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:

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.

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

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

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

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)

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.

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)

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.

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

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).

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?

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:

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)

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:

Double Slit Diffraction 2 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

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, ...