Respuesta :
Answer:
a)   I= Iâ (cos²θ - cosâ´Î¸)   b) 75.5Âş
Explanation:
a) For this exercise we must use Malus's law
     I = Iâ 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
     Iâ = Iâ cos² θ
* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90Âş 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
       θâ = 90- θâ = θ
fiate that in the exercise we must take a reference system and measure everything with respect to this system.
     I = Iâ cos² θ'
   Â
we substitute
     I = (Iâ 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= Iâ (cos²θ - cosâ´Î¸)
b) to find when the intensity is maximum,
we can use that we have an extreme point when the drift is zero
     [tex]\frac{dI}{d \theta}[/tex] = 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 Â
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 90Âş 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 75.5Âş
