Question

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?

Answer 1

I₂ = 143.79

Step-by-step explanation:

To solve this problem, work them in two parts. A first one where we look for the intensity of the incident light in the set and a second one where we silence the light transmuted by the other set,

Let's start with the set of three curling irons

Beautiful light falls on the first polarized is not polarized, therefore only half the radiation passes

I₁ = I₀ / 2

this light reaches the second polarized and must comply with the Mule law

I₂ = I₁ cos² tea

The angle between the first polarized and the second is Tea = 29.0º

I₂ = I / 2 cos² 29

The light that comes out of the third polarized is

I₃ = I₂ cos² tea

the angle between the third - second polarizer is

tea = 58-29

tea = 29th

I3 = (I₀ / 2 cos² 29) cos² 29

indicate the output intensity

I3 = 110

we clear

I₀ = 2I3 / cos4 29

I₀ = 2 110 / cos4 29

I₀ = 375.96 W / cm²

Now we have the incident intensity in the new set of three polarizers

back to the for the first polarizer

I₁ = I₀ / 2

when passing the second polarizer

I₂ = I1 cos² 29

I2 = IO /2 cos²29

let's calculate

I₂ = 375.96 / 2 cos² 29

I₂ = 143.79