Question

A polarizing filter is aligned vertically, a second filter is aligned at an arbitrary angle θ from the vertical, and third filter is polarized at a 45 angle from the first. Unpolarized light of intensity lo -1.0 W/m2 is incident on the first. All filters are ideal 1. Find an equation for the intensity of the outgoing light as a function of 0 2. Solve for the intensity at angles between O and 180 in 5 steps. Make a table of the results. 3. Solve for the angle at which the intensity is a maximum. Answer in degrees to three significant digits.

Answer 1

1.The transmitted light intensity I varies with $\theta$ according to $I=I_{0}cos^{2}\theta cos^{2}(\theta -45^{0})$ .

2.a) For $\theta =36^{0},$ $I=I_{0}cos^{2}36^{0}cos^{2}9^{0}=0.63I_{0}$.

For $\theta =72^{0},$ $I=I_{0}cos^{2}72^{0}cos^{2}27^{0}=0.07I_{0}$.

For $\theta =108^{0}$ , $I=I_{0}cos^{2}108^{0}cos^{2}63^{0}=0.02I_{0}$ .

For $\theta =144^{0}$ , $I=I_{0}cos^{2}144^{0}cos^{2}99^{0}=0.026 I_{0}$ .

For $\theta =180^{0}$ , $I=I_{0}cos^{2}180^{0}cos^{2}135^{0}=0.5I_{0}$ .

3.For $I(\theta )$ to be maximum ,$dI(\theta )/d\theta =0$.

Now $I(\theta)=I_{0}cos^{2}\theta cos^{2}(\theta -45^{0})$ .

We obtain on differentiation :
$I_{0}cos^{2}\theta *-sin2(\theta -45^{0})+I_{0}cos^{2}(\theta -45^{0})*-sin2\theta =0$ .

On simplification this reduces to

$tan2\theta =cos^{2}\theta /cos^{2}(\theta -45^{0})$.

Now $cos(\theta -45^{0})=(cos\theta +sin\theta )/\sqrt{2}$ .

On further simplification the above reduces to

$tan2\theta =2cos^{2}\theta /(1+sin2\theta )$. i.e $sin2\theta /cos2\theta =2cos^{2}\theta /(1+sin2\theta )$ .

i.e $sin2\theta +sin^{2}2\theta =2cos^{2}\theta cos2\theta$ i.e $2sin\theta cos\theta +4sin^{2}\theta cos^{2}\theta =2cos^{2}\theta (2cos^{2}-1)=4cos^{4}\theta -2cos^{2}\theta$ . i.e $2sin\theta cos\theta +4(1-cos^{2}\theta ) cos^{2}\theta =2cos^{2}\theta (2cos^{2}-1)=4cos^{4}\theta -2cos^{2}\theta$

i.e$2sin\theta cos\theta =8cos^{4}\theta -6cos^{2}\theta$ i.e $sin\theta =4cos^{3}\theta -3cos\theta$ .

Since $sin\theta =\sqrt{1-cos^{2}\theta }$ the above equation reduces to:

$16cos^{6}\theta -96cos^{4}\theta +145cos^{2}\theta =1$.

Let $x=cos^{2}\theta$ Then we have the cubic equation:

$16x^{3}-96x^{2}+145x-1=0$.

This equation must be solved to obtain x and $\theta$ i.e the angle for which the intensity of the transmitted light  is maximum.