Finding the Solution: Solving for x in a Polarization of Light Equation

[johnsobertstamos]

Homework Statement

A series of polarizers are each placed at a 12 ∘ interval from the previous polarizer. Unpolarized light is incident on this series of polarizers. How many polarizers does the light have to go through before it is 1/9 of its original intensity?

Relevant Equations

I = Io (cos theta)^2
Iintensity polarized = 1/2 Intensity unpolarized

I set up the equation

.5((cos 12)^2)^x = 1/9

Solving for x gets me 34.037

34 + 1 = 35

I've entered answers of both 34 and 35 and both have been marked as wrong. Does anybody know what I'm doing wrong here?

 

[berkeman]

Welcome to the PF.

How many polarizers does the light have to go through before it is 19 of its original intensity?

19 what of its original intensity? 19%?

 

[johnsobertstamos]

Welcome to the PF.

19 what of its original intensity? 19%?

Sorry, should be 1/9

 

[johnsobertstamos]

Ah, okay thanks.

Do both of those equations apply to the first and all subsequent polarization stages?

The bottom one applies to just the first, reducing the intensity by half. The top one applies to the rest of the polarization stages.

 

[berkeman]

.5((cos 12)^2)^x = 1/9

Solving for x gets me 34.037

34 + 1 = 35

I get a different value of x for the solution of that equation, where x is the number of polarizers after the first one. Could you show your work so we can check it please? Thanks.

I(x) = \frac{1}{2} [cos^2(12)]^x

 

[berkeman]

The bottom one applies to just the first, reducing the intensity by half. The top one applies to the rest of the polarization stages.

Yes, sorry, I deleted my post after I made it because I realized that you had already factored that in. Please check out the reply that I just made. Thanks.

 

[gneill]

"How many polarizers does the light have to go through before it is 1/9 of its original intensity?"

Is that the exact phrasing of the original problem statement? Perhaps it says something equivalent to "less than 1/9 of the original intensity"? In that case the ".037" in the resulting calculation might play a role... Just a thought.

 

[johnsobertstamos]

"How many polarizers does the light have to go through before it is 1/9 of its original intensity?"

Is that the exact phrasing of the original problem statement? Perhaps it says something equivalent to "less than 1/9 of the original intensity"? In that case the ".037" in the resulting calculation might play a role... Just a thought.

That's the phrasing given in the problem. I also forgot to mention it asks for the answer in the form of an integer.

 

[johnsobertstamos]

I get a different value of x for the solution of that equation, where x is the number of polarizers after the first one. Could you show your work so we can check it please? Thanks.

I(x) = \frac{1}{2} [cos^2(12)]^x

1) .5((cos 12)^2)^x = 1/9

2) ((cos 12)^2)^x = 2/9

3) x ln((cos 12)^2) = ln (2/9)

4) x = (ln (2/9)) / ln((cos 12)^2)

5) x = 34.037

 

[johnsobertstamos]

Still at a complete loss for what I'm doing wrong here. My math checks out every time.

 

[berkeman]

2) ((cos 12)^2)^x = 2/9

3) x ln((cos 12)^2) = ln (2/9)

I think you should be using log(), not ln()...

 

[johnsobertstamos]

I think you should be using log(), not ln()...

It gets me the same answer of 34.037 Any differences in the equations we have set up?

 

[johnsobertstamos]

I'm inputting

x = log(2/9) / log((cos 12)^2)

x = 34.037

 

[berkeman]

Hmm, I found an error by a factor of 2 just now in my previous calculation, and now I get the same as you: 1+34=35. Can you check with a TA or the instructor?

 

[TSny]

".037" in the resulting calculation might play a role... Just a thought.

Right. With 35 filters I get that the resulting intensity is still a wee bit above 1/9 the initial intensity.

 

[johnsobertstamos]

Right. With 35 filters I get that the resulting intensity is still a wee bit above 1/9 the initial intensity.

Ah so maybe it's 36 then

 

[berkeman]

That seems like the most likely option at this point. Can you check it?

 

[johnsobertstamos]

That seems like the most likely option at this point. Can you check it?

Yep 36, thank you everybody for your help. Much appreciated!