Intensity of light through two polarizers

In summary: Then, you know that the second polarizer will only allow a certain percentage of that light to pass through, which you can also calculate using that equation. The total intensity of light transmitted by the second film is 1.5 W/m^2.In summary, the intensity of unpolarized light of 3 W/m^2 is reduced by half when passing through one polarizing film. The second film, with its axis 60 degrees apart from the first, will only allow a certain percentage of the already reduced light to pass through, resulting in a total intensity of 1.5 W/m^2.
  • #1
Mddrill
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Homework Statement



Unpolarized light of intensitry 3 W/m^2 goes through two polarizing films. Their axes are 60 degrees apart. What is the intensity of light transimitted by the second film

Homework Equations


##I_2 = I_1*cos^2 phi##
Where I_1 is the intensity of light after going through the first film and I_2 is the intensity of light after going through the second film.

I also read online that the intensity of light going through one polarized lens is one half the original intensity, I don't know if I should use that since we never learned it in class and its not in the lab manual

The Attempt at a Solution


I don't know whether to just use the first equation and plug the 3W/m^2 in for I_1, or to use both equations(use half of 3 W/m^2 as I_1), which I don't think their expect since I was only given the first equation.
 
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  • #2
The problem clearly states that you start with unpolarized light. Why could you then assume that it all goes through the first polarizer?
Mddrill said:
I also read online that the intensity of light going through one polarized lens is one half the original intensity, I don't know if I should use that since we never learned it in class and its not in the lab manual
Can't you figure that factor out with the equation you are given?
 
  • #3
The equation I was given was I_2=I_1cos^phi. How would I figure ou the second equation from that?

I just want to make sure I'm not missing something because I feel like I should be able to solve the problem with just the equation I'm given.
 
  • #4
Mddrill said:
The equation I was given was I_2=I_1cos^phi. How would I figure ou the second equation from that?
For unpolarized light, you can consider that all polarizations are equiprobable. So you can calculate, on average, how much of that light will pass through the first polarizer using that equation.
 

FAQ: Intensity of light through two polarizers

What is the purpose of using two polarizers to measure the intensity of light?

The use of two polarizers allows for the measurement of the intensity of light that has passed through a polarizing filter. By placing the second polarizer at a specific angle relative to the first, the amount of light that passes through can be controlled and measured.

How does the angle between the two polarizers affect the intensity of light passing through?

The angle between the two polarizers determines the amount of light that can pass through. When the two polarizers are aligned, maximum light intensity can pass through, while when they are perpendicular, no light can pass through.

Can the intensity of light passing through two polarizers be greater than the original intensity of the light?

No, the intensity of light passing through two polarizers can never be greater than the original intensity of the light. The polarizers can only reduce the intensity of light passing through by absorbing or reflecting some of the light.

What factors can affect the intensity of light passing through two polarizers?

The intensity of light passing through two polarizers can be affected by the angle between the polarizers, the material and quality of the polarizers, and the intensity of the original light source.

How is the intensity of light through two polarizers measured?

The intensity of light passing through two polarizers can be measured using a light intensity meter or by using the equation I = I0 cos2θ, where I0 is the original light intensity and θ is the angle between the polarizers.

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