Solution to Fresnel Diffraction Problems (Part 2)

 1) Consider a circular aperture with center at 0, and n zones comprised of essentially equal areas a _i .  

a) If the n zones are even estimate the total amplitude. 

b) If the n zones are odd, estimate the total amplitude.   

Solns.

a) Let n be even  then we have for the sum of amplitudes:

a  =  a ₁ /2  -  ( a ₁ /2  +   a ₂  -  a ₃ /2)  -  ( a ₃ /2  +   a ₄  -   a ₅ /2)  - .....-    a _n /2

= a ₁ +   a  ₂ /2  +  ( a ₂ /2  + a ₃ -  a ₄ /2) +  (a ₄/2  +  a ₅ -  a ₆ /2)

 + ......+  a _n-1 /2 +  a _n

Because the amplitudes for any two adjacent zones are very nearly equal we can write:   a ₁   »    a ₂    and:  a_n-1  »    a _n

Then we have:

a  =  a ₁ /2  -   a_n /2

b)  Let n be odd then we have for the sum of amplitudes:

a  =  a ₁ /2  +  ( a ₁ /2  -  a ₂  +  a ₃ /2)  +  ( a ₃ /2  -  a ₄  +  a ₅ /2)  + .....+    a _n /2

= a ₁ -  a  ₂ /2  -  ( a ₂ /2  - a ₃ +  a ₄ /2) -  (a ₄/2  -  a ₅ +  a ₆ /2)

 - ......- a _n-1 /2 +  a _n

Because the amplitudes for any two adjacent zones are very nearly equal we can write:   a ₁   »    a ₂    and:  a_n-1  »    a _n

Then we have:

a  =  a ₁ /2  +   a_n /2

2) Show how the obliquity factor: 

cos (n, r) - cos (n, r')  =  cos (n, r) + 1

And explain why.

Soln.

We assume again in conformance with the presentation that the diagram below applies:

Since from this diagram:    r' is constant and cos (n, r') = -1

Then: cos (n, r) - cos (n, r') =  cos (n, r) - (-1) =  

cos (n, r) + 1

3) Let the first 10 even zones of a zone plate be covered. Find: a) The whole wave front amplitude,  b) the whole wave front disturbance magnitude, c) the relative amplitude at point P, and d) the intensity at point P.

Solns.:

a)  If the first ten even zones are exposed, then this leaves the amplitudes:  A ₂, A ₄ ,  A ₆,   ... A ₂₀

The sum of which »   10 A ₂  »   10 A ₁ 

The whole wave front amplitude is then:  ½ A₁ 

b) The whole wave front disturbance magnitude is:  ½ U₁

c)  The relative amplitude at point P is twenty times as great as when the plate is removed.

d)  The intensity at point P, from the inverse square law for light, is then   400 times as great as when the plate is removed.

4) The average distance of P from a zone is 15 mm.  If the obliquity angle q  is p/15  and the applicable path difference  D =  0. 15 mm, find the amplitude of the 13th zone.

Soln.:

The amplitude of the mth zone can be expressed: 

A _m  = k (s _m  / d _m  ) (1   +   cos q )

Where k = const., d _m is the average distance to P,   and q  is the angle at which light leaves the zone. 

We have then: d _m  =   15mm

m = 13,    s _{m     ~}   Ö m    » Ö 13  ,    q  =  p/15

Then:

A ₁₃  »   k (Ö 13/ 15 mm ) (1   +   cos p/15 ) 

But the applicable path difference  D =  0. 15 mm, which means  this must be factored into the computations. This means we recast d _m  = b + D

And now: (s _m  /  d _m )  = (  b + D )/   ( b + D )  =  1

(s ₁₃  /  d ₁₃ )  =  1

Then:  A ₁₃  = k (s ₁₃  / d ₁₃  ) (1   +   cos  p/15 )

Or:

A ₁₃  = k  (1   +   cos  p/15 )