Intensity of light through two polarizers

  • Thread starter Thread starter Mddrill
  • Start date Start date
  • Tags Tags
    Intensity Light
AI Thread Summary
Unpolarized light with an intensity of 3 W/m^2 passes through two polarizing films with their axes at 60 degrees apart. The intensity after the first polarizer can be calculated as half of the original intensity, resulting in 1.5 W/m^2. The intensity after the second polarizer is determined using the equation I_2 = I_1 * cos^2(phi), where phi is the angle between the polarizers. By substituting the values, the final intensity transmitted through the second film can be calculated. The discussion emphasizes the importance of understanding how unpolarized light behaves when passing through polarizers.
Mddrill
Messages
23
Reaction score
0

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.
 
Physics news on Phys.org
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?
 
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.
 
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top