Optics - Three polarizers with 1 moving

[DataGG]

Homework Statement

Thee linear polarizers are in sequence. Let first and last polarizer be crossed (perpendicular to each-other) and the middle polarizer rotate with angular frequency $\omega$. Show that under such circumstances, $I$ is given by: $$I = I_0 \frac{1}{16}(1-\cos (4\omega t))$$.

The light that strikes the first polarizer is unpolarized.

Please, see the following link (yes, my paint skills are horrible! I tried to make it in inkspace, but I don't know how to work with that program yet!): https://imgur.com/UwQShhx

Homework Equations

Malus Law, which is given by $I(\theta) = I_0 \cos^2 (\theta)$

The Attempt at a Solution

If you saw the following link, you noticed two angles, $\theta$ and $\alpha$.

I think $\theta=\omega t$ and $\alpha = 90 - \omega t$

So, after the first polarizer:

$$I_1= \frac{I_0}{2}$$

After the second polarizer:

$$I_2= I_1 \cos^2 (\omega t)$$

After the third polarizer:

$$I_3 = I_2 \cos^2 (90-\omega t) \Leftrightarrow I_3 = I_1 \cos^2 (\omega t) \cos^2 (90-\omega t) $$

I can play around with that but I never arrive at the $I = I_0 \frac{1}{16}(1-\cos (4\omega t))$.

Any ideas?

 

[Doc Al]

After the third polarizer:

$$I_3 = I_2 \cos^2 (90-\omega t) \Leftrightarrow I_3 = I_1 \cos^2 (\omega t) \cos^2 (90-\omega t) $$

I can play around with that but I never arrive at the $I = I_0 \frac{1}{16}(1-\cos (4\omega t))$.

Any ideas?

You need to monkey around with some trig identities. Step 1: Get rid of that $90 - \omega t$ term. (Get everything in terms of just $\omega t$.) Then you are just two identities away from the answer.

 

[DataGG]

You need to monkey around with some trig identities. Step 1: Get rid of that $90 - \omega t$ term. (Get everything in terms of just $\omega t$.) Then you are just two identities away from the answer.

Hey Doc, thanks for answering.

I did do that in paper, which I'll copy to latex below:

$$ I_3 = I_1 \cos^2 (\omega t) \cos^2 (90-\omega t) = I_1 \cos^2 (\omega t)[\cos (90) \cos (\omega t) - \sin (90) \sin (\omega t)]^2 = I_1 cos^2 (\omega t) (-sin^2 (\omega t)) $$

Oh, btw, I forgot to say that the light that strikes the first polarized is unpolarized. Will edit OP.

EDIT: Sorry, that latex line is long. Kind of ruins the forum template for me. Don't know how to make it so the equal signs are aligned in different lines..

 

[Doc Al]

I did do that in paper, which I'll copy to latex above:

$$ I_3 = I_1 \cos^2 (\omega t) \cos^2 (90-\omega t) = I_1 \cos^2 (\omega t)[\cos (90) \cos (\omega t) - \sin (90) \sin (\omega t)]^2 = I_1 cos^2 (\omega t) (-sin^2 (\omega t)) $$

OK, except for a sign error.

Keep going.

 

[DataGG]

OK, except for a sign error.

Keep going.

Indeed, there is a sign error.

$$I_3 = I_1 cos^2 (\omega t) (sin^2 (\omega t)) = \frac{I_0}{2} \cos^2 (\omega t) [1-cos^2(\omega t)] = \frac{I_0}{2} [\frac{1}{2}(1+ \cos (2 \omega t)][1-\frac{1}{2}(1+ \cos (2 \omega t)] $$

Then comes

$$I_3= \frac{I_0}{4}(1+ \cos (2 \omega t)) (\frac{1}{2}(1- \cos (2 \omega t)) = \frac{I_0}{8} (1 - cos^2 (2 \omega t))$$

And now I'm not sure what to do.. Maybe expand the square? I'm not seeing anything..

EDIT: Actually, expanding the square might probably work. I'll then have another co-sin term squared, and If I do the substitution I used above, a 1/16 will appear and.. ya, let me try it on paper..

 

[Doc Al]

Indeed, there is a sign error.

$$I_3 = I_1 cos^2 (\omega t) (sin^2 (\omega t)) = \frac{I_0}{2} \cos^2 (\omega t) [1-cos^2(\omega t)] = \frac{I_0}{2} [\frac{1}{2}(1- \cos (2 \omega t)][1-\frac{1}{2}(1- \cos (2 \omega t)] $$

Then comes

$$I_3= \frac{I_0}{4}(1- \cos (2 \omega t)) (\frac{1}{2}(1- \cos (2 \omega t)) = \frac{I_0}{8} (1 - cos (2 \omega t))^2 $$

And now I'm not sure what to do.. Maybe expand the square? I'm not seeing anything..

EDIT: Actually, expanding the square might probably work. I'll then have another co-sin term squared, and If I do the substitution I used above, a 1/16 will appear and.. ya, let me try it on paper..

You have another sign error: When you express $\cos^2 \omega t$ in terms of $\cos 2\omega t$. Fix that and you're almost home.

 

[DataGG]

I made a sign mistake there, again! Will correct asap

edit: damn, you were faster!

edit2: Fixed the mistake, I think.

 

[Doc Al]

I made a sign mistake there, again! Will correct asap

edit: damn, you were faster!

edit2: Fixed the mistake, I think.

Believe it or not, you still have a sign error. In your second line.

 

[DataGG]

humm

 

[Doc Al]

humm

Consider the very last factor in your first line, which is correct. When you simplified that factor in the second line, you messed up a bit.

 

[DataGG]

I'm an idiot! Alright, I now finally got it. Edited the post and below is the solution:

$$I_3 = \frac {I_0}{8}[1- \cos^2 (2 \omega t)] = \frac {I_0}{8} (1 - \frac {1}{2} - \frac {\cos (4 \omega t}{2})$$

and then finally,

$$I_3 = \frac {I_0}{16} (1- \cos (4 \omega t))$$

Thank you Doc Al!

 

[Doc Al]

I'm an idiot! Alright, I now finally got it. Edited the post and below is the solution:

$$I_3 = \frac {I_0}{8}[1- \cos^2 (2 \omega t)] = \frac {I_0}{8} (1 - \frac {1}{2} - \frac {\cos (4 \omega t}{2})$$

and then finally,

$$I_3 = \frac {I_0}{16} (1- \cos (4 \omega t))$$

Thank you Doc Al!

:thumbs: (And you are welcome!)