Assume a point source P illuminating a large aperture centered at O, a distance a away.
What is the intensity at P', a distance a' way from O?
Consider a point Q in the aperture. The distance between O and Q is s.
Light travels from Q to P' because of diffraction. PQ differs from PO by
δ, and P'Q differs from P'O by δ’.
We have a2 + s2 = (a + δ)2,
a'2 + s2 = (a' + δ')2.
If s << a and a', then δ ≈ s 2/(2a), δ' ≈ s2/(2a').
The total path difference between POP' and PQP' is Δ = s2/(2a') + s2/(2a).
Let us define radii s1, s2, s3,
..., such that, Δm = mλ/2.
The annuli defined by two successive radii are called Fresnel half-period zones. The field arising at P' from one annulus is, overall, π out of phase with the field from the neighboring annulus.
The radius of sm is found from mλ/2 = sm2/(2a') + sm2/(2a),
sm = [maa'λ/(a+a')]½,
and therefore the area of any annulus is
Am = π(sm+12 - sm2) = πaa'λ/(a + a'),
independent of m or s.
For given geometry, the areas of all the Fresnel zones are equal. Each
zone contributes approximately the same amplitude to the field at P' if the
aperture is uniformly illuminated.
Suppose the aperture to be a circle with N zones. Because the alternate zones
contribute fields that are out of phase by π, the total amplitude at P' is
AP' = A1 - A2 + A3 - A4 + ..., ± AN,
where the sign of AN depends on whether N is odd or even. If N is odd, AP' is
approximately equal to A1 and if if N is
even, AP' is approximately 0.
Fresnel zone plates are constructed by blocking either the even-numbered or the odd-numbered zones.
Assume N is even and the even zones are blocked.
Then AP' = A1 + A3 + A5 + ....
+ AN-1 ≈ NA1/2.
For the intensity at P' we have where Ip' ≈ N2I1/4.
Rewriting mλ/2 = sm2/(2a') + sm2/(2a)
as
1/a + 1/a' = mλ/sm2 = 1/(s12/λ)
we find the lens equation for the zone plate with a focal length f = s12/λ
where s12 = aa'λ/(a + a').
P' is real the image of P. A zone plate is thus an imaging device whose focal
length depends on wavelength and on the geometry of the zone plate.
The zone plate also produces produces fainter images at f' = f/3, f/5, ... . For f' = f/3, for example, a' is smaller, and each Fresnel zone is divided into 3 rings. Contributions from two of the rings cancel but one of the rings contributes to a fainter image. However, if the zone plate is constructed so that the opacity varies in a gradual, in a sinusoidal manner, the fainter images do not appear.
Now let us choose P' at a distance b = -a'' < a to the left of O. (Note: a'' is negative.)
Then the path difference between P'Q and PQ is P'Q - PQ = b + s2/(2b) -
(a + s2/(2a)).
The path difference between P'O and PO is b - a.
So the additional path difference between P'Q and PQ compared to the center path
difference is
Δ = s2/(2b) - s2/(2a).
We again define radii s1, s2, s3,
..., such that, Δm = mλ/2.
The radius of sm is found from mλ/2 = sm2/(2b)
- sm2/(2a),
sm = [mabλ/(a - b)]½,
and therefore the area of any annulus is
Am = πabλ/(a - b),
independent of m or s.
If the source location were at P’, then the field arriving at one annulus
would be π out of phase with the field arriving from the neighboring annulus.
If we construct a zone plate with N zones and block all even-numbered zones,
then a source at P’ observed from the right would have an intensity Ip'
≈ N2I1/4.
Rewriting mλ/2 = sm2/(2b) - sm2/(2a)
as
1/b - 1/a = mλ/sm2 = 1/(s12/λ),
or
1/a + 1/a'' = 1/(-s12/λ),
we again find the lens equation for the zone plate with a negative focal length f' =
-s12/λ, where s12 = abλ/(a - b).
P' is virtual the image of P. For the same zone plate f'= -f'.