Question

This is one of a pair of problems that treat the same situation. This one uses a quantum mechanical analysis, while the other uses a classical wave analysis. We encourage you to compare the words used in the two. An unpolarized photon beam is incident on the usual 2-slit apparatus, with a screen behind. One slit has a linear polarizer aligned in the H direction, while the other has a linear polarizer rotated by 45°. 1) What is the ratio of minimum to maximum photon probability densities in the interference pattern on the screen? min/max =

Answer 1

Answer:

The ratio is  $\frac{I_{min}}{I_{max}} = 0.029$

Explanation:

Let assume that the intensity of the unpolarized  photon beam is  $I_o$

   The through the linear polarizer is mathematically represented as

       $I_1 = I_ocos^2(\theta )$

Here $\theta = 0$ given that the polarizer is  linear

   So

        $I_1 = I_o$

The intensity of the $I_o$  emerging from the polarizer oriented 45° to the horizontal is  

         $I_2 = I_o cos^2(45)$

       $I_2 = \frac{I_o}{2}$

The maximum photon probability density is  mathematically represented as

          $I_{max} = ( \sqrt{I_1} + \sqrt{I_2})^2$

=>      $I_{max} = ( \sqrt{I_o} + \sqrt{\frac{I_o}{2} })^2$

The minimum photon probability density is  mathematically represented as

        $I_{max} = ( \sqrt{I_1} - \sqrt{I_2})^2$

        $I_{max} = ( \sqrt{I_o} - \sqrt{\frac{I_o}{2} })^2$

The ratio of minimum to maximum is mathematically represented as

          $\frac{I_{min}}{I_{max}} = \frac{ I_o - \frac{I_o}{I_2} }{I_o + \frac{I_o}{I_2}}$

         $\frac{I_{min}}{I_{max}} = \frac{I_o (1 - \frac{1}{\sqrt{2} } ) }{I_o(1 + \frac{1}{\sqrt{2} })}$

        $\frac{I_{min}}{I_{max}} = \frac{(\sqrt{2} - 1 )^2}{(\sqrt{2} + 1 )^2}$

       $\frac{I_{min}}{I_{max}} = \frac{0.17157}{5.8284}$

       $\frac{I_{min}}{I_{max}} = 0.029$