Question

Two polarizers A and B are aligned so that their transmission axes are vertical and horizontal, respectively. A third polarizer is placed between these two with its axis aligned at angle θ with respect to the vertical.Part A)Assuming vertically polarized light of intensity I0 is incident upon polarizer A, find an expression for the light intensity I transmitted through this three-polarizer sequence.Express your answer in terms of the variables I0 and θ.I = ?Part B)Calculate the derivative dI/dθ.Express your answer in terms of the variables I0 and θ.

Answer 1

Answer:

Explanation:

Let the angle between the first polariser and the second polariser axis is θ.

By using of law of Malus

(a)

Let the intensity of light coming out from the first polariser is I'

$I' = I_{0}Cos^{2}\theta$     .... (1)

Now the angle between the transmission axis of the second and the third polariser is 90 - θ. Let the intensity of light coming out from the third polariser is I''.

By the law of Malus

$I'' = I'Cos^{2}\left ( 90-\theta \right )$

So,

$I'' = I_{0}Cos^{2}\theta Cos^{2}\left ( 90-\theta \right )$

$I'' = I_{0}Cos^{2}\theta Sin^{2}\theta$

$I'' = \frac{I_{0}}{4}Sin^{2}2\theta$

(b)

Now differentiate with respect to θ.

$I'' = \frac{I_{0}}{4}\times 2 \times 2 \times Sin2\theta \times Cos 2\theta$

$I'' = \frac{I_{0}}{2}\times Sin 4\theta$