Answer:
Explanation:
Intensity of unpolarised light = I₀
intensity after passing through first polariser = I₀ / 2
Angle between first and second polariser is φ so
intensity after passing through second polariser
= (I₀ / 2) cos²φ
Now angle between second and third polariser
= 90 - φ
intensity after passing though third polariser
= (I₀ / 2) cos²φ cos²( 90 - φ)
= (I₀ / 2) cos²φ sin²φ
= (I₀ / 8) 4cos²φ sin²φ
= (I₀ / 8) sin²2φ
The intensity after the third polarizer will be:
I₃ = (I₀/8)*sin^2(2φ)
For non-polarized light that passes through any polarizer, we say that the intensity is reduced to its half.
Original intensity = I₀
After the first polarizer, the intensity will be:
I₁ = I₀/2.
Now, when it passes through a polarizer such that the difference in angles with the polarization is x, the new intensity will be:
I₂ = I₁*cos^2(x).
The angle between the second and the first polarizer is φ, then we have:
I₂ = (I₀/2)*cos^2(φ).
Now we also know that the first and the last polarizer are crossed (so there is an angle of 90°). Then if we define θ as the angle between the second and the third polarizer, we will have that:
φ + θ = 90°
then:
θ = 90° - φ
The intensity after the third polarizer will be:
I₃ = (I₀/2)*cos^2(φ)*cos^2(90° - φ)
And we know that:
cos(90° - φ) = sin(φ)
Then we can rewrite:
I₃ = (I₀/2)*cos^2(φ)*sin^2(φ)
But we want a single trigonometric function, then we use the relation:
cos^2(φ)*sin^2(φ) = sin^2(2φ)/4
And replacing that, we get:
I₃ = (I₀/2)*sin^2(2φ)/4 = (I₀/8)*sin^2(2φ)
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