is the angle between the transmission axis of the last filter and the polarization direction of the light that strikes the last filter?

 

[Levi Tate]

45 degrees, I think.

 

[Levi Tate]

I am going to do all the problems in the book until i figure this out, I'm really determined not to be this bad at physics.

 

[TSny]

45 degrees, I think.

That's right. So if you set the middle filter at 45^o you should be able to find the final intensity in terms of the initial intensity I_o.

For comparison, also see what you get if the middle filter is set at 30^o.

 

[Levi Tate]

Alright, I am going to try this again here, thank you so much.

 

[Levi Tate]

It turns out that the most light that comes through is if each angle is 60 degrees. I still don't fully understand the logic though.

 

[Levi Tate]

I definitely don't think this class is introductory physics. Here is the answer sheet, it was number 2 that we had in question here. I mean, it's possible that my teacher is wrong, but he has his Phd in optics.

Most of the text got messed up because I didn't have the right edition of Word, but number 2 is perfectly intact.

 

[TSny]

It turns out that the most light that comes through is if each angle is 60 degrees. I still don't fully understand the logic though.

I don't think it's 60 degrees. Remember, the three angles must add to 90 degrees. But, the correct answer is not what I originally thought! My intuition was faulty. I mistakenly thought that it would be best to align the first filter in the $\hat{i}$ direction (the initial polarization direction of the light). But that's not so. Sorry for not seeing that.

As in the solution that you posted as an attachment, let $\theta_1$ be the angle of the first filter with respect to polarization of the incident light, $\theta_2$ the angle of the second filter with respect to the first filter, and $\theta_3$ the angle of the third filter with respect to the second. We have the condition that the sum of the three angles must be 90^o.

The final amplitude of the electric field will be

$E_{final} = E_{initial} cos\theta_1 cos\theta_2 cos\theta_3$ .

Note that this expression is completely symmetric in the three angles. So, if there is a unique answer for the angles that maximize $E_{final}$, then the answer must be that all three angles are equal. Using the fact that the angles add to 90^o then gives the answer. But we had to assume the answer is unique.

Your posted solution finds the answer graphically.

Another way to get the answer is using calculus to find the angles that maximize $E_{final}$. Letting $\theta_3 = 90^o - \theta_1 -\theta_2$ you have

$E_{final} = E_{initial} cos\theta_1 cos\theta_2 cos(90^o - \theta_1 -\theta_2)$

or,

$E_{final} = E_{initial} cos\theta_1 cos\theta_2 sin(\theta_1 +\theta_2)$

A maximum occurs where the partial derivatives of $E_{final}$ with respect to $\theta_1$ and $\theta_2$ are both equal to zero. You can show that this will lead to the condition that $\theta_1 = \theta_2$.

You could then start over letting $\theta_1 = 90^o - \theta_2 -\theta_3$ to write $E_{final} = E_{initial} cos\theta_2 cos\theta_3 sin(\theta_2 +\theta_3)$ and show that maximum requires $\theta_2 = \theta_3$.

Hope this helps.

 

[Levi Tate]

I meant 30 degrees up there. I am going to try this here, your explanation of the solution is better than my teachers. Thank you so much.