In: Physics
Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 29.0? and 55.0?, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 75.0W/cm2 after it passes through the stack.
1. If the incident intensity is kept constant, what is the intensity of the light after it has passed through the stack if the second polarizer is removed?
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Three ideal polarizing filters are stacked, with the polarizing
axis of the second and third filters at 29.0and 58.0, respectively,
to that of the first. If unpolarized light is incident on the
stack, the light has intensity 110 after it passes through the
stack.
If the incident intensity is kept constant, what is the intensity
of the light after it has passed through the stack if the second
polarizer is removed?
Here's the formula you'll be using for this problem: malus's
law
I = Icos^2(theta)
When a ray goes through an ideal polarizer it gets cut in half. So
you know that when the unpolarized light goes thorough the first
polarizer it will be :
I = I/2 , let's call this I(1) for now.
Now you have the light travel through a second polarizer. The
intensity of this light is now:
I = I(1)cos^2(29). We'll call this I(2)
Finally the light will travel past the third polarizer, and the
intensity will be given by this:
I = I(2)cos^2(58-29). We'll Call this I(3)
You are given I(3) so what you have to do is solve for I2 and I1
backwards starting from I3.
110 = I2cos^2(58-29) = (I2 = 110/cos^2(58-29)) = 143.80
143.80 = I1cos^2(29) = (I1 = 143.80/cos^2(29)) = 187.98
So 187.98 is the intensity of the light after it passes through the
first polarizer. Since the second one is removed , leaving the
third one to do the rest of the light blocking, all you have to do
is take the intensity of light after the first polarizer, and use
Malus's law to calculate the new intensity with 3rd polarizer's
angle.
Inew = 187.98cos^2(58) = 52.3