Final reviewInterference and diffraction

Phys 322Lecture 37

Reminder: Please complete the online course evaluation

Final examFinal Exam, Saturday December 14, 1:00-3:00 pm, in PHYS 223.Material covered: Hecht Chapters 2-10 (except 5.8, 5.9, 6.3-6.5, 8.9-8.13, 10.3-10.5).Please bring two formula sheets (letter-size paper), scientific calculator, ID, etc.

Notes:1. The final exam is comprehensive, but the emphasis will be on the last part of

the material (Ch. 9-10)2. Two crib sheets (8 ½" x 11") are permitted.3. There will be 4-5 problems, one with qualitative questions and the rest

quantitative problems.

In a double slit interference experiment, the distance D from the slits to a screen is 1 m, the slit separation is 1 mm, and the wavelength of the incident plane wave is 500 nm.

a) Find the difference in the positions of two consecutive maxima near the center of the interference pattern.

Da

y

Find the separation between the 6th and 8th maxima located on each side of the central fringe (the 6th is on the positive side and the 8th fringe is on the negative side).

Review problem 1

a

s

Interferometers9.3 Wavefront-splitting interferometers9.3.1 Young’s experiment (1801)Young’s originality: Obtaining spatially coherent light from a pinhole, and splitting the wavefront of the spatially coherent light with two pinholes.

Optical path length difference:r2 – r1 = a sin θ = ay/s ( If s>> a, s>> y )

Position of bright fringes:

m = 0, ±1, ±2, … is the order number.asmy

am

m

m

ma msin

Space between fringes:asy

Intensity distribution of the fringes:

sayIrrkIII 2

0212

02

0 cos42

)(cos42

cos4

ay/s

In a double slit interference experiment, the distance D from the slits to a screen is 1 m, the slit separation is 1 mm, and the wavelength of the incident plane wave is 500 nm.

a) Find the difference in the positions of two consecutive maxima near the center of the interference pattern.

93

small angle approximation

1.0 (500 10 ) 0.510

Dya

my m mmm

D = s = 1 ma

y

Find the separation between the 6th and 8th maxima located on each side of the central fringe (the 6th is on the positive side and the 8th fringe is on the negative side).

~ 6 ( 8) 14 ~ 7D Dy mma a

A soap film surrounded by air has an index of refraction of 1.34. If a region of the film appears bright red (0 = 633 nm) in normally reflected light, what is the minimum thickness of the film there?

Review problem 2

9.4 Amplitude-splitting interferometers9.4.1 Dielectric films – double-beam interferenceApplication: Optical coating reduces or enhances reflection of optical devices through thin-film interference effects.

Fringes of equal inclinationConsider the first two reflections (others are weak).Optical path length difference:

tft

tf

t

f

itt

fi

t

f

f

dndndn

dndn

ACndn

ADnBCABn

cos2cos

sin2cos2

sintan2cos2

sincos2

)(

2

11

1

dA

B

C

D

P

S

i

t

nf

n1

n2

Phase difference:Assume nf >n1, nf >n2. There is a phase shift of between external and internal reflections (when the incident angle is not large).

tf d

nk cos

4

Bright spot: 4

)12(cos ft md

Dark spot:

42cos f

t md

A soap film surrounded by air has an index of refraction of 1.34. If a region of the film appears bright red (0 = 633 nm) in normally reflected light, what is the minimum thickness of the film there?

0

cos (2 1) eqn. 9.364

ft

ff

d m

n

0 633118

4 4(1.34)f

nmd nm

n

A double-slit diffraction pattern is formed using a HeNe laser light at 632 nm. Each slit has a width of 0.080 mm. The pattern reveals that the third-order interference maxima are missing from the pattern.(a) What is the slit separation?(b) What is the irradiance of the first two interference fringes, relative to the zeroth-order

maximum?

Review problem 3

Coherent line source:y

x

P (x,y)

dy

r

-D/2

D/2dy

rikryxE

D

DL )exp(),(2/

2/

L is the source strength per unit length.This equation changes a diffraction problem into an integration (interference) problem.

10.2 Fraunhofer diffraction10.2.1 The single slit

y

x

P (x,y)

y

r

-D/2

D/2

R

2sin)0()(

II

sin)2/( ),exp(sinsin)2/(

]sin)2/sin[()exp(

)]sin(exp[

)exp(),(

2/

2/

2/

2/

kDikRRD

kDkDikR

RD

dyR

yRik

dyrikryxE

L

L

D

DL

D

DL

In amplitude, r is approximated by R

In phase, r is approximated by R-y sinFraunhofer condition

The phase is the same as a point source at the center of the slit

2sin)0()(

II

0.047 0.016

maxima ,...47.3 ,46.2 ,43.1 ,0 ,tanminima ...3 ,2 , ,0sin

0)sincos(sin2)0(

sin)0()(

3

2

IddI

II

10.2.2 The double slit

z

x

P (x,z)R-a sin

R

ab )]sin(exp[)exp(sin

)()(),(

)]sin(exp[)(2/

2/

2/

2/

aRikikRbC

dzzFCdzzFCzxE

zRikzFba

ba

b

b

2

2

0 cossin4)(

II

sin)2/(sin)2/(

kbka

The result is a rapidly varying double-slit interference pattern (cos2) modulated by a slowly varying single-slit diffraction pattern (sin2/2).

(©WIU OptoLab)

22

/sin)/sinsin(sincos)0()(

bbaII

single-slit diffractiontwo-slit interferenceEnvelopeFringes

Question:Which interference maximum coincides with the first diffraction minimum?

bam

bma

sinsin

“Half fringe” (split fringe) may occur there.

2

2

0 cossin4)(

II

sin)2/(sin)2/(

kbka

A double-slit diffraction pattern is formed using a HeNe laser light at 632 nm. Each slit has a width of 0.080 mm. The pattern reveals that the third-order interference maxima are missing from the pattern.(a) What is the slit separation?(b) What is the irradiance of the first two interference fringes, relative to the zeroth-order

maximum?

Review problem 3

For f = (sin )/

What is the relative irradiance of the subsidiary maxima in a three slit Fraunhofer diffraction pattern?

Review problem 4

10.2.3 Diffraction by many slits

)2exp(1)2exp(1)exp(sin

])1(2exp[...)4exp()2exp(1)exp(sin

)(...)(

)()(),(

)]sin(exp[)(

2/)1(

2/)1(

2/2

2/2

2/

2/

2/

2/

iNiikRbC

NiiiikRbC

dzzFCdzzFC

dzzFCdzzFCzxE

zRikzF

baN

baN

ba

ba

ba

ba

b

b

z

x

P (x,z)

R-a sin

R

ab

R-2a sin

22

20

sinsinsin)(

N

NII

Principle maxima:Minima (N-1):

Subsidiary maxima (N-2):

... ,2 , ,0

)1(... ,3 ,2 , N

NNNN

2

)32( ..., ,25 ,

23

NN

NN

64sin

sinsin 22

Nba

N

What is the relative irradiance of the subsidiary maxima in a three slit Fraunhofer diffraction pattern?

32 2N

2 2

2

(0) sin sin( )sin

I NIN

sinfor small , ~ 1

2

( ) 1 1~(0) 9

II N

You want to distinguish between =500.2 and 500.3 nm light with a grating at normal incidence to the light and with a maximum angle for third-order diffraction of 5. What is the minimum width of the grating?

Review problem 5

GratingsDiffractive grating: An optical device with regularly spaced array of diffracting elements.Transmission gratings and reflection gratings.

Grating equation: i

m

a

i

m

a

ma im )sin(sin

Blazed grating: Enhancing the energy of a certain order of diffraction.Blaze angle: Specular reflection: mir 2 i

r

0

a

specular

0th

Grating spectroscopy:N-slit interference:

cos2

2

)sin)(sin2/(sin

sin)(2

0

Na

N

ka

NII

i

Angular width due toinstrumental broadening

Angular dispersion:m

mim a

mddma

cos)sin(sin

dm

d

Limit of resolution:

)sin(sin

coscos

coscos2

minminmin

min

criterion sRayleigh'

im

mm

m

m NamN

am

am

dd

NaNa

min

Angular width

Angular dispersion

min

min

Resolving power

Free spectral range:

mmm

ma fsri

)sin(sin

sinm =1

m =3

m =2

fsr

Question: Why does the resolving power increase with increasing order number and with increasing number of illuminated slits?

You want to distinguish between =500.2 and 500.3 nm light with a grating at normal incidence to the light and with a maximum angle for third-order diffraction of 5. What is the minimum width of the grating?

Review problem 5

1. T or F: Zone plates are used to change the polarization of an electromagnetic wave.2. T or F: The blaze on a reflection grating can improve the amount of energy in a desired

order of diffraction.3. T or F: The resolving power of a grating spectrometer used in a particular diffraction

order depends only on the number of lines illuminated (not wavelength or grating period).

4. T or F: The central peak of the Fraunhofer diffraction from two narrow slits separated by spacing h has the same width as the central diffraction peak from a single slit with width x = h.

5. T or F: The Fraunhofer diffraction pattern appearing at the focus of a lens varies in angular width, depending on the focal length of the lens used.

6. T or F: Fraunhofer diffraction can be viewed as a spatial Fourier transform (or inverse transform if you prefer) on the field at the aperture.

7. T or F: It is always possible to completely eliminate reflections with a single-layer antireflection coating as long as the right thickness is chosen for a given real index.

8. T or F: A half-wave plate rotates light initially linearly polarized at an angle with respect to the fast axis through a total angle 2.

9. T or F: A Michelson interferometer can be used to measure the spectral intensity of light I ().

10. T or F: A Michelson interferometer can be used to measure the wavelength of light.

Sample qualitative questions

T or F: Vertically polarized light illuminates a Young’s double-slit setup and fringes are seen on a distant screen with good visibility. A half wave plate is placed in front of one of the slits so that the polarization for that slit becomes horizontally polarized. Here’s the statement: The fringes at the screen will shift position but maintain their good visibility.

More qualitative questions

Effects of finite coherent length:

• Light from each slit has a coherent length lc.For sunlight lc is about 3.

• The waves from two slits can only interfere ifr2 – r1 <lc.

• The contrast of the fringes degrades when the amount of the overlap between uncorrelated wavegroups increases.

PA B

B'A'

lc

Other Wavefront-splitting interferometers

Fresnel’s double mirror: Slits S1 and S2 act as virtual coherent sources. They are images of slit S in two mirrors.

= r2 – r1, y = s/a

PM1

M2

S1

S2

S

a

r1

r2

s

y

Shield

Fresnel double prism: Interference between light refracted from the upper and the lower prisms. Two virtual coherent source S1 and S2 are produced by the prisms.

Question: Where are S1 and S2?

S1

S2

Sa

Lloyd’s mirror: Interference between light from source S and its image S ' in the mirror. Glancing incidence causes a phase shift of . The fringes are complementary to Young’s.

sayII2

0 sin4

S

S'

as

y

Extended source: All rays inclined at the same angle arrive at the same point.Fringes of equal inclination: Arcs centered on the perpendicular from the eye to the film.

Haidinger fringes: The fringes of equal inclination viewed at nearly normal incidence. Concentric circular bands.

t

f dn

cos2

sin 2

Fringes of equal thicknessFizeau fringes: Contours from a non-uniform film when viewed at nearly normal incidence.

x

n1

n2

nf

Thickness: d = xInterference maximum: 2nf dm = (m + ½)

Fringe space: x = f /24

)12( ,4

)12( fm

fm mxmd

Newton’s rings: Interference pattern from air film between glass surfaces.

xd

Thickness:

Bright rings: 2nf dm = (m + ½)

Dark rings:

RxddRdx2

)2(2

2

Rmxmd fmf

m

21 ,

4)12(

Rmx fm

22

2sin x

Rn

Beam splitter

Movable mirror

(©WIU OptoLab)

S S1 S2

M1 M2

P

cos2d

cos4 d

Optical path length difference:

Phase difference:

Dark fringes: md m cos2

Multiple beam interference9.4.2 Mirrored interferometersMichelson interferometer1) Compensation plate: Negates dispersion from the beam splitter2) Collimated source: Fringes of equal thickness3) Point source: Interference of spherical waves4) Extended source: Fringes of equal inclination

9.6 Multiple beam interferenceInterference between multiple reflection and multiple refractions:

td

S

i

nf

0E

rE0 ''0 ttrE ''30 ttrE ''50 ttrE

'0ttE 20 ''rttE 4

0 ''rttE

2

212

rrFCoefficient of finesse:

Airy function:2sin

121

2sin

12

1

2sin

121

1

cos21)1(

11

1)1(

'1'

...)'''1('1' ,'

cos4

cos2

22

2

22

2

22

2

24

222

2

2

2

20

20

362420

2

rr

rr

II

II

rr

rrr

err

II

errE

erttE

erererttEErttrr

dn

k

dn

i

t

i

r

ii

t

ii

iiit

tf

tf

)2/(sin1)2/(sin

)2/(sin11

2

2

2

FF

II

FII

i

r

i

t

)2/(sin11 )( 2

F

A

)2/(sin11

12

2

2

2

FII

rrF

i

t

it II /

/

F = 0.2 (r2 = 0.046)F = 1 (r2 = 0.17)F = 200 (r2 = 0.87)

9.6.1 The Fabry-Perot interferometer1) High resolving power2) Prototype of laser cavity

S P

dFF

FA

21sin2

21

)2/(sin11 )(

12/1

2

Half-width of transmission:

Finesse:F42 2/1

22 Ff

Chapter 10 DiffractionFraunhofer diffraction10.1 Preliminary considerationsDiffraction: The deviation of light from rectilinear propagation.There is no significant physical distinction between interference and diffraction.Huygens-Fresnel Principle: Every unobstructed point of a wave front serves as a source of spherical wavelets. The amplitude of the optical field at any point beyond is the superposition of all these wavelets, considering their amplitudes and phases.

Fresnel (near field) diffraction: Fraunhofer (far field) diffraction: Both the incoming and outgoing waves approach being planar. R > a2/. R is the smaller of the two distances from the source to the aperture (with size a) and from the aperture the to observation point.Mathematical criteria: Path difference is a linear function of the aperture variables.

S

P

aR1

R2S

a

P

a sin

10.2.4 The rectangular apertureCoherent aperture:

Aperture

)exp( dSrikrE A

2

2

2222

222

222

/)(1

/)(21

/)(2/)(1

)()(

RZzYyR

RZzYyR

RZzYyRzyR

zZyYXr

ZYXR

dS=dydz P(Y,Z)rR

x

y

z

Y

Za

b

Aperture

/)(exp)exp(),( dSRZzYyikR

ikRZYE A

,'

'sin'

'sin)exp(

/)(exp)exp(),(2/

2/

2/

2/

RikRab

dydzRZzYyikR

ikRZYE

A

a

a

b

bA

Rectangular aperture:

RkbYRkaZ

2/',2/'

22

''sin

''sin)0(),(

IZYI

22

''sin

''sin)0(),(

IZYI

kbRm YRkbY 2 ... ,2 ,2/' Y minimum:

kaRm ZRkaZ 2 ... ,2 ,2/' Z minimum:

10.2.5 The circular apertureImportance in optical instrumentation: The image of a distant point source is not a point, but a diffraction pattern.

2

12

222

12

0 0

2

0 0

0

2

0 0

Aperture

/)/(4

)/(2)exp(

)/(2)exp(

cos/exp)exp(

)cos(/exp)exp(

/)(exp)exp(),(

)cos(coscossinsin

RkaqRkaqJ

RAE

RkaqJkaqRa

RikR

dRqkJR

ikR

ddRqikR

ikR

ddRqikR

ikR

dSRZzYyikR

ikRZYE

qqqZzYy

A

A

aA

aA

aA

A

21)(lim

)()]([

])cos(exp[2

)(

)cosexp(21)(

1

0

1

2

0

2

00

uuJ

uJuuJudud

dvvumviiuJ

dvviuuJ

u

mm

mm

m

m

Bessel functions:.

21

2

1

sin)sin(2)0(

/)/(2)0()(

kakaJI

RkaqRkaqJII

P(Y,Z)R

x

y

z

Y

Z

q

a

Circular aperture

21

sin)sin(2)0()(

kakaJII

J0(u)

J1(u)

u

)0(/)( II

sinka

3.83

0.018

q1

Df

aRq 22.1 2

22.11

Radius of Airy disk:

83.3/ 0)/( 11 RkaqRkaqJ

DP

f

10.2.6 Resolution of imaging systemsOverlap of two incoherent point sources:

D P1S1

S2

P2

Dfq 22.11

Rayleigh’s criterion for bare resolution:The center of one Airy disk falls on the first minimum of the other Airy disk.

Angular limit of resolution: D 22.1min

Our eyes: '1mm 2

nm 550min

Wavelength dependence: CD DVD

Question: Comparing the circular and the square aperture, why does the square aperture produce a smaller diffraction pattern? (/D vs. 1.22 /D)