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KengaRu [80]
3 years ago
5

Unpolarized light of intensity 30 W/cm2 is incident on a linear polarizer set at the polarizing angle θ1 = 22 ∘. The emerging li

ght then passes through a second polarizer that is set at the polarizing angle θ2 = 162 ∘. Note that both polarizing angles are measured from the vertical. What is the intensity of the light that emerges from the second polarizer?
Physics
1 answer:
Snezhnost [94]3 years ago
6 0

Answer:

I₂ = 8.80 W/cm²

Explanation:

To solve this exercise we need to apply the Malus's law.

First, when the light gors through the first polarizer which is linear, intensity is always cut in half, therefore:

I₁ = I / 2   (1).

Applying this, we have:

I₁ = 30 / 2 = 15 W/cm²

Now, we can calculate the intensity of the light, after it passes through the second polarizer. The Law of Malus states that light intensity will be:

I₂ = I cos²Ф   (2)

Where Ф, is the difference between the angle of the first polarizer and the second. In this way we have:

Ф = α - β

Ф = 162 - 22 = 140°

Now, applying Malus's law we have:

I₂ = 15 cos²140°

<h2>I₂ = 8.80 W/cm²</h2>

Hope this helps

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<em>yay !</em>

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