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Q. Unpolarised light of intensity 30.0W/m^2 passes through a polariser inclined at 35 degrees to the vertical. What is the trnasmitted intensity?
I know that I have to use Malus law, this is my working:
As it is unpolarised, I half 30/2=15
I=I0cos^2θ
= 15cos^2(35)
= 10W/m^2 but the answer is 15W/m^2
Can anyone please help...
As you know, Malus' law is this:Quote:
Unpolarized light of intensity 30.0W/m^2 passes through a polarizer inclined at 35 degrees to the vertical. What is the transmitted intensity?
I know that I have to use Malus law, this is my working:
As it is unpolarised, i half 30/2=15
I=I0cos^2θ
= 15cos^2(35)
= 10W/m^2 but the answer is 15W/m^2
where θ is the angle between the plane of polarization of the light and the polarizer. To use Malus' law, one typically uses two polarizers. The difference in the angles between their planes of polarization is θ. You feed unpolarized light to the first polarizer and polarized light comes out -- at half the intensity of the incident light.
This is a trick question. The 35° here is a red herring, probably meant to throw you off . Regardless of the position of the polarizer, the average value of cos(θ) is 1/2 and for unpolarized light (through a single polarizer -- because once it gets through the polarizer, it is no longer unpolarized light), Malus law is (you actually knew this, you just didn't realize that there was nothing to do after you got this far):
Wow, thanks for that. I was able to complete most of my other questions but then I got stuck on these two q's.
Q. Two sheets of Polaroid are placed on top of each other with an angle φ between their preferred
directions. What would be the ratio of transmitted to incident intensity for initially unpolarised light if
φ = 0.0°, 30.0°, 60.0°, 90.0°?
I'm not sure how to go about this question, do I use Malus law I=I0cos^2θ because we have unpolarised light?
AND the other q. was:
Q. Randomly polarised light passes through two polaroid sheets which are placed, one on top of the
other, with their axes at 90.0°.
(a) What fraction of the incoming light passes through the combination?
(b) If a third polaroid sheet is inserted between the two, with its axis at 45.0° to both, what
fraction of the incident light now passes?
The q. put me off when it said that the polaroid sheets are to be placed on top of each other. Help please...
1. Remember that the first one cuts the intensity of the incident light by 1/2 as it turns it into polarized light. As I explained, this is the usual way of running the experiment. The second one cuts it again by Malus' law. Io/2 times cos^2(φ)Quote:
Originally Posted by anony12
2. Same idea. With the first sheet, the incident light is cut in half. Except that the angle is 90 degrees and cos(90) degrees = 0. You get extinction. With a second sheet, at 45 degrees, you get 1/2 of the light through that and with a third sheet at 45 degrees cuts it down by another 1/2. You'll end up with 1/2 * 1/2 * 1/2 = 1/8 of the incident light.
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