12 rr Difference of Optical Path Length

min2

)12(

maxsin

m

md

Interference Two waves

One wave Many waves2

max...3,2,1 2

)12(

min...3,2,1 2

2

center at the fringeBright 0

sin

mm

mmma

Diffraction

22)2/1(

m

II m

)2/1(

m

AA m

Intensity in single-slit diffraction

sin0 N

a

sin22

N

ao

sin

2aN

2sin2

RA

2sin2

mA

sin2let a

2

2

m

sin

II

sin

mA

0 mII Maximum at center

ma sin 0I Minimum

)2

1(sin ma

22)2/1(

m

II m Maximum

Central maximum

Diameter d

light

d

22.1

Width of the bright fringe at the center:

dffr 22.1

a

fxM

2

2

aM

22.1min d

Two objects are just resolved when the maximum of one is at the minimum of the other.

ability resolution/1 RR

not resolved resolved just resolved

R

12 rr sind

sin22

d

Double-slit Interference and Diffraction Combined

factor ceinterferen cos2

sin where sin 2

m

aII

Each slit

factorn diffractio sin 2

cos sin 22

m

II

Combined

sin

2

d

A

A

A

A

cos2' AA 2cosmII

2sin21

RE

R

mE

R

sin

2

2sin

1 mm EEE

Double-slit Interference and Diffraction Combined

sin)(

2

sin)(2 adad

sin

2

sin2 aa

sinsin)(

2

aad

sind

2/sin2 1 EE

)(

)22

sin(2

sin

2

cos

cos

cossin

2 mE cos)sin

( 22

mII

factor ceinterferen cos2 factorn diffractio

sin 2

The width of the central fringe is dependent on λ/a.

width the,a

missing fringes

sin

d

22

m cos sin

II

sin

a

if m=d/a

the mth maximumd

m sin

the first minimuma

'sin

' ,'sinsin

A slit of width a = 0.5 mm is illuminated by a monochromic light. Behind the slit there placed a lens (f = 100 cm), and the first maximum fringe is observed at a distance of 1.5 mm from the central bright fringe on the focal plane (screen). Find the wavelength of the light, and the width of the central bright fringe.

1 ,2

)12(sin :max mma

Example

nm5003

nm1500

)12(

2

fm

ax

mm220 a

fx

f

xsin

A typical astronomical telescope has a lens with a dimension of 10 m. Find the minimum resolving angle. Considering we use this telescope to observe the moon surface. What is the separation of two objects can just be resolved? (take wavelength of 540 nm)

89

106.610

1054022.122.1

DR

km 10 3.8 be moon to theandearth the

of surfacesbetween distance Taken the5

m52

106.6108.3 88

Rmemoon Lx

Example

Example Two slits with separation of d = 5.5 mm is illuminated by a monochromic light. There are 21 fringes in the central maximum. What is the width of each slit?

The missing fringe is 11th.

a=d/11=0.5mm

11sin d

sina

Chapter 43 Gratings and Spectra

1-slit

2-slit

Multiple slits?

Gratings Diffraction

Diffraction Interference

slit 1 slits 5

slits 6

slits 20

slits 2

slits 3

N is increasing, the maximum become sharper

,...)2,1,0( sin mmd

The location of principal maxima is determined by:

Principal maxima

d is slip separation, also is called grating constant (grating spacing).

sin

2d

sin

2NdN '2m

min'

sin N

md

),2 ,1'( m

.sin d

.2 m

min

max

sin

d

Nd

2d

Nd

00sin

N is increasing, the is decreasing, the maxima become sharper

md sin

,...)2,1,0( sin mmd

The location of principal maxima is determined by:

Principal maxima

d is slip separation, also is called grating constant (grating spacing).

sin

2d

sin

2NdN '2m

min'

sin N

md

),2 ,1'( m

.sin d

.2 m

sin

d

Nd

2d

Nd

00sin

N is increasing, the δθ is decreasing, the maxima become sharper

md sin)sin( d

)sincoscos(sin d

cossin dd N

m

Nd

cos

Nd

cosNd

Nd

md sin

Secondary maxima

There are some minima between two principal maxima.

Superposition of some phases

d

sin

cosNd

Secondary maxima

Nd

3 slipsThere are 2 minima between two principal maxima.

4 slips

Δφ=0, 2π, ….

Δφ=π/2 Δφ=3π/2Δφ=π

Δφ=0, 2

Δφ=4π/3

1

23

Δφ=2π/3

123

There are 3 minima between two principal maxima.

If the grating has N slips, there are N-1 minima between two principle maxima.

If the grating has N slips, there are N-2 secondary maxima between two principle maxima.

N is increasing, the δθ is decreasing, the maxima become sharper

N=5

Homework:P977-978 18, 24, 27,29P979 7

Multiple slits

Gratings

slit 1 slits 5

slits 6

slits 20

slits 2

slits 3 N is increasing, the maximum become sharper

min

max

,2

)12(

,sin

Am

Amd

two slits

3 slipsThere are 2 minima between

two principal maxima.

4 slipsThere are 3 minima between

two principal maxima.

If the grating has N slips, there are N-1 minima between two principle maxima.

If the grating has N slips, there are N-2 secondary maxima between two principle maxima.

N is increasing, the maxima become sharper

Intensity distribution

sin, dL sa

ΔL sin2

,

dsa

sin

2, dNN saN

)sin

sin(2

dNRE

saN ,

ER

sinmE E

sind

sin

)sin(NEm

mE

2

2

2

2

0 sin

)(sinsin NII

factor ceinterferen

factorn diffractio

sin

)sin(sin NE)

sinsin(

)2

sin(22

,

dNE

Esa

m

)

sinsin(

)sin

sin(

dNdEm

N is increasing, the maxima become sharper

ΔL

sind

2 2

2 2

sin sin ( )

sinm

NI I

factor ceinterferen

factorn diffractio

sina

max ,sin md

max ,sin md

,'sin mdNN

'

N

πmβ min

min

N is increasing, the maxima become sharper

md sin

'sina

madif

11sin md

21'sin ma

am

mdif

2

1

mth fringe is missing

m1th fringe is missing

ΔL

sind

2

2sin

mII

2

2

2

2

sin

)(sinsin NIm

factor ceinterferen

factorn diffractio

sina

If the grating has N slips, there are N-1 minima between two principle maximum, and there are N-2 secondary maximum

Width of the maximum

cosNd

Nd

Diffraction gratings

Transmission grating

Reflection grating

Kinds of gratings

,...2,1,0

sin

m

md

cos

: widthAngle

Nd Nd

Example

d

Wave length

When waveis mth max

What is for (m+1)th max

mdd sin30sin 0

)1(sinsin mdd

030sinsin dd

)5.0(sin 1

d

Example: A grating (N=5000) is illuminated by two monochromic lights with wave lengths of 600 and 400 nm respectively. The mth principal maximum of the former light is meet the (m+1)th principal maximum of the later at 3 cm from the central fringe on the screen. The focus length of the lens is 50 cm. Find the grating constant d, and the typical width of the principle fringes.

1sin md

1sin mf

ydd

2)1(sin md

2)1( 21 mmm

m102 51 y

mfd

cos

1

Nd

Nd1 6106

m3m103 6 fy

m2' y

62 104' Nd

Diffraction Grating ,...2,1,0

sin

m

md

0 ,0 m

Diffraction Grating ,...2,1,0

sin

m

md

Spectra of light0 ,0 m

Dispersion and resolving power

Dispersion To display the light with different wavelength

The ability of a grating is determined by two intrinsic properties of the grating:

Resolving power

12

(1)

12

(2)the width of the line

1 2

D

d

d

md sin sind

m cosm

d

cosd

m

Dm ,

R

cosNd

cos

m

d

mN

RN , Rm ,

central1st (m=1) 2nd

(m=2)

3rd (m=3)

A grating N=9600, with width W=3.00cm,

line spectrum of mercury vapor (=546 nm).

nm3125N

Wd

41080.4 NmR

Example

What is dispersion Din 3nd order (m=3)

What is resolving power Rin 5th order (m=5)

546nm 3, m

D

nm011.0

R

6.31)(sin ,sin 1-

d

mmd

d

m sin

dd

md cos

cos

d

m

d

d

/nm3.87arcmin/nm0646.0

Dispersion and resolving power

nm6.589 nm,0.589 nm,2500 21 d

6.13)(sin ,sin 11-1

d

mλmd

ddcos ,sind

m

d

mλ

Example

?D 1,

m

0.97 6.13cos os 1 c

0.014 cos 1d

m5

1

104 cos

d

m

d

d

Dispersion

? 1, m

A grating with width W=4.15 cm. To resolve =415.496 nm and =415.487 nm in the second order, what is N? At what angle is 2nd maximum? Whis the width of 2nd maximum?

nm1800N

Wd

NmR

Example

m

N

5.27)(sin ,sin 1-

d

mmd

23000

cosNd

rad51013.1

009.02

496.415

Gratings

Spectrum Structure of crystal

Constructive interference: 2dsin m d 0.5nm in NaCl

d

1st maximum will be at 9°

Crystal solid

dsin

=0.17nm

X-ray Diffraction

As a gratingscrystal

2nd maximum will be at 20°

3rd maximum will be at 31°

1sin ( ) 2

m

d

4th maximum 5th maximum = 500nm ?

impossible

Bragg’s law

,...3,2,1

sin2

m

md

Wavelength:

nm 0.1 of valueTypical

d

d

Laue experiment

Laue spots

Determination of crystalline structures

A X-ray with λ=0.10 nm to 0.14nm, A crystal with a0=0.263nm,

nm263.00 ad

md sin2

Example

If a diffraction beam is to exist, when =450

a0

37.0m

3m 0.123nm

What is m,

Laue experiment

A X-ray with λ=0.110 nm, A crystal with a0=0.563nm,

Example

If a diffraction beam is to exist, 2=?

a0

0

0

0

0

0

7.242 ,10 ,9.562 ,9

2.772 ,8 ,7.932 ,7

2.1082 ,6 ,5.1212 ,5

0.1342 ,4 ,9.1452 ,3

5.1572 ,2 ,0.1692 ,1

mm

mm

mm

mm

mm

)563.02

110.0(cos)

2(cos 11

m

d

m

2 mθd 'sin2

mθd )90sin(2 0

mθd cos2

nm4.020

ad md 'sin2

a0

d

)4.02

110.0(cos)

2(cos 11

m

d

m

A X-ray with λ=0.110 nm, A crystal with a0=0.563nm,

Example

If a diffraction beam is to exist, 2=?

2

5.372 ,7

8.682 ,6 ,1.932 ,5

3.1132 ,4 ,3.1372 ,3

1.1482 ,2 ,2.1642 ,1

0

0

0

m

mm

mm

mm

mθd )90sin(2 0

X-ray diffraction

X-ray

sample

Debye method

2

I

2d·cos=m

A X-ray with λ=0.110 nm, the incidence angle is 600, assuming the scattering is from the dashed planes, find the unit cell size a0 05.26

000 5.335.2660

Example

a0

d2

1tan

md 05.33sin2

05.33sin2

md

00 5.26sin

da

nmam 22.0 ,1 0

00 5.33sin5.26sin2

m

nmam 44.0 ,2 0

ExercisesP996-997 7, 11, 17, 19, 27, 33