Solutions To Fraunhofer Diffraction Problems (1)

 1)  Sketch the intensity diagram for a single slit diffraction grating and assign relative amplitudes and intensities for the first maximum. 

Solution: 

The sketch would be:

 

And the relative amplitude at the center (m= 0) so l/ d= 0 is 1, and the intensity = (1)² = 1. The relative amplitude at the first maximum can be obtained fairly easily to a good approximation. Assigning angle values  we find for the same diagram:

Given the amplitude A_o = 1, then the approximate value of the next intensity can be found using:

I_n = (q ²) ^-1 I_o      

Similarly for amplitude: A_n = (q ) ^-1 A_o      

 Then if q = 3p/2:

A₁ = (3p/2)^-1 (A_o  ) =  (2/3p)  A_o    = 0.212 A_o

And for the intensity at first maximum.:

 I₁ = [(3p/2)²]^-1 (I_o  ) =  [(9p²/4)]^-1  I_o   

I₁ = [(4/9p²)]  I_o    =   0.045 I_o

2)  Show that for the three first odd (half integral) wave positions in double slit diffraction, that the amplitude         I _av   = 0.  

Solution:

The three first odd (half integral) wave positions - and amplitudes- can be obtained from the double slit diffraction diagram:

So that: for each of:

q  = p /2,  q  = 3p /2,  q  = 5p /2

We have: I _o    = 0  i.e.  for each odd integral  q ,    Or

d sin q  =    l/ 2 ,    3l/ 2   ,  5l/ 2 

There will correspond:

I _o (l/ 2)  = 0  

I _o (3l /2)  =  0

I _o (5 l /2)  = 0 

Then:

I _av    = I _o (l /2) +  I _o (3l /2) + I _o (5l /2)  =  0

And:  I _av   = 0.