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

The ratio is
`(I_(min))/(I_(max)) = 0.029`

Step-by-step 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 = (I_o)/(2)`

The maximum photon probability density is mathematically represented as

`I_(max) = ( √(I_1) + √(I_2))^2`

=>
`I_(max) = ( √(I_o) + \sqrt{(I_o)/(2) })^2`

The minimum photon probability density is mathematically represented as

`I_(max) = ( √(I_1) - √(I_2))^2`

`I_(max) = ( √(I_o) - \sqrt{(I_o)/(2) })^2`

The ratio of minimum to maximum is mathematically represented as

`(I_(min))/(I_(max)) = ( I_o - (I_o)/(I_2) )/(I_o + (I_o)/(I_2))`

`(I_(min))/(I_(max)) = (I_o (1 - (1)/(√(2) ) ) )/(I_o(1 + (1)/(√(2) )))`

`(I_(min))/(I_(max)) = ((√(2) - 1 )^2)/((√(2) + 1 )^2)`

`(I_(min))/(I_(max)) = (0.17157)/(5.8284)`

`(I_(min))/(I_(max)) = 0.029`