南开大学光学工程内部课件Nov 16th

If the primary wave was simply to propagate from S to P, it is
E
0 r0
e
i [t k ( r0 )]
The two equations must be exactly the same.
So we introduce a /2 phase difference between the primary wave and the secondary wave to make the two equations so.
2 It is approximately equal to one half of the contribution from the first zone, i.e.
E
E1
E
K 1 A e
i [t k ( r0 )]
r0
i
Fresnel Diffraction
Fresnel Diffraction
If m is odd, the series can be reformulated in two ways, either as:
E E1 2 ( E1 2 2 E2 E3 2 )( Em 2 E3 2 E4 E5 2 )
... (
To map the rest of the pattern, we now move the sensor along any line perpendicular to the axis. For off-axial point P, some zones are partially obscured. At certain position, the sum of all partial zones “saw” by P is zero. As P move radically outward, we detects a series of relative maxima and minima.
Fresnel Diffraction
If m is even, since Km0, and each adjacent contribution is nearly equal,
E ( E1 E 2 ) ( E 3 E 4 ) ... ( E m 1 E m ) 0
Fresnel Diffraction
Off the axis, a whole series of concentric bright and dark rings will surround the central spot.
பைடு நூலகம்
i [t k ( r0 )]
r0
i
Fresnel Diffraction
The sum of the optical disturbance from all m zones at P is
E E1 E 2 E 3 ... E m E1 E 2 E 3 ... E m
r0 If the aperture has a radius R, to a good approximation, the number of zones within it is simply
R 2
A ( r0 ) R
2
A

r0
r0
Fresnel Diffraction
Fresnel Diffraction
We can gain a better appreciation of the actual size of the things we are dealing with by computing the number of zones in a given aperture. The area of each zone is:
Fresnel Diffraction

Circular obstacles
If the opaque obstacle, be it a disk or sphere, obscures the first m zones, then
E Em 1 Em 2 ... Em
E
E2 2

Em 2
Fresnel Diffraction
We conclude from above that
E E1 2 Em 2
This same result is obtained in the case when
El ( El 1 El 1 ) / 2
Fresnel Diffraction
with experimental observations.
K ( )
i 2
(1 cos )
Fresnel Diffraction

Circular apertures —— Spherical waves
Consider that a monochromatic spherical wave incident on a screen which has a small hole.
Fresnel Diffraction
Fresnel’s procedure applied to a point source can be used as a semiquantitative method with which to study diffraction at a circular aperture. First we consider the irradiance arriving a very small sensor placed at point P on the symmetry axis. Let us assume that the sensor at P “sees” an integral number of zones m, filling the aperture.
El ( El 1 El 1 ) / 2
equation (a) becomes
E E1 2 Em 2
Fresnel Diffraction
From equation (b) we have
E E1 E2 2 E m 1 2 Em
Since K() goes from 1 to 0 over a great many zones, we can neglect any variation between adjacent zones, i.e. │E1│= │E2│, │Em-1│= │Em│. So
dS d 2 ( sin )
Fresnel Diffraction
r ( r0 ) 2 ( r0 ) cos
2 2 2
So
dS 2
We have
E l ( 1)
r0
rdr. Constant!!!
l 1
2 Kl A e
│Em│now approaches zero because Km0. Repeating the same procedure as in circular hole yields
E E m 1 2
Fresnel Diffraction
There is a bright spot everywhere along the central axis except immediately behind the circular obstacle (as P moves to the disk, increases, Km+10 and the irradiance gradually falls off to zero). The spot is called Poison’s Spot. If the disk is large, the (m+1)th zone is very narrow and any irregularities in the obstacle’s surface may seriously obscure that zone. For Poison’s spot to be readily observable, the obstacle must be smooth and circular.
dE K
A
r
e
i [t k ( r )]
dS
A is the source strength per unit area. The obliquity factor must vary slowly and may presumed constant over a single Fresnel zone
When the last term │Em│ corresponds to an even m, the same procedure leads to
E
E1 2

Em 2
Fresnel Diffraction
For a unobstructed wavefront, │Em│= 0
because K() for mth zone is 0. So
If m is not an integer, i.e. a fraction of a zone appears in the aperture, the irradiance at P is somewhere between zero and its maximum value.
Fresnel Diffraction
So I0.
Fresnel Diffraction
If m is odd