Q: Unpolarized light of intensity Io=950 W/m^2 is incident unpon three polarizer
ID: 1436523 • Letter: Q
Question
Q:
Unpolarized light of intensity Io=950 W/m^2 is incident unpon three polarizerz. The axis of the first polarizer is vertical. The axis of the second polarzer is rotated at an angle theta=55 degrees from the vertical. The axis of the third polarizer is horizontal.
Io=90 W/m^2
theta= 55 derees
a) what is the intensity of the light in W/m^2 after it passes through the first polarizer?
b) What is the intensity of the light in W/m^2 after it passes the second polarizer?
c) what is the intensity in W/m^2 of the light after it passes through the third polarizer?
D) Write an equation for the intensity of light after it passes through all three polarizers in terms of the intensity of unpolarized light entering the first polarizer Io and the angle of the second polarizer relative to the first, given that the first and third polarizer are crossed (90 degree between them). Use trigonometiric identities to simplify and give the results in terms of single trigonometric function of 2 theta
E) what is the smallest angle theta 2, in degrees, from the vertical that the second polarizer could be rotated such that the intensity of light passing through all three polarizers I3= 28W/m^2
Explanation / Answer
(a)
After the beam passes through the first polarizer, it is polarized and its intensity is cut in half, I1 = Io/2
I1 = 475 W/m^2
(b)
Using Maul's law -
Intensity of Light after passinf through Second polarizer,
I2 = I1*cos^2()
I2 = 475 * cos^2(55)
I2 = 156.2 W/m^2
(c)
Intensity of Light after passinf through Third polarizer,
I3 = I2*cos^2()
I3 = 156.2 * cos^2(90-55)
I3 = 156.2 * cos^2(35)
I3 = 104.8 W/m^2
(d)
I = Io/2 * cos^2() * cos^2(90 - )
I = Io/2 * cos^2() * sin^2()
We know, Cos() * sin() = 1/2*sin(2)
I = Io/8 * sin^2(2)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.