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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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Answer:

a) p_i=1.568\hat{i}+0.752 \hat{j}

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Explanation:

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m_2=0.47\ kg

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Initial momentum:

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p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}

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p_i=\sqrt{1.568^2+0.752 ^2}

p_i=1.739\ kg.m.s^{-1}

From the conservation of momentum:

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c)

Vertical component of final velocity:

v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

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