Answer:
The ratio is [tex]\frac{I_{min}}{I_{max}} = 0.029[/tex]
Explanation:
Let assume that the intensity of the unpolarized photon beam is [tex]I_o[/tex]
The through the linear polarizer is mathematically represented as
[tex]I_1 = I_ocos^2(\theta )[/tex]
Here [tex]\theta = 0[/tex] given that the polarizer is linear
So
[tex]I_1 = I_o[/tex]
The intensity of the [tex]I_o[/tex] emerging from the polarizer oriented 45° to the horizontal is
[tex]I_2 = I_o cos^2(45)[/tex]
[tex]I_2 = \frac{I_o}{2}[/tex]
The maximum photon probability density is mathematically represented as
[tex]I_{max} = ( \sqrt{I_1} + \sqrt{I_2})^2[/tex]
=> [tex]I_{max} = ( \sqrt{I_o} + \sqrt{\frac{I_o}{2} })^2[/tex]
The minimum photon probability density is mathematically represented as
[tex]I_{max} = ( \sqrt{I_1} - \sqrt{I_2})^2[/tex]
[tex]I_{max} = ( \sqrt{I_o} - \sqrt{\frac{I_o}{2} })^2[/tex]
The ratio of minimum to maximum is mathematically represented as
[tex]\frac{I_{min}}{I_{max}} = \frac{ I_o - \frac{I_o}{I_2} }{I_o + \frac{I_o}{I_2}}[/tex]
[tex]\frac{I_{min}}{I_{max}} = \frac{I_o (1 - \frac{1}{\sqrt{2} } ) }{I_o(1 + \frac{1}{\sqrt{2} })}[/tex]
[tex]\frac{I_{min}}{I_{max}} = \frac{(\sqrt{2} - 1 )^2}{(\sqrt{2} + 1 )^2}[/tex]
[tex]\frac{I_{min}}{I_{max}} = \frac{0.17157}{5.8284}[/tex]
[tex]\frac{I_{min}}{I_{max}} = 0.029[/tex]
