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myrzilka [38]
3 years ago
12

Three polarizing filters are stacked with the polarizing axes of the second and third at 45.0∘ and 90.0∘, respectively, with

that of the first. (a) If unpolarized light of intensity I0 is incident on the stack, find the intensity and state of polarization of light emerging from each filter. (b) If the second filter is removed, what is the intensity of the light emerging from each remaining filter?
Physics
1 answer:
USPshnik [31]3 years ago
3 0

Answer:

a) the intensity of the polarization of light emerging from each filter is <em>I₀/2, </em>0.25 <em>I₀ </em>and 0.125 <em>I₀ </em>

b) the intensity of light emerging from the each remaining filter is <em>I₀ / 2 </em>and <em>0.</em>

Explanation:

a) The intensity of the incident light is <em>I₀ </em>and therefore the intensity of light transmitted through the first filter is

<em>I₁ = I₀ / 2</em>

The polarizing direction will be parallel to the axis of the first filter intensity of light transmitted through the second filter:

<em>I₂</em> = <em>I₁</em> cos²(θ₁)

<em />

<em>I₂</em> =  (<em>I₀ </em>/ 2) cos²(45°)

<em>I₂</em> = 0.25 <em>I₀</em>

<em />

<em />

Therefore, the polarizing direction will be at 45° to the axis of the first filter. The light intensity transmitted through the filter will be

<em>I₃</em> = <em>I₂</em> cos²(θ₂ - θ₁)

<em>I₃ </em>= <em>I₂</em> cos²(90 - 45)

<em />

<em>I₃ </em>= <em>I₂</em> cos²(45)

<em>I₃ </em>= 0.25 <em>I₀ </em>cos²(45)

<em>I₃ </em>= 0.125 <em>I₀ </em>

<em />

Therefore, the intensity of the polarization of light emerging from each filter is <em>I₀/2, </em>0.25 <em>I₀ </em>and 0.125 <em>I₀ </em>

<em />

b)  the intensity of light transmitted through the first filter is

<em>I</em>₁ = <em>I₀ </em>/ 2

when the second filter is removed.

The intensity of light that is transmitted through the third filter is :

<em>I₃' </em>= <em>I</em>₁ cos²(90)

<em>I₃'  </em>= (<em>I₀ </em>/ 2)(0)

<em>I₃'  </em>= 0

Therefore, the intensity of light emerging from the each remaining filter is     <em>I₀ / 2 </em>and <em>0.</em>

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