Question

Imagine you have two identical perfect linear polarizers and a source of natural light. Place them one behind the other and position their transmission axes at 0 and 50 degrees respectively. Now insert between them a third linear polarizer with its transmission axis at 25 degrees. If 1000 W/cm2 of natural light is incident, how much will emerge with and without the middle polarizer in place?

Answer 1

Answer:

Without the third polarizer inserted in the middle of the first and second polarizer the irradiance of light that would emerge from the second polarizer is

             $I_2=321.39 W/cm^2$

With the insertion of the third polarizer  at the middle of the first and second polarizer the irradiance of light that would emerge from the second polarizer is

              $I_4 = I(\theta_d_1) = 337.3 W/cm^2$

Explanation:

From the question we are told that the

         The angle between the transmission axis of first linear polarizer and the vertical axis  is $\theta_1 = 0^o$

                   The angle between the transmission axis of second linear polarizer and the vertical axis  is $\theta_2 = 50^o$  

                   The angle between the transmission axis of third linear polarizer and the vertical axis  is $\theta_2 = 25^o$  

         The  irradiance of the incident natural light is $I = 1000 W/cm^2$

Generally a  linear polarizer divides the irradiance of a natural light by 2

  So For the first polarizer the irradiance of the natural light would become

              $I_1 = \frac{I}{2}$

Substituting values

             $I_1 = \frac{1000}{2}$

                 $=500 \ W/cm^2$

Now looking at the question we can deduce that the angle between the transmission axis of the and second polarizer is

               $\theta_d = \theta _2 - \theta_1$

Substituting values

               $\theta_d = 50^o - 0^o$

                    $=50^o$

According to Malus Law the irradiance of light that would come out from  the second polarizer is obtained by this mathematical expression

      $I(\theta_d) = I_1 cos^2(\theta_d)$

Substituting values

       $I_2 = I(\theta_d) = 500 *cos (50)$

           $=321.39 W/cm^2$

 When the third polarizer is inserted between the first and second polarizer, we have that

  The angle between the first  polarizer and the third polarizer is mathematically evaluated as

               $\theta_d_1 = \theta_3 - \theta_1$

Substituting the values

             $\theta_d_1 = 25^o -0^o$

                  $= 25^o$

According to Malus Law the irradiance of light that would come out from  the third  polarizer is obtained by this mathematical expression

              $I(\theta_d_1) = I_1 cos^2(\theta_d_1)$

              $I(\theta_d_1) = I_1 cos^2 (25)$

Substituting values

             $= 500 cos^2(25)$

              $I_3 = I(\theta_d_1) = 410.69 W/cm^2$

The angle between the third polarizer and the  second polarizer is  mathematically evaluated as

               $\theta_d_2 = \theta_2 - \theta_3$

Substituting the values

             $\theta_d_2 = 50^o -25^o$                    [Note the third polarizer is placed at the

                  $= 25^o$                                The middle of the first and second

                                                             polarizer

According to Malus Law the irradiance of light that would come out from  the second  polarizer is obtained by this mathematical expression

              $I(\theta_d_2) = I_3 cos^2(\theta_d_2)$

              $I(\theta_d_1) = I_3 cos^2 (25)$

Substituting values

             $= 410.69 cos^2(25)$

              $I_4 = I(\theta_d_1) = 337.3 W/cm^2$