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Three polarizing filters are stacked, with the polarizing axis ofthe second and third filters at angles of 22.2^\circ and 68.0^\circ, respectively, to that of thefirst. If unpolarized light is incident on the stack, the light hasan intensity of 75.5 W/cm^2 after it passes through thestack.

a) If the incident intensity is kept constant, what is theintensity of the light after it has passed through the stack if thesecond polarizer is removed?

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Answer:

I₂ = 25.4 W

Step-by-step explanation:

Polarization problems can be solved with the malus law

I = I₀ cos² θ

Let's apply this formula to find the intendant intensity (Gone)

Second and third polarizer, at an angle between them is

θ₂ = 68.0-22.2 = 45.8º

I = I₂ cos² θ₂

I₂ = I / cos₂ θ₂

I₂ = 75.5 / cos² 45.8

I₂ = 155.3 W

We repeat for First and second polarizer

I₂ = I₁ cos² θ₁

I₁ = I₂ / cos² θ₁

I₁ = 155.3 / cos² 22.2

I₁ = 181.2 W

Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized

I₁ = I₀ / 2

I₀ = 2 I₁

I₀ = 2 181.2

I₀ = 362.4 W

Now we remove the second polarizer the intensity that reaches the third polarizer is

I₁ = 181.2 W

The intensity at the exit is

I₂ = I₁ cos² θ₂

I₂ = 181.2 cos² 68.0

I₂ = 25.4 W

User Ayush Raj Singh
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