Question

Polarizers 1 and 3 have their axes of polarization indicated by the black solid lines, perpendicular to each other. Polarizer 2 has an axis of polarization that has an angle of θ = 50.9° with respect to the polarization axis of polarizer 1. What percentage of the initial (unpolarized) light intensity passes through?

Answer 1

Answer:

percentage of unpolarized light intensity,

$\frac{I"'}{I} = 8.52%$

Given:

Polarization angle for the 2nd polarizer, $\theta = 50.9^{\circ}$

Solution:

If I be the intensity of unpolarized light, then the intensity of light passing through the first polarizer, I':

$I' = \frac{I}{2}$

Now, the intensity of light through polarizer 2, I":

$I" = I'cos^{2}\theta = \frac{I}{2}cos^{2}\theta$

Also Polarizer 1 and 3 are perpendicular to each other:

Therefore, $\theta' = 90^{\circ} - 50.9^{\circ} = 49.1^{\circ}$

Intensity of light passing through polarizer 3:

$I"' = I" cos^{2}\theta' = \frac{I}{2}cos^{2}\theta cos^{2}\theta'$

$I"' = \frac{I}{2}cos^{2}50.9^{\circ}\times cos^{2}49.1^{\circ}$

I"' = 0.0852 I

Now, percentage of unpolarized light intensity:

$\frac{I"'}{I}\times 100 = 0.0852\times 100 = 8.52%$