Question

Consider a sequence of ideal polarizing filters, each with its axis making the same angle with the axis of the previous filter. These polarizers rotate the plane of polarization of a light beam by a total angle of 45◦ .

a) In order to have no more than a 10.0% reduction of intensity, what is the minimum number of polarizers needed? Hint: The equation you will set up to solve this will be transcendental, so it is impossible to solve with pen and paper. You should graph both sides of the function separately and see where they intersect.

b) What is the angle between adjacent polarizers?

Question 2: Consider a sequence of ideal polarizing filters, each with its axis making the same angle with the axis of the previous filter. These polarizers rotate the plane of polarization of a light beam by a total angle of 45°. a) In order to have no more than a 10.0% reduction of intensity, what is the minimum number of polarizers needed? Hint: The equation you will set up to solve this will be transcendental, so it is impossible to solve with pen and paper. You should graph both sides of the function separately and see where they intersect. b) What is the angle between adjacent polarizers?

Answer 1

Answer -2

   (a) Let the first sheet have its axis at angle $\small \theta$ to the original plane of polarization, and let each further sheet have its axis turned by the same angle.

The firs sheet passes intensity,

  $\small I_{1}=I_{max}cos^{2}\theta$

The second sheet passes intensity,   

12 = Imar costo

and the nth sheet lets through,

$\small I_{max}cos^{2n}\theta \geq 0.90I_{max}$

Where,

$\small \theta =\frac{45^{\circ}}{n}$

(a) Trying different integers,

For n = 5,

  $\small cos^{2\times 5}\frac{45^{\circ}}{5} \geq 0.885$

   For n=6,

     $\small cos^{2\times 6}\frac{45^{\circ}}{6} \geq 0.902$

So, minimum number of polarizers needed,

   n=6

(b) The angle between adjacent polarizers is,

$\small \theta =\frac{45^{\circ}}{n}=\frac{45^{\circ}}{6}=7.50^{\circ}$