Question

         Two polarizers are oriented at 58.0° to one another. Light polarized at a 29.0° angle to each

          polarizer passes through both. What reduction in intensity takes place?

          If instead the orientation of the polarizers is 40°to one another and that only 15% of the light gets through both of them, what was the initial polarization direction of the incident light?

          

Answer 1

Initialy The light polarizez at 29^{\circ} from first poloroid. Let I_o be the unpolorized intensity of light.

Intensity of light after passing first poloroid

$I_1=I_ocos^{2}29^{\circ}\\I_1=0.765I_o$

Now the intensity of light after passing through second polarizers

$I_2=I_1cos^2(58^{\circ}-29^{\circ})\\I_2=I_1cos^229^{\circ}\\I_2=0.765I_o\times 0.765=0.585I_o$

Reduction in intensity of light= $\dfrac{I_o-0.585I_o}{I_o}=\dfrac{0.415I_o}{I_o}\times 100=41.5$ %

Hence The intensity of light is reduced by 41.5%

If the orientation of the polarizers is 40°to one another and that only 15% of the light gets through both of them,

let the initial angle made by incident light be $\theta$

Then, After passing through second polarizer

$\Rightarrow\dfrac{15I_o}{100}=I_1cos^2(40-\theta)~~~~~~~-(1)$

Intensity of light passing from first polarizer

$I_1=I_ocos^2\theta~~~~~~~~-(2)$

From eqs.(1) and (2)-

$\dfrac{15}{100}=cos^2\theta cos^2(40-\theta)$

$\dfrac{1}{2}\times 2cos\theta cos(40^{\circ}-\theta)=\sqrt{0.15}$

$cos(\theta+40-\theta)+cos(\theta-40+\theta)=2\times 0.3872$

$\Rightarrow cos40+cos(2\theta-40)=0.7644\\\Rightarrow cos(2\theta-40)=0.7644-0.7664\\\Rightarrow 2\theta-40=cos^{-1}0\\\Rightarrow 2\theta=40+90\\\Rightarrow\theta=\dfrac{130}{2}$

$\Rightarrow \theta=65^{\circ}$

Hence the initial incidence angle is $65^{\circ}.$