1. Fraunhofer diffraction from Circular apertures:
Lecture 20
Fraunhofer diffraction from Circular apertures:
For circular aperture of area A, we redefine dE0 = EA dA
Thus, amplitude of field at P (arrangement as in slide 4 in Lecture 19) is then given by x s R
Elemental area of rectangular strip of area dA = x ds
Length x at height s is
(R = aperture radius)
The integral above becomes:

2. Circular apertures:
Using substitution: v = s/R and  = kR sin  ; we can rewrite as
This standard definite integral takes value of:
J1() = first-order Bessel function of the first kind, given by
From the series expansion, the ratio J1()/ has limit ½ as   0; and this Bessel function oscillates with its amplitude decreasing as  gets larger.

3. Circular apertures (Bessel function):
This is the order of Bessel function that we are interested in

4. Zeroes of Bessel function Jn(x) occur at x =
Circular apertures (Bessel function):
Zeroes of Bessel function Jn(x) occur at x = s
n=0
n=1
n=2
n=3
n=4
n=5 1
2.405
3.832
5.135
6.379
7.586
8.780 2
5.520
7.016
8.147
9.760
11.064
12.339 3
8.654
10.173
11.620
13.017
14.373
15.700 4
11.792
13.323
14.796
16.224
17.616
18.982 5
14.931
16.470
17.960
19.410
20.827
22.220 6
18.071
19.616
21.117
22.583
24.018
25.431 7
21.212
22.760
24.270
25.749
27.200
28.628 8
24.353
25.903
27.421
28.909
30.371
31.813 9
27.494
29.047
30.571
32.050
33.512
34.983
s = order of zero

5. I0 = irradiance at   0 or at  = 0.
Circular apertures (Bessel function):
Thus irradiance for circular aperture of diameter D can now be written as:
(20-1)
I0 = irradiance at   0 or at  = 0.
From the Table of zeroes of Bessel function, the 1st zero of J1() occurs at  = 3.832, thus, central maximum of irradiance falls to zero when
(20-2)
Comparing functions J1(x)/x and (sin x)/x :
both approaches a maximum when x  0, thus, their irradiances is greatest at center of pattern ( = 0)
their pattern is symmetrical about optical axis through center of circular aperture
at 1st minimum, m = 1 for slit pattern (in m = b sin ) analogous to m’ = 1.22 for circular aperture

6. The Airy disc is the diffracted “image” of the circular aperture.
Circular apertures (Bessel function): I0 5 5
corresponds to 
Airy disc
Central maximum is a circle of light that corresponds to the zeroth order of diffraction called the Airy disc.
The Airy disc is the diffracted “image” of the circular aperture.
The pattern has a rotational symmetry about the optical axis
From (20-2) given by D sin  = 1.22, the far-field angular radius of this Airy disc is approximated (sin   ) to:
(20-3)

7. Resolution :
A telescope with a round objective is subject to diffraction effects as with a circular aperture.
sharpness of the primary image of distant star is then limited by diffraction
this image occupies the region of Airy disc
This inevitable diffraction blurring in the image restricts the resolution of the instrument (in terms of being able to produce distinct images for distant object points)  S1 S2
Diffraction-limited images of two point objects formed by a lens. As long as the Airy disc are well separated, the images are well resolved.
When S1 and S2 are too close, their image patterns overlap, and it will be difficult to resolve as distinct object points.

8. The Rayleigh criterion for just resolvable images requires
the centers of the image patterns to be not less than the angular radius of the Airy disc (i.e. the distance between the first diffraction minimum of the image of one source point and the maximum of another)
Resolved
Unresolved
Rayleigh criterion
()min
Therefore, the limit of resolution is:
(20-4)
(D may be the diameter of the objective lens of telescope)

9. Quantitative example:
Suppose each lens on a pair of binoculars has diameter 35 mm. How far apart must two stars be before they are theoretically resolvable by either of the lenses?
1.92105 rad  1.1103  = 0.066’ = 4” of arc (or 4 arcseconds);
550 nm has been chosen as the average wavelength for visible light (ranges from 400 nm – 700 nm)
If the stars are near the center of our galaxy, a distance d of around 30,000 light years, then their actual separation s is approximately
s = d min = (30,000)(1.92105) = 0.58 light years
If we detect them by their long-wavelength radio waves (say 1 km), replacing the lenses with dish antennas, the resolution is less.

10. Single slit Circular aperture
Rayleigh criterion for single slit differs from that of circular aperture
Single slit
Circular aperture
In the case of a microscope, the minimum separation xmin of two just-resolved objects near the focal plane of its objective lens (diameter D, focal length f) is:
min
xmin A B
Radius of Airy disc
Minimum angular resolution of a microscope f
Ratio D/f  numerical aperture (typical value = 1.2) thus xmin  ;
That’s why UV, X-ray and electron (shorter ) microscopes have high-resolution

11. Resolution limits due to diffraction in the human eye
min I1 I2
pupil
retina
Eye lens
screen
aperture
Diffraction by eye with pupil as aperture limits the resolution of objects subtending angle min
Night vision (pupil is larger at ~ 8mm) has higher resolution than daylight vision. Unfortunately, there isn’t enough light to take advantage of this.
At bright noon, pupil diameter ~2 mm; (taking average wavelength of 550 nm) theoretical resolvable angular distance is
Thus for 2 lines at 1 mm apart, they must not be farther than (103/3.36104 = 2.98 m) away in order to be barely-discernable (theoretically).

12. Resolution limits due to diffraction in the telescope
A telescope has an objective of 50 cm diameter. Its angular limit of resolution at wavelength 550 nm is:
If it is used to see two objects on the surface of the Moon (Earth-Moon distance = 3.844108 m), the separation of the two objects such that they are theoretically resolvable is:
If the objective lens has focal length 2 m, the corresponding distance between the images of the objects on the focal plane of the objective is:
The pupil of your eye is 4 mm in diameter when you view the images on the objective. What is the farthest position of your eye from the focal plane of the objective such that you are just able to resolve them?

13. Double Slit Diffraction
In the case of double slit, we modify the limits of integration of the equation for the field at point P on the screen for single slit (Eq. 19-4) and expressing for the amplitude only:
(20-5) b a
Specification of slit width and separation for double slit diffraction
Integration and substitution of the limits yields:

14. Using Euler’s equation, we get:
Double Slit Diffraction
Then substituting a term involving slit width b and a term involving slit separation a expressed as:
(20-6)
will result in:
Using Euler’s equation, we get:
(20-7)
Thus, the irradiance is:
(20-8)

15. Rewriting the double-slit interference term:
Double Slit Diffraction
Therefore:
(21-5)
where
Double-slit interference
Single-slit diffraction
Maximum irradiance in pattern centre for Double slit is 4 X that of single slit
Rewriting the double-slit interference term:
Compare to expression for Young’s double-slit:

16. Double Slit Diffraction2 4 6 8
10
12   
Interference (solid line) and diffraction (dashed line) functions for double-slit Fraunhofer diffraction [slit separation 6 X slit width (a = 6b)]
4I0 6
12 
Resultant irradiance for the double slit above
Diffraction pattern due to single-slit
Diffraction pattern due to double-slit

17. Diffraction minima occur for  = m where m = 1, 2, …,
Double Slit Diffraction
Diffraction minima occur for  = m where m = 1, 2, …,
or when at condition: m = b sin 
Interference maxima occur for  = p where p = 0, 1, 2, …,
or when at condition: p = a sin 
When diffraction minima coincide (at the same ) with interference fringe maxima, the fringe goes missing. Condition for missing orders:
(20-9)
(20-10) or
(20-11)
This is achieved when slit separation is a perfect integral of the slit width, i.e., a = nb
In the example in previous slide a = 6b, thus, the missing orders are when p = 6, 12, ...