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Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that its transmission axis makes an angle with the transmission axis of the first sheet. (a) Derive an expression for the intensity of the transmitted light as a function of (b) Show that the intensity transmitted through all three sheets is maximum when\

User Shawn Aukstak
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1 Answer

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29 votes

Answer:

a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º

Step-by-step explanation:

a) For this exercise we must use Malus's law

I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

I = I₁ cos² θ'

we substitute

I = (I₀ cos² tea) cos² (θ - 90)

cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

I = Io cos² θ sin² θ

1= cos²θ+ sin²θ

sin²θ = 1 - cos²θ

I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero


image = 0

\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

cos θ - 2 cos³ θ = 0

cos θ ( 1 - 2 cos² θ) = 0

The zeros of this function are in

θ = 90º

1-2cos²θ =0 cos θ = 0.25 θ = 75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

User Sellarafaeli
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