Tuesday, June 7, 2022

Solutions to Fresnel Diffraction Problems (Part 3)

 1) Without plotting a Cornu spiral, find the hypothetical value of D  for such a spiral plotted for the diffraction pattern of a single slit of  width 0.80mm, assuming Fresnel zone parameters a = 40.0 cm and b = 50.0 cm with red light  of 6400 Å.  From this result obtain the relative intensity I.


Soln.

Dv =   S Ö{ 2 (a + b)/  abl  }  = 

0.08 cm Ö{ 2 (40 cm + 50 cm )/ (40cm) (50 cm) 6.4   x  10 -5  cm

Dv = 3.0      

And  from table of Fresnel integrals:

 (Dx)  =    0.6058 – 0.0000 = 0.6058

(Dy)  =    0.4963 – 0.0000 = 0.4963

A2   =   (Dx) 2  +   (D y) 2

I    » A2   =   (0.6058) 2    +   (0.4963) 2

 I    »   0.613


2) A student is given the top section of a Cornu spiral to analyze for an exam:


Using this, obtain the relative amplitude for the particular diffraction pattern.  From this find the relative intensity.

Soln.

From the diagram, on inspection of v along the spiral,  we see:

(D v)   =   (1.4 - 0.9)  =  0.5

Obtain the  (Dx,  Dy)  limits from the table of Fresnel integrals for these values, e.g.

(Dx)  =   0.5431 -   (0.7648) =  (-0.222)

(Dy)  =  0.7135  -   0.3398   =   0.374

A2   =   (Dx) 2  +   (D y) 2

The relative amplitude is:

A =  Ö (Dx) 2  +   (D y) 2

=    Ö (-0.2222  +   (0.3742

=   0.435  

And the relative intensity is:

I    » A2   = (0.435) 2    = 0.189 

3) Plot the graph for a Cornu spiral for a single slit diffraction pattern at intervals of D= -0.10 to 3.0 and Dx from 0 to 0.90. On the graph draw a chord from x = 0.40 to x = 0.75 and from this estimate Dalong the Cornu arc.  Hence or otherwise obtain the relative amplitude and the intensity.

Soln.  

The  portion of the Cornu spiral of interest is shown below:


This is extracted from the full spiral graphed below:


From the given information and the graph segment (upper right quadrant of Cornu spiral) we obtain: 




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