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Polarization I = Io*cos^2 theta

  • Thread starter DAP1MP13
  • Start date
8
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Two polarizers are oriented at 38.8o to each other and plane-polarized light is incident on them. If only 14.6 percent of the light gets through both of them, what was the initial polarization direction of the incident light?

Ok, I'm not exactly sure what this problem is asking for, do they want an angle or what?

So basically the only equation given for this section is I = Io*cos^2 theta.
I did ((cos*38.8)^2)((cos*38.8)^2) = .369....

I don't see how I can solve the problem with just this, I thought maybe I can relate it to Snell's eq., but there is no index (n) given.
Also I know that I-final = 14.6% = .146, but once again I don't really know what to do with this.

Any hints and help will be greatly appreciated.
Thanks....
 
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Doc Al

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That equation, the law of Malus, tells you what percentage (of the intensity) of a polarized light beam gets through a polarizer set at an angle of [itex]\theta[/itex] to the incident beam's plane of polarization. (And realize that when a beam passes through a polarizer, the part that survives is totally polarized along the axis of the polarizer.)

You figured out the percentage that got throught the second polarizer. So what must have been the percentage that got through the first? (You know the net percentage that got through.)
 
8
0
Thanks Doc Al

So the net amount that got through both polarizers was .369, but it says that .146 got through.
The amount that went throught the first would just be (cos38.8)^2 = .607.

I still don't know how this would help me to get the answer....
Through 1st polarizer => .607
Through 2nd polarizer => .369 or .146
How would this information tell you what was the initial polarization direction of the incident light?
 
8
0
Can I get some more help on this please?

Thanks
 
88
0
DAP1MP13 said:
Two polarizers are oriented at 38.8o to each other and plane-polarized light is incident on them. If only 14.6 percent of the light gets through both of them, what was the initial polarization direction of the incident light?

Ok, I'm not exactly sure what this problem is asking for, do they want an angle or what?

So basically the only equation given for this section is I = Io*cos^2 theta.
I did ((cos*38.8)^2)((cos*38.8)^2) = .369....

I don't see how I can solve the problem with just this, I thought maybe I can relate it to Snell's eq., but there is no index (n) given.
Also I know that I-final = 14.6% = .146, but once again I don't really know what to do with this.
there are 2 angles involved in this problem.
first, the 2 polarizers are rotated relative to one another by angle θ2=38.8°
then, the plane polarized incident light is rotated relative to the first polarizer by angle θ1.
the total net transmission thru both polarizers is given to be 0.146, so the transmission equation is:
I = I0cos21)cos22)
or rearranging and placing known values into the equation:
I/I0 = 0.146
= cos21)cos22)
= cos21)cos2(38.8°)

cos21) = (0.146)/cos2(38.8°)
cos(θ1) = SQRT{(0.146)/cos2(38.8°)}

Now solve for θ1 to find the incident light angle (relative to first polarizer).
 

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