Question

Unpolarized light of intensity I0=950 W/m2 is incident upon three polarizers. The axis of the first polarizer is vertical. The axis of the second polarzer is rotated at an angle of 55 degrees from the vertical. The axis of the third polarizer is horizontal.

Write an equation for the intensity of light after it has passed through all three polarizers in terms of the intensity of unpolarized light entering the first polarizer I0 and the angle of the second polarizer relative to the first, given that the first and third polarizers are crossed (90° between them). Use trigonometric identities to simplify and give the results in terms of a single trigonometric function of φ = 2θ.

Answer 1

Answer:

The equation for the intensity of light after it has passed through all three polarizers is:

                                                   $I_{3}=\frac{1}{8}I_{0}sin^{2}(\phi)$                                              

Explanation:

When unpolarized light goes through the first polarizer, whose axis of transmission is vertical, the intensity is cut by half:

                                                        $I_{1}=\frac{1}{2} I_{0}$

When the light passes through the second polarizer, whose axis of transmission is rotated an angle $\theta=55^{o}$ from the axis of transmission of the first polarizer, using Malus's law we have that:

                                            $I_{2}=I_{1}cos^{2}(\theta)=\frac{1}{2}I_{0}cos^{2}(\theta)$

When the light passes through the third polarizer, whose axis of transmission is horizontal, using once again Malus's law we have that:

                                $I_{3}=I_{2}cos^{2}(\frac{\pi}{2}-\theta)=\frac{1}{2}I_{0}cos^{2}(\theta)cos^{2}(\frac{\pi}{2}-\theta)$

We use the next trigonometric identities:

•           $cos^{2}(\frac{\pi}{2}-\theta)=sin^{2}(\theta)$

•          $sin^{2}(2\theta)=4sin^{2}(\theta)cos^{2}(\theta)$

Manipulating the last equation we get that:

                                                $I_{3}=\frac{1}{8}I_{0}sin^{2}(2\theta)$

Replacing  $\phi=2\theta$ we have that:

                                                $I_{3}=\frac{1}{8}I_{0}sin^{2}(\phi)$

     

Answer 2

The equation for the intensity of light is mathematically given as

$I_{3}=\frac{1}{8}I_{0}sin^{2}(\phi)$

What is the equation for the intensity of light?

Question Parameter(s):

I0=950 W/m2

at an angle of 55 degrees

Generally, the equation for the unpolarized light through a first polarizer is mathematically given as

   I1=0.5 I0

For the light through the second polarizer,

$I_{2}=I_{1}cos^{2}(\theta)\\ I_{2}=\frac{1}{2}I_{0}cos^{2}(\theta)$

For the light through the Third polarizer,

$I_{3}=I_{2}cos^{2}(\frac{\pi}{2}-\theta)\\ I_{3}=\frac{1}{2}I_{0}cos^{2}(\theta)cos^{2}(\frac{\pi}{2}-\theta)$

In conclusion, using  trigonometric identities we have

$I_{3}=\frac{1}{8}I_{0}sin^{2}(2\theta)$

Hence

$I_{3}=\frac{1}{8}I_{0}sin^{2}(\phi)$

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