diffraction modified aug2011

$: Diffraction Modified Aug2011$
This sunset picture illustrates how diffraction of light in the atmosphere

can produce spectacular images. High ice crystals create the coloured

fringes and atmospheric layers with different properties distort the shape

of the sun.

DIFFRACTION DIFFRACTION DIFFRACTION DIFFRACTION

OF LIGHTOF LIGHTOF LIGHTOF LIGHT

Diffraction

3

Topics

� Diffraction and wave theory of light

� Single-slit diffraction

� Intensity in single-slit diffraction

� Diffraction at a circular aperture

� Double-slit interference and diffraction

combined

� Multiple slits

� Diffraction gratings

� Dispersion and resolving power

� X-ray diffraction

Text Book:PHYSICS VOL 2 by Halliday, Resnick and Krane (5th Edition)

MIT- MANIPAL

What is the cause of diffraction?

Bending of light around the obstacle or an small aperture

What is the result?

�Light waves do not travel in straight line and cast sharp shadowsOf obstacles kept in the path.

�The intensity does not become zero exactly where a geometricShadows starts; but it gradually decreases to zero in a small distance

Definition

The phenomenon of bending of light around the edges of obstacles or slits,and hence its encroachment into the region of geometrical shadow isknown as diffraction.

What is diffraction?

According to Huygens wave theory of light,

Light travels in the form of waves and produces wave fronts in a

medium.

Each and every point on a particular wave front vibrates in the same

phase.

Each point on a particular wave front behaves like a secondary source,

sending out secondary waves.

These secondary waves starting out from different secondary sources on

a wave front can interfere.

This interference is called diffraction

The phenomenon of bending of light around the edges of obstacles or

slits, and hence its encroachment into the region of geometrical shadow is

known as diffraction.

For diffraction effects to be noticeable, the size of the object causing

diffraction should have dimensions comparable to the wavelength of light

falling on the object.

6

Fig: Diffraction pattern of razor blade viewed in

monochromatic light

BE-PHYSICS-DIFFRACTION-2010-11

DIFFRACTION AND WAVE THEORY OF LIGHT

MIT- MANIPAL

7BE-PHYSICS-DIFFRACTION-2010-11

Diffraction pattern occurs when

coherent wave-fronts of light fall on

opaque barrier B, which contains an

aperture of arbitrary shape. The

diffraction pattern can be seen on

screen C.

When C is very close to B a geometric

shadow is observed because the

diffraction effects are negligible.


MIT- MANIPAL

8


BE-PHYSICS-DIFFRACTION-2010-11

In fraunhofer diffraction, both the incident and

emergent wave-fronts are plane (the rays are

parallel) i.e., both the source and the screen are

effectively at infinite distances, from the

aperture causing diffraction.

Fraunhofer diffraction is a special limiting

case of the more general Fresnel diffraction.

In laboratory Fraunhofer diffraction is

realized by using converging lenses for

conversion of spherical wavefront into plane

wavefront and vice versa.

MIT- MANIPAL


Fresnal Diffraction: The incident wave

fronts are spherical or cylindrical. i.e.,

the source of light is at a finite

distance from the diffracting aperture.

The screen on which the diffraction

pattern is displayed is also at a finite

distance from the diffracting aperture.


MIT- MANIPAL

SINGLE-SLIT DIFFRACTION


All the diffracted rays arriving at P0 are in-phase.

Hence they interfere constructively and produce maximum (central

maximum) of intensity I0 at P0.

MIT- MANIPAL

11

At point P1,

path difference between r1 and r2

is (a/2) sinθ

λθ

λθ

=

=

sin

2sin

2

minimum,first for condition theSo

aor

a


This is satisfied for every pair of rays, one of which is from upper half of the slit and

the other is a corresponding ray from lower half of the slit.

12


At point P2,

path difference between r1 and r2

is (a/4) sinθ

λθλ

θ 2sin2

sin4

minimum, secondfor condition theSo

== aora

This is satisfied for every pair of rays, separated by a distance a/4.

...3,2,1, msin

minima, mfor condition thegeneral,In TH

±±±== λθ ma

NOTE: There is a secondary maximum approximately half way between each

adjacent pair of minima.

MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 13


Problem: SP42-1

A slit of width ‘a’ is illuminated by white light. For what value of ‘a’ does the

first minimum for red light (λ = 650nm) fall at θ = 15o?

Ans: a = 2510nm = 2.51µm



Problem: SP42-2

In SP42-1, what is the wavelength ‘λ’ of the light whose first diffraction

maximum (not counting the central maximum) falls at 15o, thus coinciding

with the first minimum of red light?

Ans: a sin θ = 1.5λ. So, λ=433nm.



Problem: E42-5

A single slit is illuminated by light whose wavelengths are λa and λb, so

chosen that the first diffraction minimum of λa component coincides with

the second minimum of the λb component.

(a) What is the relationship between the two wavelengths?

(b) Do any other minima in the two patterns coincide?

Ans: 2 : 1 , ma = mb/2


Questions to be Answered…

1. Distinguish between Fresnel and Fraunhofer Diffraction. [4]

2. Write a short note on wave fronts. [2]

3. Write a short note on Huygens wave theory [2]

4. Discuss the diffraction due to single-slit. Obtain the locations of the

minima and maxima qualitatively.[5]

5. Obtain an expression for the intensity in single-slit diffraction pattern,

using phasor-diagram. [5]

17

INTENSITY IN SINGLE – SLIT DIFFRACTION

• Aim is to find an expression for the intensity of the entire pattern as a

function of the diffraction angle.(using phasors)

• Imagine slit of width ‘a’ is divided into N parallel strips, each of width δx and

produces a wave of the same amplitude δE0 at P.

• The phase difference between two waves arriving at point P from two points

on the slit (with separation δx) is,

θδλπ

=φ∆ sinx2

BE-PHYSICS-DIFFRACTION-2010-11MIT- MANIPAL

18


Phasor showing

a) Central maximum (θ = 0)

b) A direction slightly shifted from

central maximum

c) First minimum

d) First maximum beyond the

central maximum

(corresponds to N = 18)


Some Phasor representations

19


2 where

sinEE ,Or

2sin

2

EE

Combining,

R

E Also

2sinR2E

diagram, From

m

m

m

φ=α

αα

=

φφ

=

=φ

φ=

θ

θ

θ

MIT- MANIPAL

20


3,.....2,1,m wherem

0sin minima,for eqn., above theFrom

intensity max. theis wheresin

sinE intensity The

2

m

2

m

2

2

m

2

±±±==⇒

=

∝Ι

Ι=Ι

=∝Ι

παα

αα

αα

θ

θθ

mE

E

MIT- MANIPAL

21


φ is the phase difference between rays

from the top and bottom of the slit.

So we can write,

θλπφ

α

θλπ

φ

sin2

So,

sin2

a

a

==

=


Since α = mπ =

∴∴∴∴ a sin θθθθ = mλλλλ

θλπ

sina

22


The intensity distribution in single-slit

diffraction for three different values of

the ratio a/λ

MIT- MANIPAL

NOTE: The wider the slit, the narrower is the central diffraction peak


Problem: SP42-3

Calculate, approximately, the relative intensities of the

maxima in the single slit Fraunhofer diffraction pattern.


2

m

sin

Ι=Ιαα

θ Where α = (m+ 1/2 )π

Ans: 0.045 (m=1), 0.016 (m=2), 0.0083(m=3) etc.

Hint:


Problem: SP42-4

Find the width ∆θ of the central maximum in a single slit

Fraunhofer diffraction. The width can be represented as the

angle between the two points in the pattern where the

intensity is one-half that at the center of the pattern.


Hint: Iθ = 1/2 Im

Use iterative technique for finding α

Ans: α = 1.39156, then θ =5.1 degree. So, ∆θ = 10.2 degree

2

m

sin

Ι=Ιαα

θ


Problem: E42-11

Monochromatic light with wavelength 538 nm falls on a slit with width

25.2µm. The distance from the slit to a screen is 3.48m. Consider a point

on the screen 1.13cm from the central maximum. Calculate (a) θ (b) α (c)

ratio of the intensity at this point to the intensity at the central

maximum.


Ans: θ = 0.00325 = 0.1860 , α = 0.478 rad, Iθ/Im = 0.926


Questions to be Answered Today…

1. Discuss qualitatively diffraction at a circular aperture. [2]

2. Explain Rayleigh’s criterion for resolving images due to a

circular aperture. [2]

27

DIFFRACTION AT A CIRCULAR APERTURE

DIFFRACTION PATTERN DUE TO A CIRCULAR APERTURE

Note: Diffraction effects often limit the ability of telescopes and other optical instruments to form precise images

28


� The mathematical analysis of diffraction by a circular aperture shows that the

first minimum occurs at an angle from the central axis given by

aperture. ofdiameter theis d where

22.1sind

λθ =

slit width theis a wheresin

isn diffractioslit singlein minimumfirst for equation The

a

λθ =

NOTE: In case of circular aperture, the factor 1.22 arises when we divide the

aperture into elementary Huygens sources and integrate over the aperture.

29

Rayleigh’s criterion for optical resolution: The images of two closely

spaced sources is said to be just resolved if the angular separation of the two

point sources is such that the central maximum of the diffraction pattern of one

source falls on the first minimum of the diffraction pattern of the other.

d

is

dR

λθ

θ

λθ

22.1

as dappoximate becan it small, very since

22.1sin

R

R

1

=

= − θθθθR is the smallest angular

separation for which we

can resolve the images of

two objects.

a. Well resolved

b. Just resolved

c. Not resolved

MIT- MANIPAL 30

Problem: SP42-5

A converging lens 32mm in diameter has a focal length f of 24

cm. (a) What angular separation must two distant point objects have to

satisfy Rayleigh’s criterion? Assume that λ = 550nm. (b) How far apart are

the centers of the diffraction patterns in the focal plane of the lens?


Ans: 2.1 x 10-5 rad, 5µm

MIT- MANIPAL 31

Problem: E42-21

The painting contains small dots (≈2 mm in diameter)

of pure pigment, as indicated in figure. The illusion of

colour mixing occurs because the pupils of the

observer’s eyes diffract light entering them. Calculate

the minimum distance an observer must stand from

painting to achieve the desired blending of colour.

(wavelength = 475nm, diameter of pupil = 4.4mm)


Ans: 1.32 x 10-4 rad, 15m

The two head lights of an approaching automobile are 1.42 m apart . At what

A) angular separation and

B) maximum distance will the eye resolve them?

Assume a pupil diameter of 5 mm and a wavelength of 562 nm. Also assume that

the diffraction effects alone limit the resolution.

Solution: Given data ;

Pupil diameter, a = 5 mm, Wavelength, λ = 562 nm,

Head lights distance, y = 1.42 m Angular separation, ө = ?

Resolving D =?

A. Angular separation required for the resolution is

өR = (1.22λ/a)

=1.37 x 10-4 rad

B. өR = y/D

=1.42/D =1.37x10 -4 rad.

D=1.04 X 10 4 m

Problem: E42.15

Problem:E42.19If superman really had X-ray vision at 0.12nm wavelength and

a 4.3mm pupil diameter, at what maximum altitude could be

distinguish villains from heroes assuming the minimum detail

required was 4.8cm?

Solution: Given data ;

Pupil diameter, a = 4.3 mm, Wavelength, λ = 0.12 nm,

Distance, y = 4.8c m

Resolving D =?

Angular separation required for the resolution is

өR = (1.22λ/a) = y/D

D= {(4.8x10-2m)(4.3x10-3m)}/{1.22(0.12x10-9m)}

D= 1.4x106 m

33

Diffraction in double slit interference?

• We neglected the diffraction effects during the discussion of double slit interference. Why?

• Does the effect of diffraction on double slit interference depends on the slit width ’a’ ?

� We assumed ‘a’ <<<< <<<< λλλλ

� Note that , for such narrow slits, the central part of the screen on which

light falls is uniformly illuminated by the diffracted waves from each slit.

When such waves interfere, they produce interference fringes of uniform

intensity.

MIT- MANIPAL 34

The intensity distribution in single slit diffraction for three

different values of ‘a’

MIT- MANIPAL 35

36

DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED

Note: Here, Fringe width depends on ‘d’ not ‘a’

But, intensity variation?

37

Fig: Interference fringes for a double slit with slit seperation d=50λ. Three

different slit widths are shown

38

Interference

Diffraction

Interference + Diffraction

( )ββββcos

cos

cos

cos

2222INT m,INT , I I =θ

2

Ι=θ αααα

αααα sin

sin

sin

sin

DIF

DIF

DIF

DIF

m,m,m,m,

DIF ,I

( )2

2

βΙ=θ αααα

ααααsin

sin

sin

sin

cos

cos

cos

cos

mmmmI


MIT- MANIPAL

Where ββββ = ππππdsinθθθθ/λλλλ

Where αααα = ππππasinθθθθ/λλλλ

39


Each of the two slits is divided into N zones. Electric field at P is found by adding

the phasors. There is phase difference of ∆φ = φ/N between each of the N phasors

where φ is the phase difference between 1st phasor and Nth phasor.

MIT- MANIPAL

40


Adding all the phasors, we get the resultant E1 due to the first slit. ξ is the phase

difference between the light waves at the point P, emitted from bottom edge of the

first slit and top edge of the second slit. E2 is the resultant due to the second slit. Eθ is

the resultant of E1 and E2.

41

βθλπ

=φ+ξ

θλπ

=φ

θ−λπ

=ξ

φ+ξ=

φ+ξ−

π=

δ

φ+ξ−π=δ

π=ξ+φ

+δ+φ

δ=θ

iswhichsind2

,getwe,eqnaboveofsidesbothtosina

2Adding

sin)ad(2

and

)A.........(2

cos22

sin2

sinAlso

)(or

22where

2sinE2E

,figuretheFrom

1

42


2

2

m

m1

m1

sin)(cos

cossin

)E2(E,ie2

sinE2E

sinEE

,slitonetoduePatamplitudeelectricthe

,havewe,ndiffractioslitglesinFrom

cos2

sin

,getwe),A(eqninthisngSubstituti

αα

βΙ=Ι∴

β

αα

=δ

=∴

αα

=

−

β=δ

θ

θθ

SINGLE-SLIT DIFFRACTION PATTERN

DOUBLE-SLITINTERFERENCE PATTERN

MIT- MANIPAL 43

Problem: SP42- 6


In a double slit experiment, the distance D of the screen

from the slits is 52cm, the wavelength is 480nm, slit

separation d is 0.12mm and the slit width a is 0.025mm.

a) What is the spacing between adjacent fringes?

b) What is the distance from the central maximum to the

first minimum of the fringe envelope?

Ans: 2.1 mm , 10mm

MIT- MANIPAL 44

Problem: SP42- 7

What requirements must be met for the central maximum

of the envelope of the double-slit interference pattern to

contain exactly 11 fringes?

Hint: condition required is 6th minimum of the interference should coincide

with the first minimum of the diffraction

Ans: d/a = 11/2

MIT- MANIPAL 45

Problem 42-E26

(a) Design a double slit system in which the fourth fringe, not

counting the central maximum, is missing.

(b) What other fringes , if any, are also missing?

Ans: (a). If d = 4a there will be no fourth interference maximum!

(b) mi = 4md

MIT- MANIPAL 46

Problem: E42-29

(a) How many complete fringes appear between the first

minima of the fringe envelope on either side of the central

maximum for a double-slit pattern if λ = 557 nm, d = 0.150

mm, and a = 0.030 mm?

(b) What is the ratio of the intensity of the third fringe to the

side of the center to that of the central fringe?

Ans: 9, 0.255

MIT- MANIPAL 47

MULTIPLE SLITS

� Thomas Young used his double slit interference technique for the first

measurement of wavelength of light.

� But, Accuracy of measurement depends on accurate determination of

the location of the fringes.

� As we increase the number of slits, the bright fringes continue to

become narrower, and the precision of wavelength measurement

continues to improve.

� Sharpness of fringes depends on the number of slits in the grating!

48MIT- MANIPAL

Intensity pattern for (a) Two-slit (b) Five-slit grating

(diffraction effect is neglected)

MULTIPLE SLITS

49

Condition for principal maxima,

d sin θθθθ = m λλλλ

where d is the separation between

adjacent slits.

� Location of principal maxima is

independent of number of slits.

�Note that if light passing through

any pair of adjacent slits is in phase

at a particular point on the screen,

then light passing through any pair

of slits, even non adjacent ones, is

also in phase at that point.

MULTIPLE SLITS

50

Width of the maxima: Case 1: Central maximum(Proving of sharpening of the principal maxima as N increased)

� The pattern contains central maximum with minima on either side.

� At the location of central maximum, the phase difference between the

waves from the adjacent slits is zero.

� At minima, the phase difference is such that,

slits ofnumber theis N where2

N

πφ =∆

� Corresponding path difference is,N2

Lλ

=φ∆

πλ

=∆

51

� Also we know,

N2L

λ=φ∆

πλ

=∆

Nd

Ndsin

sindN

sindL

0

0

0

0

λ≈δθ

λ=δθ

δθ=λ

δθ=∆

From the equation, we can infer that, for given λ and

d, if we increase the number of slits (N), then the

angular width of principal maximum decreases. ie, the

principal maximum becomes sharper.

52

Width of the maxima: Other principal maxima

For the mth principal maximum at θby a grating: d sinθθθθ = m λλλλ.

For the first minimum at θθθθ + δδδδθθθθafter the mth principal maximum,

( )NNNNλλλλmmmmλλλλθθθθθθθθs

in

sin

sin

sin

dddd +=δ+

MINIMUM AT θ +δδδδθ

mth PRINCIPAL MAXIMUM AT θ

MULTIPLE SLITS

MINIMUM AT θ +δδδδθ

mth PRINCIPAL MAXIMUM AT θ 53

( )NNNNλλλλmmmmλλλλθθθθθθθθs

in

sin

sin

sin

dddd +=δ+

Nmsin coscos sind

1

λ+λ=

δθθ+δθθ

δθ321321

( ) Nmcos dsin d λ+λ=δθθ+θ321}

( ) Nmdm λλδθθλ +=+ cos

θλ

=δθcos d N

The principal maximum become sharper as

number of slits (N) increases

ANGULAR HALF WIDTH OF mTH

PRINCIPAL MAXIMUM AT θ

Width of the maxima: Other principal maxima


Problem: SP43- 1

MULTIPLE SLITS

A certain grating has 104 slits with a spacing of d = 2100 nm.

It is illuminated with yellow sodium light (λ = 589 nm). Find

(a) the angular position of all principal maxima observed

and (b) the angular width of the largest order maximum.

Ans: 16.30 (m=1), 34.10 (m=2), 57.30 (m=3) [total 7 principal maxima]

δθ = 5.2 x 10-5 rad = 0.00300

MIT- MANIPAL 55

Answer the following questions:

• Discuss qualitatively the diffraction due to multiple slits (eg,

5 slits). [4]

• Obtain an expression for the width of the central maximum

in diffraction pattern due to multiple slits. [4]

• Obtain an expression for the width of a principal maximum

at an angle in diffraction pattern due to multiple slits. [4]

MIT- MANIPAL 56

Problem: E43-5

Light of wavelength 600 nm is incident normally on a

diffraction grating. Two adjacent principal maxima occur at

sin θ = 0.20 and sin θ = 0.30. The fourth order is missing. (a)

what is the separation between adjacent slits? (b) What is the

smallest possible individual slit width? (c) Name all orders

actually appearing on the screen with the values derived in

(a) and (b).

Ans: 6µm, 1.5 µmC) The visible orders would be integer values of m except for when m is

multiple of four( m=1,2,3, 5,6,7, 9,……….)

MIT- MANIPAL 57

Problem: E43- 3

With light from a gaseous discharge tube incident normally on a

grating with a distance 1.73 µm between adjacent slit centers, a

green line appears with sharp maxima at measured

transmission angles θ = ±17.6°, 37.3°, -37.1°, 65.2° and -65.0°.Compute wavelength of the green line that best fits the data.

Solution:

m θθθθ sinθθθθ m θθθθ sinθθθθ

1 17.6 0.302 -1 -17.6 -0.302

2 37.3 0.606 -2 -37.1 -0.603

3 65.2 0.908 -3 -65.0 -0.906


Ans: slope= sin θ/m = λ/d

So, λλλλ= 522 nm.

59

DIFFRACTION GRATINGS

� The diffraction grating, a useful device for

analyzing light sources, consists of a large number

of equally spaced parallel slits.

� A transmission grating can be made by cutting

parallel grooves on a glass plate with a precision

ruling machine. The spaces between the grooves

are transparent to the light and hence act as

separate slits.

� A reflection grating can be made by cutting

parallel grooves on the surface of a reflective

material. The reflection of light from the spaces

between the grooves is specular, and the reflection

from the grooves cut into the material is diffuse.

MIT- MANIPAL

61MIT- MANIPAL


Sample spectra of visible light emitted by a gaseous source



Problem: SP43-2

A diffraction grating has 104 rulings uniformly spaced

over 25.0mm. It is illuminated at normal incidence by

yellow light from sodium vapor lamp which contains two

closely spaced lines of wavelengths 589.0nm and

589.59nm. (a) At what angle will the first order

maximum occur for the first of these wavelengths? (b)

What is the angular separation between the first order

maxima of these lines?

Ans: 13.60 , 0.0140

MIT- MANIPAL 63

Problem: E43-9

Given a grating with 400 rulings/mm, how many orders of

the entire visible spectrum (400-700nm) can be produced?

Solution: d= 1mm/400To find the number of orders of the entire visible spectrum that will be

present, we need only consider the wavelength which will be on the outside

of the maxima. That will be the longer wavelengths, so we only need to look

at the 700 nm behaviour.

Use d sin θ = mλ, assume maximum angle 900

Ans: m=3

MIT- MANIPAL 64

Problem: E43-11

White light (400 nm < λ < 700 nm) is incident on a grating .

Show that, no matter what the value of the grating spacing d,

the second- and third-order spectra overlap.

Solution: If the second-order spectra overlaps the third-order, it is because

the 700 nm second-order line is at a larger angle than the 400 nm third-order

line.

Using dsinθ = mλ, sinθ = mλ/d

(2 x 700)/d > (3x400)/d

∴ sinθ2 , 700nm > sinθ3 , 400nm

MIT- MANIPAL 65

Questions to be answered today…

Obtain an expression for dispersion by a diffraction grating.[3]

Obtain an expression for resolving power of a diffraction

grating. [3]

66

DISPERSION AND RESOLVING POWER

linesspectral ofwavelengthbetweenDifference

linesspectralbetweenseparationAngularDispersion =

ΔλΔλΔλΔλ

ΔθΔθΔθΔθ

DDDD =

The ability of a grating to produce spectra that permit precise measurement of

wavelengths is determined by two intrinsic properties of the grating,

(1) Dispersion (the separation ∆θ∆θ∆θ∆θ b/w spectral lines that differ in wavelength ∆λ∆λ∆λ∆λ)

(2) Resolving power (the width or sharpness of lines)

Dispersion is useful quantity in distinguishing wavelengths that are close to each

other, a grating must spread apart the diffraction lines associated with the various

wavelengths.

MIT- MANIPAL

Dispersion

67

Dispersion

d sinθθθθ = m λλλλ

Differentiating the above equation,

θθθθcos

cos

cos

cos

ddddmmmm

ΔλΔλΔλΔλ

ΔθΔθΔθΔθ

==D

So, To achieve higher dispersion we must use a grating of smaller

grating spacing and work in higher order m .

ΔλΔλΔλΔλ

ΔθΔθΔθΔθ

DDDD =

MIT- MANIPAL

d cos θθθθ ∆θ∆θ∆θ∆θ = m ∆∆∆∆λλλλ


68

� Ability of the grating to resolve two nearby spectral lines so that

the two lines can be viewed or photographed as separate lines.

� To resolve lines whose wavelengths are close together, the lines

should be as narrow as possible.

For two close spectral lines of wavelength λ1 and λ2, justresolved by the grating, the resolving power is defined as

λ∆λ

=R 21 λ−λ=λ∆2

21 λ+λ=λ

MIT- MANIPAL

Resolving power


Note: ∆λ is the smallest wavelength difference that is just resolved

69

θθθθcos

cos

cos

cos

ddddmmmm

ΔλΔλΔλΔλ

ΔθΔθΔθΔθ

==Dθ

λ=θ∆

cos dN

θ=

λ∆

θ

λ

cosd

mcosdN

mNR =λ∆λ

=

Resolving power increases with increasing N

We have,

Putting second equation in first equation,

Resolving power

70MIT- MANIPAL

Intensity patterns of two close

lines due to three gratings A, B, C.

N = 5000

d = 10 µm

R = 5000

D = 0.1 rad/µm

N = 5000

d = 5 µm

R = 5000

D = 0.2 rad/µm

N = 10000

d = 10 µm

R = 10000

D = 0.1 rad/µm


71

A grating has 9600 lines uniformly spaced over a width

3cm and is illuminated by mercury light.

a) What is the expected dispersion in the third order, in

the vicinity of intense green line (λ = 546nm)?

b) What is the resolving power of this grating in the

fifth order?

Soln: d = W/N = 3125nm, use dsinθθθθ = mλλλλ and

θ=31.60 , 1.13 x 10-3 rad/nm, R=Nm=4.8 x 104

MIT- MANIPAL

Problem: SP43-3


θθθθcos

cos

cos

cos

ddddmmmm

ΔλΔλΔλΔλ

ΔθΔθΔθΔθ

==D

72

A diffraction grating has 1.20 X 104 rulings uniformly spaced

over a width W = 2.50cm. It is illuminated at normal incidence

by yellow light from a sodium vapor lamp. This light contains

two closely spaced lines of wavelengths 589.0 nm and 589.59

nm. (a) At what angle does the first maximum occur for the first

of these wavelengths? (b) What is the angular separation

between these two lines (1st order)? (c) How close in

wavelength can two lines be (in first order) and still be resolved

by this grating? (d) How many rulings can a grating have and just

resolve the sodium doublet line?

Ans: 16.40 , 2.95 x 10-4 rad, 1.2 x 104 , 0.049nm, R= 998, N = 998

MIT- MANIPAL

Problem: SP43-4


73

The sodium doublet in the spectrum of sodium is a pair of

lines with wavelengths 589.0 and 589.6 nm. Calculate the

minimum number of rulings in a grating needed to

resolve this doublet in the second-order spectrum.

Soltn:

N ≈ 500

MIT- MANIPAL


Problem: E43-17

74

In a particular grating, the sodium doublet is viewed in third

order at 10.2° to the normal and is barely resolved. Find (a)

the ruling spacing and (b) the total width of grating.

Soltn: use dsinθθθθ = mλλλλ, Assume λ=589nm, find d

d= 9.98 µm .

Assume R=1000 = Nm, N=333. So, 333(9.98µm) = 3.3 mm.

MIT- MANIPAL


Problem: E43-21

Wilhelm Roentgen

• Wilhelm Roentgen

• German physicist Wilhelm Roentgen won

the 1901 Nobel Prize in physics.

Roentgen, who was the first Nobel

laureate in physics, won the award for

his discovery of a type of short-wave

radiation popularly known as X rays

�X-rays are electromagnetic radiations of very short wavelength

ranging from 0.1 Å (0.01 nm) to 100 Å (10 nm).

�X-rays can be produced when kinetic energy of fast moving

electrons is transformed into energy of electromagnetic waves.

78

X-RAY DIFFRACTION

� For the observation of diffraction phenomenon by grating, the

grating space should have the dimension of the wavelength of

the wave diffracted. (ie., d≈ λ)

� Since the x-ray wavelength and the inter-planar spacing in

crystals are of the same order, a crystal can be a suitable grating

for observing the diffraction of x-rays.

x-ray diffraction producingLaue’s pattern

X-ray tube

79

X-RAY DIFFRACTION

� When a x-ray beam is incident on a sample of a

single crystal, diffraction occurs resulting in a

pattern consisting of an array of symmetrically

arranged diffraction spots, called Laue’s spots.

� The single crystal acts like a grating with a grating

constant comparable with the wavelength of x-rays,

making the diffraction pattern distinctly visible.

� Since the diffraction pattern is decided by the

crystal structure, the study of the diffraction

pattern helps in the analysis of the crystal

parameters.

A Laue pattern of a

single crystal.Each dot represents a

point of constructive

interference.

80

X-RAY DIFFRACTION

NaCl crystal (a0 =0.563nm)

A plane through a crystal of NaCl

Unit Cell: Smallest unit from which the

crystal may be built up by repetition in 3

dimensions

Sir William Henry Bragg

British physicist SirWilliam Henry Braggwon the Nobel Prizein physics in 1915.

He developed X-raycrystallography whichwas used to studycrystal structure.

82MIT- MANIPAL

X-RAY DIFFRACTIONBragg planes

� In every crystal, several sets of parallel planes called the Bragg planes can

be identified.

� Each of these planes have an identical and a definite arrangement of

atoms.

� Different sets of Bragg planes are oriented at different angles and are

characterized by different inter planar distances d.

� In this given case,

5

0ad =

83MIT- MANIPAL

θ�� Glancing angle. ie., anglebetween the incident x-ray beamand the reflecting crystal planes.

� For constructive interferenceof diffracted x-rays the pathdifference for the rays fromthe adjacent planes, (abc inthe figure) must be an integralnumber of wavelength.

ie.,

2d sin θθθθ = n λλλλ

MIT- MANIPAL 84

Here, The total path difference = 2dsinθSo, for constructive interference, 2d sin θθθθ = n λλλλ

85

Note :

� The directions(but not the intensities) of all the diffracted x-ray beams

that can emerge from a crystal is determined by the geometry of three

dimensional lattice of diffracting centers.

� The intensities of the diffracted beams emerging from a crystal depend

on the diffracting characteristics of the unit cell.

� The diffracting characteristics of a unit cell depend on how the

electrons are distributed throughout the volume of the cell

So,

� By studying the directions of diffracted x-ray beams, we can learn the

basic symmetry of the crystal.

� By studying the intensities we can learn how the electrons are

distributed in a unit cell.

86

X-RAY DIFFRACTION

(a) Electron density contour of an organic molecule

(b) A structural representation of same molecule

� The x-rays are diffracted by the electron concentrations in the

material. By studying the directions of diffracted x-ray beam,

we can study the basic symmetry of the crystal.

� By studying the intensity, we can learn how the electrons are

distributed in a unit cell.

87

X-RAY DIFFRACTION

Problem: SP43-5

At what angles must an x-ray

beam with wavelength = 0.110

nm fall on the family of planes

in figure if a diffracted beam is

to exist? Assume material to

be sodium chloride (a0 =

0.563nm)

Soltn: 2d sin θθθθ = n λλλλand

Ans: 12.60 (n=1), 25.90 (n=2),40.90 (n=3) and 60.80 (n=4)

5

0ad =


X-RAY DIFFRACTION

Problem: E43-25

A beam of x-rays of wavelength 29.3 pm is incident on a

calcite crystal of lattice spacing 0.313 nm. Find the smallest

angle between the crystal planes and the beam that will

result in constructive reflection of the x-rays.

Soltn: We are looking for the smallest angle; this will correspond to the

largest d and the smallest m.

That means m = 1 and d = 0.313 nm.

Ans: 2.680


X-RAY DIFFRACTION

Problem: E43-33

First order Bragg scattering

from a certain crystal occurs

at an angle of incidence of

63.8°, (ref. figure).

Wavelength of x-rays is

0.261nm. Assuming that the

scattering is from the dashed

planes, find unit cell size a0.

Soltn: a0 = √2d, θ = 63.80- 450

d= 0.4049nm, a0 = 0.5726nm

MIT- MANIPAL 90

QUESTIONS – DIFFRACTION

Discuss the diffraction due to single-slit. Obtain the

locations of the minima and maxima qualitatively. [5]

Obtain an expression for the intensity in single-slit

diffraction pattern, using phasor-diagram. [5]

Calculate, approximately, the relative intensities of the first

three secondary maxima in the single-slit diffraction

pattern. [4]

Discuss qualitatively diffraction at a circular aperture. [2]

MIT- MANIPAL 91


Explain Rayleigh’s criterion for resolving images due to a

circular aperture. [2]

Obtain an expression for the intensity in double-slit

diffraction pattern, using phasor-diagram. [5]

Discuss qualitatively the diffraction due to multiple slits

(eg, 5 slits). [4]

Obtain an expression for the width of the central

maximum in diffraction pattern due to multiple slits. [4]

MIT- MANIPAL 92


Obtain an expression for the width of a principal

maximum at an angle in diffraction pattern due to

multiple slits. [4]

Obtain an expression for dispersion by a diffraction

grating. [3]

Obtain an expression for resolving power of a diffraction

grating. [3]

Discuss Bragg’s law for X-ray diffraction. [3]


ANSWERS

E42-1: 690 nm

E42-11: 0.186°, 0.478 radian, 0.926

E42-16: 36.2 m

E42-19: 1400 km

E42-21: 15 m

E42-26: (a) d =4a (b) Every 4th fringe

E42-29: (a) 9 (b) 0.255

E43-3: 523 nm

E43-5: (a) 6 µm (b) 1.5 µm (c) m = 0, 1, 2, 3, 5, 6, 7, 9

E43-9: 3

E43-17: 491

E43-21: (a) 9.98 µm (b) 3.27 nm

E43-25: 2.68 degree

E43-33: 0.5726 nm

diffraction modified aug2011

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