Diffraction

$Page 1: Diffraction$
Optics 2----by Dr.H.Huang, Department of Applied Physics 1

The Hong Kong Polytechnic University Diffraction

Introduction:

Diffraction is often distinguished from interference on that: in diffraction phenomena, the interfering beams originate from a continuous distribution of sources; in interference phenomena the interfering beams originate from a discrete number of sources.

If both the source of light and observation screen are effectively far enough from the diffraction aperture so that the wavefronts arriving at the aperture and observation screen may be considered plane, it is called Fraunhofer, or far-field, diffraction. When the curvature of the wavefront must be taken into account, it is called Fresnel, or near-field, diffraction.



Fraunhofer Diffraction at Single Apertures:

Each interval contributes spherical wavelets at P of the form,

tkrip e

r

dEdE

0

trkip e

r

dEdE

00

tkskriLtrkiLp e

r

dsEe

r

dsEdE

sin

0

00

tkrib

b

iksLp edse

r

EE

0

2

2

sin

0

2

2

0

2 sin

IEI p

sin2

1kb



Fringe Pattern:

Dark fringe:

The second, third and fourth maxima of the diffraction pattern occur at =1.43, 2.46 and 3.47, respectively.

mkb sin2

1

b

fmy

The central maximum represents essentially the image of the slit on a distant screen.

The angular width of the central maximumIs

The linear width of the central maximum is

b

2

b

LLW

2



Rectangular Slits:

2

2

2

2

0

sinsin

II

b

fmy

a

fnx

b

a

a



Circular Slits:

The far-field angular radius of Airy disc is,

2

10

)(2

J

II

sin2

1kD

22.1sin D

3.832

Airy Disc

D

22.1



Rayleigh’s Criterion:

For a microscope,

The ratio D/f is the numerical aperture.

D

22.1)( min

Dffx

22.1minmin



Example:Two stars have an angular separation of 44.7310-7 radian. Find the minimum diameter of the telescope objective which can just resolve the stars in light of 550 nm wavelength.

Example:Calculate the minimum angular subtense of two points which can be just resolved by an eye with a 6 mm diameter pupil in light of 555 nm wavelength.

cm151073.4422.1 7min d

d

radian10129.122.1 4min

d



Diffraction by Small Particles:

Babinet’s Principle

1 and 2 are complementary apertures.

Suppose that monochromatic plane wavefronts are incident normally on 1 and the diffracted light is imaged on a screen.

In a direction to the normal let the magnitude of the electric vector be E1. Replace 1 with 2 and let the magnitude of the

electric vector be E2. Apparently,

E1+E2=0 and E1=E2

Therefore,

It means that the diffraction patterns for 1 and 2 are identical.

21 II



Fraunhofer Diffraction at Two Slits:

ba

ba

iskba

ba

iskLR dsedse

r

EE

21

21

sin21

21

sin

0

2

2

0 cossin

4

II

sin2

1kb sin

2

1ka

sin

cos2

sincoscos 222 aka

a=2b



Fraunhofer Diffraction at Two Slits:

a=6b

2

2

0 cossin

4

II



Diffraction at Many Slits:

2

1

212

212

sin212

212

sin

0

N

j

baj

baj

iskbaj

baj

iskLR dsedse

r

EE

22

0 sin

sinsin

N

II

N=8 and a=3b

ma sinprincipal maxima at



Diffraction Grating:

ma mi sinsin



Free Spectral Range:—the non-overlapping wavelength range in a particular order.

The non-overlapping spectral region is smaller for higher orders.

The wavelength are better separated as their order increases. This property is described by angular dispersion, or dispersive power of a grating,

Linear dispersion is,

Resolving power of a grating is defined as,

Using the Rayleigh’s criterion, suppose the number of grating grooves is N, we have

ma sin

d

dD m

ma

mD

cos

fDd

df

d

dy m

min

R

mNR



Example:A grating has 4000 grooves or lines per centimeter. Calculate the dispersive power in the second order spectrum in the visible range.

We take the mean wavelength to be 550 nm.

Example:Find the number of lines (grooves) required on a grating to just resolve the two sodium lines, 1=589.592 nm and 2=588.995 nm, in the second order spectrum of a grating.

rad/m109.8cos

10.26sin2;m105.24000

1

5

2

26

a

mD

mama m

494

2

min

21min21

NmNR



Fresnel Diffraction:

ikrp e

r

dEdE

0

daEdE L0

rikSL e

r

EE

dae

rr

EdE rrikS

p

darr

eF

ikEE

rrikS

p 2

201

rr

02 rr

20

NrrN



...

...

4321

34

2321

aaaa

eaeaeaaA iii

11 aA

212 aaA

3213 aaaA

Two conclusions

(1) If N is small, there is large changes in the resultant phasor AN as the contribution

from each new zone is added. The resultant amplitude seems to oscillate between magnitudes that are larger and smaller than the limiting value of a1/2. As the

aperture gradually increases, one can see oscillations between bright and dark in a fixed position of the screen.

(2) If N is large, as in the case of unlimited aperture, the resultant amplitude is half that of the first contribution zone, a1/2.



Fresnel zone plate:very other Fresnel zone is blocked

The zone plate radii are approximately given by, 0NrRN



Diffraction by Straight Edges :Use cylindrical waves.



Example:Plane wave of =550 nm are incident normally on a circular aperture of radius mm. Does a bright or a dark spot appear at the point P on the axis 4 m from the hole? If the intensity of the incident light is I0, calculate the intensity at P.

Example:A 4 mm diameter circular hole in an opaque screen is illuminated by plane waves of wavelength 500 nm. If the angle of incidence is zero, find the positions of the first two intensity maxima and the first intensity minimum along the central axis.

The first two maxima will occur when N=1 and 3, respectively. The first minimum occurs when N=2.

11

0

2

1

1

0154321

0

2

0

442

spotbrightnumberodd5

IIE

E

I

IEEEEEEE

r

RNNrR N

N

m4:2

m67.2:3m8:1

2

22

N

RrN

N

RrN

N

RrN



Example:Plane waves of 550 nm wavelength are incident normally on a narrow slit of width 0.25 mm. Calculate the distance between the first minima on either side of the central maximum when the Fraunhofer diffraction pattern is imaged by a lens of focal length 60 cm.

Example:Plane waves (=550 nm) fall normally on a slit 0.25 mm wide. The separation of the fourth order minima of the Fraunhofer diffraction pattern in the focal plane of the lens is 1.25 mm. Calculate the focal length of the lens.

Example:Light from a distant point source enters a converging lens of focal length 22.5 cm. How large must the lens be if the Airy disc is to be 10-6 m in diameter? =450 nm

mm64.22

b

ffW

cm10.78

88sin

2

1

WbW

fb

kb

cm7.2410

44.210

22.122

66

fD

Dff



Example:A telescope objective is 12 cm in diameter and has a focal length of 150 cm. Light of mean wavelength 550 nm from a star is imaged by the objective. Calculate the size of the Airy disc.

Example:Assuming Rayleigh’s criterion can be applied to the eye, how far apart must two small lights be in order to be just resolved at a distance of 1000 m? Take the pupil diameter as 2.5 mm, the wavelength to be 555 nm, and the eye’s refractive index 1.333. Assume a single surface model eye with the pupil at the surface.

mm017.022.1

22 D

ff

minmin

min 2sin

2sin

22.122.1

nnnDD i

im

cm1.2722.1

minmin D

LLnLx i

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Diffraction

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