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1 Phys 2310Fri. Nov. 18, 2011 Today’s Topics Begin Chapter 10: Diffraction Reading for Next Time

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Phys 2310Fri. Nov. 18, 2011 Today’s Topics. Begin Chapter 10: Diffraction Reading for Next Time. TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: A A A A A A A. Reading this Week. By Fri.: Begin Ch. 10 (10.1 – 10.3) General Considerations, - PowerPoint PPT Presentation

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Page 1: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

1

Phys 2310Fri. Nov. 18, 2011 Today’s Topics

• Begin Chapter 10: Diffraction

• Reading for Next Time

Page 2: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

2

Reading this Week

By Fri.:Begin Ch. 10 (10.1 – 10.3) General Considerations,

Fraunhofer Diffraction, Fresnel Difraction

Page 3: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

3

Homework this Week

Chapter 9 Homework

Chapter 9: #7, 8, 10, 14, 26 (due Monday Nov. 28)

Chapter 10: #8, 25, 30, 33 (due Friday,Dec. 1)

Page 4: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

4

Chapter 10: Diffraction

• General Considerations– There is no real physical distinction between diffraction and interference– Huygens-Fresnel Principle

• Modification to Huygens Principle:Every unobstructed point of a wave-front, at any given instant, serves as a source for

spherical, secondary wavelets (with the same frequency as that of the primary wave). The amplitude of the optical field at any point beyond is the superposition of all these wavelets (considering both their amplitudes and phases). This is really a quantum mechanical effect since there is otherwise no physical explanation.

– Opaque Obstructions• When an aperture is large compared to , the effects of the boundary is minimal

since the fraction of waves affected is small and vice versa.• Formally, diffraction occurs as a result of the boundary conditions for Maxwell’s

equations. The math is difficult but is solvable for a few special cases.• Aperture: imagine a set of fictitious, non-interacting oscillators distributed over

the opening. Electrons at the edge interact with the E-field and dampen the EM wave.

– Fraunhofer and Fresnel Diffraction• Fresnel diffraction: when the distance between aperture and screen/detector is

small. Shape of wavefront is important.• Fraunhofer diffraction: when the distance between aperture and screen/detector is

large (R > a2/). This allows the assumption of plane waves.

Page 5: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

5

Chapter 10: Diffraction

• General Considerations– Light from Several Coherent Oscillators

• E-field will add according to amplitude and phase:

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Page 6: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

6

Chapter 10: Diffraction

• Fraunhofer Diffraction– Small angle: r ~ R – ysin~ R - y– Example of a single slit

• Position of maxima depend on

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Page 7: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

7

Chapter 10: Diffraction

• Fraunhofer Diffraction from2 Slits– In this case the E-field is the sum of that from each slit:

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Page 8: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

8

Chapter 10: Diffraction

• Diffraction by Many Slits– Now generalize to N slits:

distinct. more andbrighter get maxima individual theincreases N as that figure thefrom Note

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Page 9: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

9

Chapter 10: Diffraction• Diffraction Grating

– For a slit separation of a the location of each order is:

– a sinm = m (notedependence)Thus white-light produces a spectrum at each order

(m).The angular dispersion can be computed via

differentiation.A diamond is used to cut groves into glass and

aluminized to act as little mirrors.It can be used in reflection without aluminizing

– Grating spectroscopy:• A collimating lens can be used to produce an input

beam with i = constant. The grating equation:a(sinm – sini) = m

• The dispersed light from the grating can be imaged using a lens acting as a camera. Thus the camera “sees” light entering at different field angles according to wavelength. Resulting image is a series of “slit images” displaced according to wavelength.

• See text for application examples.

Page 10: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

10

Chapter 10: Diffraction• Diffraction from a Square

Aperture– Similar to a single slit but we

now integrate in 2-d

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Page 11: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

11

Chapter 10: Diffraction

• Diffraction from a Circular Aperture– Similar but we integrate over the aperture in azimuth angle and r.

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Page 12: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

12

Chapter 10: Diffraction

• Implications of Diffraction in Optical Systems– Diffraction Limits the

Resolution of Optical Systems• The larger the aperture the

smaller the core and the “Airy Rings”

• The larger the wavelength the larger the core and the “Airy Rings”

• Regardless of magnification the resolution of given aperture is limited.

– Difficult for Hubble to see planets around nearby stars.

min = 1.22 /D

Page 13: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

13

Chapter 10: Diffraction

• Fresnel Diffraction– When either the screen or the source is at a finite distance plane waves are

insufficient and Huygens wavelets are located along a curved surface. The math is much more complicated since we must specify the directionality of the secondary wavelets from the source itself [K() = ½(1+cos)]. The secondary wavelets around an annulus are in phase relative to the source and hence the phase at P is t-k(+r).

490)-488 (pg.result thecompute toused becan addition phasor or series geometricA

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Page 14: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

14

Fresnel’s Half-Period Zones

• Fresnel’s Approach– Consider a series of zones s1,

s2, s3 … around point O, each /2 further from point P

– Consider phase difference D

even) is m (if 22

and odd) is m (if 22

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Page 15: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

15

Fresnel Diffraction from a Circular Aperture

• Imagine zones within a circular aperture– If radius corresponds to the outer

edge of first half-period zone: A = 2Ainf and intensity is 4x higher.

– If radius corresponds to the outer edge of second half-period zone:

A = A1 – A2 = 0!

– Intensity drops even though hole is bigger!

– Increasing hole size further results in periodic maxima and minima.

Page 16: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

16

Fresnel Zone Plates

• Alternately blocking either the even or odd zones using a mask (right).– Configured for a specific

source.

• Result is a lens that will image a distant source!– “Focal length” will be:

21

2 s

m

sf m

Page 17: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

17

Fresnel Diffraction from a Circular Aperture

• Now let’s add the amplitudes from small circular zones within an aperture

• Divide each half period zone into 8 subzones:

figure.

handright get the wesubzones malinfinitesiWith

aperture.circular afor curve"vibration " The figure.

handleft produces zones successive Adding

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added when and CD vector isresult theand

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Page 18: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

18

Fresnel Diffraction from a Single Slit

• Now consider a slit.– Half-period zones on

cylindrical wavefront:

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Page 19: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

19

Chapter 10: Fresnel Integrals

• Fresnel Integrals – Derivation of Fresnel

integrals:

evaluated.y numericall becan but

form closedin integrated becannot They

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dddx

Page 20: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

20

Chapter 10: Fresnel Diffraction from Edge• Fresnel Diffraction from Edge

– Start at point P where the amplitude is OZ

– Along the screen the verctor head remains fixed an the tail moves along the spiral to a maximum at b’.

– Continuing along the screen the amplitude goes through B’ and a minimum at c’.

– Secondary maxima (fringes) occur at d’.

– Ampiltude decreases and converges to OZ’.

– Scale of the pattern:

cm 0708.02

)(

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nm.. 500 and cm, 100 b a distancesLet

a

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a

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ba

abs

Page 21: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

21

Chapter 10: Fresnel Diffraction from Slit

• Fresnel Diffraction from Slit– Procedure is similar but each

side acts as an opaque edge.– For a = 100cm, b = 400cm, =

400 nm, and s = 0.02 cm, = 0.5

– Intensity at P’ found by drawing vector = 0.5 at different positions along the spiral and measuring the corresponding amplitude (A and A’)

Page 22: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

22

Chapter 10: Fresnel Diffraction

• Fresnel Diffraction from Rectangular (slit) Aperture– See Hecht sec. 10.3.8 for alternative explanation of how to

use the spiral to get the complex amplitude given the silt width (positions on the spiral).

Page 23: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

23

Chapter 10: Fresnel Diffraction

• Fresnel Diffraction from Semi-opaque Aperture

Page 24: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

24

Reading this Week

By Fri.:Begin Ch. 10 (10.1 – 10.3) General Considerations,

Fraunhofer Diffraction, Fresnel Difraction

Page 25: Phys 2310Fri. Nov. 18, 2011 Today’s Topics

25

Homework this Week

Chapter 9 Homework

Chapter 9: #7, 8, 10, 14, 26 (due Monday Nov. 28)

Chapter 10: #8, 25, 30, 33 (due Friday,Dec. 1)