Monday, October 23, 2023

Solutions To Fraunhofer Diffraction Problems

 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 approximationAssigning angle values  we find for the same diagram:


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


In = (q 2-1 Io      

Similarly for amplitude: An = (q ) -1 Ao      

 Then if q = 3p/2:

A1 = (3p/2)-1 (Ao  ) =  (2/3p)  Ao    = 0.212 Ao

And for the intensity at first maximum.:

 I= [(3p/2)2]-1 (Io  ) =  [(9p2/4)]-1  Io   

I= [(4/9p2)]  Io    =   0.045 Io

2)  Show that for the three first odd (half integral) wave positions in double slit diffraction, that the amplitude         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    = 0  i.e.  for each odd integral  q ,    Or

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

There will correspond:

(l/ 2)  = 0  

(3l /2)  =  0

(5 l /2)  = 0 

Then:

av    = I (l /2) +  (3l /2) + (5l /2)  =  0


And:  av   = 0. 

No comments: