Intensity of light through N linear polarizers

1. Feb 5, 2013

Levi Tate

1. The problem statement, all variables and given/known data

What is the intensity of the electric field that can be transmitted through N linear polarizers if the polarization of the initially linear polarized light is to be rotated 90 degrees?

2. Relevant equations

I=IoCos^2x

3. The attempt at a solution

I am seeing like a Riemann sum forming, the infinitesimal being taken as it is rotated. That is about as much as I've got, some sort of rate of change as it is rotated infinitesimally. This is a 300 level optics class and my teacher makes the homework way too hard, in my opinion.

2. Feb 5, 2013

Simon Bridge

... there should be some condition on the rotation angle on each polarizer.

Assuming they are all the same angle to each other... the first and last polarizers are $90^\circ = \pi/2$ to each other. (why?)

To understand the problem, you should play around with it for a bit:
Try it for N=2,3,4 ... and see what happens.
For N=2, there is just the first and last polarizers which are crossed - so how much light is transmitted?
What can you say about the N=2 setup in relation to the aim (rotating the polarized light through $\pi/2$ radiens?)

If $\theta$ is the angle between sucessive polarizers, then N=3 means $\theta = \pi/4$ and N=4 means $\theta=\pi/6$ (that would be 45deg and 30deg if you prefer but it's good to get into the habit of working in radiens).

So for N=3, how much light gets through?

Last edited: Feb 5, 2013
3. Feb 5, 2013

Levi Tate

Actually it turns out for N= 3 the maximum light coming through is at pi/3, somebody else very smart figured it out on this forum actually.

The answer is I=IoCos^N(θ/2/N)

According to my teacher. I just have no idea what the hell he did in the derivation of getting this answer.

Does anybody think this problem is too hard for a 300 level course?

4. Feb 5, 2013

Levi Tate

excuse me i meant pi/6 up there

5. Feb 5, 2013

Simon Bridge

The lynchpin is that, in general, if you number the polarizers 1,2,3,... then the the intensity of light after the nth polarizer is $I_n = I_{n-1}\cos(\theta_n/2)$ where $\theta_n$ is the relative angle of polarizer n from polarizer n-1.

No - it is not too hard for your level.
Electric fields are way harder.

6. Feb 5, 2013

Levi Tate

I think it's too hard I come out of introductory EM and am getting murdered by this guy, I can't even follow the solutions and I'm a very serious student. The only kids that can solve the homework are the group of kids taking quantum physics 2.

7. Feb 5, 2013

Simon Bridge

Well, I could be mistaken about what you mean by "a 300-level course" ... but this is in the High School curriculum in NZ. If you are in a college science course involving maths it is a reasonable (indeed - routine) task.

You need to practice playing around with relations and trying to understand them as physics rather than abstract equations. Like do you understand the answer? It won't do you any good if you don't.

I feel your frustration and sympathize - however there is no easy way to get these skills.

Last edited: Feb 5, 2013
8. Feb 5, 2013

Levi Tate

Apparently my country really is a bunch of idiots then.

9. Feb 5, 2013

Levi Tate

I'm sorry i'm just really really frustrated right now, i've been pouring my heart into this class and this guy came over from Cornell a month ago and is using complex variables and advanced vector calculus, stuff that the physics kids don't learn until we take that class next semester, but he doesn't care. I'm just so ****ing mad because I've been working really hard about this stuff, crying over it, and then you get the problem wrong and the solution doesn't make any sense.

Thank you for the help and all, sorry about that.

10. Feb 5, 2013

Simon Bridge

If this is the USA - you guys do have a reputation you know... but it depends really. If you get used to one way it can be rough when you have to do things another way.
When I say what happens here, I don't mean to put anybody down - I mean to highlight my POV.
This is an international forum and what seems OK for NZ may not apply in other countries.

Your prof is expecting you to be able to work stuff out for yourself. It is normal for a physics course to require math that has yet to be formally taught - you get used to it.. and, these days, you have the internet. Why in my day we used to do these things from the cradle while shovelling snow... or something.

You are actually training to be able to do problems where there are no known methods to solve them - you have to figure the method out from scratch. All this is good practice.

Meantime: builds character :D

11. Feb 5, 2013

Levi Tate

Yeah i know most of the people here are ****ing jerk offs. Let me tell you something. I'm not one of them, alright.

You make a solid point, it will build character if it doesn't drive me insane first.

12. Feb 6, 2013

Simon Bridge

That's the spirit - nil carborundum :D