Question

A beam of unpolarized light shines on a stack of five ideal polarizers, set up so that the angles between the polarization axes of pairs of adjacent polarizers are all equal. The intensity of the transmitted beam is reduced from the intensity of the initial beam by a factor of phi=0.277 . Find the angle theta between the axes of each pair of adjacent polarizers.

Answer 1

Answer:

$21.75$

Explanation:

$n$ = number of polarizers through which light pass through = 5

$\theta$ = Angle between each pair of adjacent polarizers

$I_{o}$ = Intensity of unpolarized light

$I_{n}$ = Intensity of transmitted beam after passing all polarizers

It is given that

$\frac{I_{n}}{I_{o}}= 0.277$

we know that the intensity of light after passing through "n" polarizers is given as

$I_{n} = (0.5) I_{o} Cos^{2n-2}\theta$

$\frac{I_{n}}{I_{o}} = (0.5) Cos^{2n-2}\theta$

inserting the values

$0.277 = (0.5)Cos^{2(5)-2}\theta$

$0.554 = Cos^{8}\theta$

$Cos\theta = 0.9288$

$\theta = Cos^{-1}(0.9288)$

$\theta = 21.75$