DiffractionTopics

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)

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.

DIFFRACTION AND WAVE THEORY OF LIGHT

Diffraction pattern of razor blade

viewed in monochromatic light

DiffractionFor plane waves entering a single slit, the waves emerging from the slit start spreading out, diffracting.

Huygen’s Principle: All points on a wavefront serve as point sources of spherical secondary wavelets. After time t, the new position of the wavefront will be that of a surface tangent to these secondary wavelets.

Light as a Wave

Fig. 35-2

Diffraction pattern occurswhen coherent wave-fronts oflight fall on opaque barrier B,which contains an aperture ofarbitrary shape. Thediffraction pattern can beseen on screen C.When C is very close to B ageometric shadow is observedbecause the diffraction effectsare negligible.

DIFFRACTION AND WAVE THEORY OF LIGHT

Types of Diffraction

1. Fresnel Diffraction

2. Fraunhofer Diffraction

Fresnel diffraction & Fraunhofer diffraction

Both Incident wave front are

spherical or cylindrical

Both the source and the screen

are effectively at finite

distances, from the aperture

causing diffraction.

No convex lens used

Analysis of pattern is

complicated

Both incident and emergent wave

fronts are plane.

Both the source and the screen

are effectively at infinite

distances, from the aperture

causing diffraction.

Two convex lenses are used.

Analysis of pattern is simple.

In laboratory Fraunhofer diffraction is realized by using converginglenses for conversion of spherical wavefront into plane wavefront andvice versa.

Interference

• Superposition of two separate

wave fronts originating from

two coherent source.

• The regions of minimum

intensity is perfectly dark

• Equal fringes width

• Uniform intensity

Diffraction

• Superposition of secondary

wavelets originating from different

parts of the same wave front

• The regions of minimum intensity

is not perfectly dark

• Un-equal fringes width

• Non- uniform intensity

SINGLE-SLIT DIFFRACTION

Suppose light is incident on single slit of width a, then all the

diffracted rays arriving at P0 are in-phase.

Hence they interfere constructively and produce maximum

(central maximum) of intensity I0 at P0.

When the path length difference between rays r1 and r2 is λ/2, the two

rays will be out of phase when they reach P1 on the screen, resulting in

destructive interference at P1. The starting point of r2 at the center of the

slit to point b.

At point P1, path difference between r1 and r2 is (a/2) sin

2sin

2

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.

sina

So the condition for First minimum

For second minimum at P2, divide slit into 4 zones of

equal widths a/4 (separation between pairs of rays).

Destructive interference occurs when the path length

difference for each pair is λ/2.

(second minimum)sin sin 24 2a a

Dividing the slit into increasingly larger even

numbers of zones, we can find higher order minima:

There is a secondary maximum approximately half

way between each adjacent pair of minima.

...3,2,1, m

sin

maIn General, the condition for mth minima,

Expression for Intensity in Single-Slit Diffraction

To find the net electric field Eθ

(intensity a Eθ2) at point P on

the screen.

The slit is equally divided into

N zones, each with width δx.

Each zone acts as a source of

Huygens wavelets. These

zones can be superimposed at

the screen to obtain the

intensity as a function of θ, the

angle to the central axis.

INTENSITY IN SINGLE – SLIT DIFFRACTION

sinx2

phase path length2difference difference

The phase relationships among the wavelets arriving from different

zones is given by

INTENSITY IN SINGLE – SLIT DIFFRACTION

a) Central maximum

b) A direction slightly

shifted from central

maximum

c) First minimum

d) First maximum beyond

the central maximum

( corresponds to N = 18)

Phasors in single slit diffraction,

showing conditions at

Intensity in Single – slit diffractionIf we divide the slit into

infinitesimally wide zones δx,

the arc of the phasors

approaches the arc of a circle.

The length of the arc is Em.

f is the difference in phase

between the infinitesimal

vectors at the left and right

ends of the arc. f is also the

angle between the 2 radii

marked R.

sinE

2sin

2E

m

m

E

E

f is the phase difference between rays from the top and bottom

of the slit. The path difference for these rays is asinѲ.

f is related to the path length difference across the entire slit.

RE 2/

2sin

diagram, From

2sin2

RE

)E ( m RArclength

REm

2 where

So we can write

3,.....2,1,m wherem sin or,3,.....2,1,m wherem

0sin minima,for eqn., above theFrom

intensity max. theis I wheresinI

sinE Iintensity The

2m

2

m

22m

2

a

E

E

m

sina2

So,

sina2

INTENSITY IN SINGLE – SLIT DIFFRACTION

The intensity distribution in

single-slit diffraction for three

different values of the ratio a/

The condition for minimum is

asinѲ= mλ

Phase difference is given by

α = mπ

The condition for maximum is

asinѲ= (m+ ½ )λ

Phase difference is given by

α = (m+1/2) π

NOTE:α = f /2 half the Phase diff

m=1,2,3,4,….

[1] When a monochromatic light is incident on a slit 0.022mm wide, the

first diffraction minimum is observed at an angle of 1.8o from the

direction of the incident beam. Find the wavelength of incident light.

Solution:

a sinѲ= mλ

For first minimum, m=1,

λ =a sinѲ

=(0.022mm) (sin1.80 )

λ =691nm

[2a] A slit of width a is illuminated by white light. For what value of a

does the minimum for red light ( = 650nm) fall at = 15o?

Solution:

a sinѲ= mλ

For first minimum, m=1,

a=λ/ sinѲ

=650nm/sin150

a=2.51μm

[2b] What is the wavelength ’ of the light whose first diffraction

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

coinciding with the first minimum of red light?

Solution:

The condition for maximum is ,

asinѲ= (m+ ½ )λ’

λ’ = asinѲ/ (m+ ½ )

λ’= [2.51μm x sin150)]/(1+½)

= 433nm

Light of this wavelength is violet.

[3] A single slit is illuminated by light whose wavelengths are a and b,

so chosen that the first diffraction minimum of a component coincides

with the second minimum of the b component.

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

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

Solution:

a) We have asinѲ=mλ -----(1)

a sin Ѳa = a

a sin Ѳb = 2 b

If the angles match, then so will the sine of the angles.

a = 2 b

aaba 21

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

Using Eq. (1) and We have a = 2 b

ma = mb/2

when mb is an even integer ma is an integer. Then all of the direction

minima from a are overlapped by a minima from b .

am

abba

ama

ma

bbb

2m a

[4] Calculate, approximately, the relative intensities of the maxima in the

single slit Fraunhofer diffraction pattern.Solution:

The maxima lie approximately half way between the minima and are

roughly given by α = (m+1/2) π, m=1,2,3…….

Substituting into equation IѲ= Im(sinα/ α)2

IѲ= Im [sin (m+1/2) π / (m+1/2) π]2

which reduces to

IѲ/ Im = {1 / (m+1/2) 2 π2}

This yield

IѲ/ Im = 0.045 ( m=1)

IѲ/ Im= 0.016 ( m=2)

IѲ/ Im= 0.0083 ( m=3) and so forth.

The successive maxima decrease rapidly in intensity.

[5] Monochromatic light with wavelength 538 nm falls on a slit with width

25.2m. 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) (b) a

(c)ratio of the intensity at this point to the intensity at the central

maximum

(a) This is a small angle approximation problem,

sin Ѳ ≈ tan Ѳ ≈ Ѳ ≈ Y/D

Ѳ =(1.13 x 10-2m)/(3.48 m)

=3.25 x10-3 rad

(b) The phase difference, is

α = (πa/λ)sin Ѳ

=(π x 25.2 x 10-6m/538 x 10-9 m) sin( 3.25 10-3 rad)

= 0.478 rad

(c) The intensity at a point is related to the intensity at the

central maximum by Eq.

2sin

mII

926.0

2

)478.0()478.0sin(

radrad

[6] Find the width 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. (Given a=5λ in fig)

Solution:

We have IѲ= Im(sinα/ α)2 and given IѲ =Im/2

(sinα/ α)2 =1/2

Put cal in radian mode and start from αx =1

Let us rewrite the equation as αx=√2 sin αx

αx = 1.39156 (ittertative)

αx =(πa/λ)sinѲx

Ѳx= sin-1(αx λ/ πa )

= sin-1(1.39/ 5π ) By considering a=5λ

=5.10

The width of the curve is then found from Δ Ѳ = 2 Ѳx=10.20

DIFFRACTION AT A CIRCULAR APERTURE

DIFFRACTION PATTERN DUE TO A CIRCULAR APERTURE

Light

a

Image is not a point, as expected from geometrical optics! Diffraction isresponsible for this image pattern.

Diffraction at a circular apertureIn focusing an image, a lens posses only the light that falls within its circularperimeter. Form this pint of view, a lens behaves like a circular aperture inan opaque screen, such an aperture forms a diffraction pattern analogous tothat of a single slit . Diffraction effects often limit the ability of telescopesand other instruments to form a precise images.Distant point source, e,g., star

lens

d

Image is not a point, as expected from geometrical optics! Diffraction isresponsible for this image pattern.

DIFFRACTION AT A CIRCULAR APERTURE

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. of diameter the is d whered

22.1sin

widthslit the is a wherea

sin

is ndiffractio slit single in minimum first for equation The

In case of circular aperture, the factor 1.22 arises when we divide the aperture into elementary Huygens sources and integrate over the aperture.

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

isdR

22.1

as dappoximate becan it small, very since22.1sin

R

R1

R is the smallest

angular separation for

which we can resolve

the images of two

objects.

a. Not resolved

b. Just resolved

c. Well resolved

Application:

We can resolve an object of smaller angular separation in

lens by

1. Increasing the lens diameter or

2. Using a shorter wavelength

Example: Telescope, microscope

The pointillistic painting. The Seine at Herblay by Maximilien Luceconsists of thousands of colored dots. With the viewer very close to thecanvas, the dots and their true colors are visible. At normal viewingdistances, the dots are irresolvable and thus blend.

[1] 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 = 550nm. (b) How far apart are the

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

Solution:

(a) We have

θR = 1.22 (λ/d)

= 1.22 ( 550 x 10-9/32 x10-3)

θR = 2.10 x 10-5 rad

(b) The linear seperation is ΔX=f θR

=0.24 x 2.10 x 10-5 rad

=5μm

[2] The wall of a large room is covered with acoustic tile in which small

holes are drilled 5.2mm from the center to the center. How far can a

person be from such a tile and still distinguish individual holes

assuming ideal conditions? Assume the diameter of the pupil of the

observer’s eye to be 4.6mm and the wavelength to be 542nm.

Solution : Given d = 4.6mm λ=542nm and y = 5.2mm D=?

Solution:

(a) We have

θR = 1.22 (λ/d) and θR=y/D

1.22 (λ/d)=y/D

D =y d / (1.22 λ)

=5.2 x10-3 x 4.6 x10-3 / (1.22 x 542x10-9)

D=36.2m

[3] If Superman really had x-ray vision at 0.12nm wavelength and a 4.3

mm pupil diameter, at what maximum altitude could he distinguish

villains from heroes, assuming the minimum detail required was 4.8cm ?

Solution : Given d = 4.3mm λ=0.12nm and y = 4.8cm D=?

Solution:

(a) We have

θR = 1.22 (λ/d) and θR=y/D

1.22 (λ/d)=y/D

D=y d/ (1.22 λ)

=4.8 x10-2 x 4.3 x10-3 / (1.22 x 0.12x10-9)

D=1400km

[4] 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 color. ( = 475nm, diameter of pupil = 4.4mm)Solution : Given d = 4.4mm and λ=475nm

(a) We have

θR = 1.22 (λ/d)

= 1.22 ( 475x 10-9/4.4 x10-3)

θR = 1.32 x 10-4 rad

The dots are 2 mm apart, so we want to stand a

distance D away such that

D>y/θR

= 2x 10-3 /1.32 10-4 rad

=15m

DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED

Fig. shows the double slit producing both interference and diffraction pattern

Interference

Diffraction

Interference + Diffraction

βcos 2INT m,INT , I I

2

αα sin DIF m,DIF ,I

2

2

ααsin cos m

I

(c) Two Single slits a~l

(a)Two vanishingly narrow slits a<<l

(b)Single slit a~l

In Young's double-slit expt, we assumed that the slit width a<<λ. What if this is not the case?

βcosα

αsinI 22

2

m

I

SINGLE-SLIT DIFFRACTION PATTERN

YOUNG’S DOUBLE-SLITINTERFERENCE

PATTERN

We know, Intensity (amplitude)2

Intensity of Interference pattern for two infinitesimally narrow slit is

Intensity for the diffracted wave from either slit is

sin,int

dwhereI m βcos 2intI

sin,2 awhereinI difm

s

diffI

Combined pattern leads to

Figure shows the geometry for the analysis of double slit interference

and diffraction combined. Each of the two slits is divided into N zones.

The net electric field at P is found by adding the N electric field vectors

(N phasors).

Expression for Intensity using phasor method:

Fig shows the first N phasors (corresponding to the upper slit) and

the resultant E1.There is phase difference of = /N between each of

the N phasors where is the phase difference between1st phasor and

Nth phasor.

Adding all the phasors, we get the resultant E1 due to the first slit. 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. E is the resultant of E1 and E2.

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

cos 22

sin2

sin A

Also

)(

22

2sin2 1

or

where

EE

1

2/2

sindiagram, From

EE

cos2

sin,),(

getweAeqninthisngSubstituti

sin)(2 ad

,sin

2aboveeqnofsidesbothto

aAdding

iswhichd sin

2

pathdiffPhasediff2

sin)(

2ad

From single slit diffraction, we have

the electric amplitude at P due to

one slit,

sin

1 mEE

SINGLE-SLIT DIFFRACTION PATTERN

DOUBLE-SLITINTERFERENCE PATTERN

22 sin)(cos

m

cossin)2(

mEE

2sin2 1

EE

Case 1: Interference Minima: Interference minima occurs

Case 2: Diffraction MinimaOccurs when α = ±π ,±2π ± 3π ---± pπ, where p is an integer.

21 msin

21 msin

;21 m---------

25,

23,

2

dord

ie

-)-------3,- 2, ,1(sin,sin., ppaorpaie

Case 3: Interference MaximaOccurs when β = o ± π, ± 2π, ± 3π, ------ + m π, ,

-)-------3,- 2, ,1,0(sin,sin., mmdormdie

Case 4: Missing ordersOccurs when the condition for a interference maximum and

that of a diffraction minimum are both fulfilled for the same

value of .

ie., d sin = m (m = 0,1, 2, 3,-----------)

a sin = p (p = 1, 2, 3, 4, -----------)

ie., d / a = m / p. This ratio determines the orders which are

missing.

If d /a = 2, orders 2, 4, 6, -------- are missing.

If d / a = 3, orders 3, 6, 9, -------- are missing.

d / a = 1, all orders are missing. (single slit)

....6,4,2.....3,2,1

2

2

mthenpIf

pmp

m

Thus the orders 2, 4, 6 etc of the interference maxima will be missing in the diffraction pattern.

Thus the orders 3, 6, 9 etc of the interference maxima will be missing in the diffraction pattern.

....9,6,3.....3,2,1

3

3

mthenpIf

pmp

mm=4p

M=5p

m=2p

m=3p

PROBLEMS

[1] 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?

(a)The spacing between adjacent fringes is given by

y = λD/d

=(480x10 -9) (52x10 -2)/(0.12x10-3)

=2.1mm

Solution:

Angular separation of the first minimum is

asinθ =mλ

sinθ =λ/a (m=1)

= 0.01920

Y = D tan θ = D sinθ =(52x10 -2)(0.0192)

Y = 10mm

d/a= 0.12mm/0.25mm=4.8i.e , 4+4+1=9 There are about complete 9 fringes in the central peak of the

diffraction envelope

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

minimum of the fringe envelope?

[2] What requirements must be met for the central maximum of the

envelope of the double-slit interference pattern to contain exactly 11

fringes? Solution:

The condition for the interference maximum is given by

dsinѲ=mλ--------------(1) where m=0,1,2,…………….

The condition for the diffraction minimum is given by

asinѲ=pλ --------------(2) where p=1,2,3,…………….

d/a= m/p

The condition is met if 6th Intf. maxima coincides with the 1st Diff minima.

For Intf. maximum m=6 and Diff. Minimum p=1 d/a=6

The 6th interference maximum is squelched by the diffraction minimum.

Then there are only 5 complete fringes on either side of the central

maximum. 5+5+1=11

[2] What requirements must be met for the central maximum of the

envelope of the double-slit interference pattern to contain exactly 11

fringes? Solution:

The condition for the interference minimum given by

dsinѲ= (m+1/2)/ λ----(1) where m=0,1,2,…………….

The condition for the diffraction minimum is given by

asinѲ=pλ --------------(2) where p=1,2,3,…………….

[1]/[2] d/a= (m+1/2)/p

The condition is met if 6th Intf. minimum coincides with the 1st Diff

minimum.

For sixth Intf. minimum m=5 and Diff. Minimum p=1

d/a=11/2=5.5

The slit separation d must be 11/2 times the slit width a. This condition

depends only on the ratio of d/ a and not at all on the wavelength.

[3] (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 = 557nm, d = 0.150mm, and a = 0.030mm? (b)What is the ratio

of the intensity of the third fringe to the side of the center to that of the

central fringe?

Solution:

(a) The condition for the interference maximum is given by

dsinѲ=mλ-----(1) where m=0,1,2,…………….

The condition for the diffraction minimum is given by

asinѲ=pλ --------------(2) where p=1,2,3,…………….

d/a= m/p Here p=1 for first minimum

m=d/a=(0.150)/(0.030) = 5.

The 5th interference maximum is squelched by the diffraction minimum.

Then there are only 4 complete fringes on either side of the central

maximum. 4+4+1=9

(b) We have IѲ= Im(sinα/ α)2(cos β)2

Where α=πasin Ѳ /λ and β =πdsin Ѳ /λ

For the third fringe m = 3,

dsinѲ=mλ dsinѲ=3λ (bright fringe)

Then β = πdsin Ѳ /λ= (π 3 λ) /λ

β = 3π

α=[πasin Ѳ] /λ

α =[ πa(3λ/d]/ λ since sinѲ=3λ/d

α =3 πa/d

So the relative intensity of the 3rd fringe is

IѲ= Im(sinα/ α)]2(cos β)2

I3 / Im= [sin (3 πa/d) / (3 πa/d)]2(cos 3 π)2

I3 / I m =0.255

Solution:

(a)The condition for the interference maximum is given by

dsinѲ=mλ-----(1) where m=0,1,2,…………….

The condition for the diffraction minimum is given by

asinѲ=pλ --------------(2) where p=1,2,3,…………….

d/a= m/p

d/a=4 d=4a

Hence, if d = 4a there will be no fourth interference maximum!

(b) Whenever m = 4p there will be a missing maximum.

i.e 4,8,12,16….. Interference maximum will be the missing

[4] Design a double slit system in which the fourth fringe, not

counting the central maximum, is missing. What other fringes , if

any are also missing?

MULTIPLE SLITS

Multiple slit arrangement

will be the interference

pattern multiplied by the

single slit diffraction

envelope. This assumes

that all the slits are

identical.

Condition for principal

maxima,

d sin = m

m=0,1,2,3,…

where d is the separation

between adjacent slits.

Location of principal

maxima is independent of

number of slits.

MULTIPLE SLITS

Intensity pattern for (a) Two-slit diffraction (b) Five-slit diffraction(diffraction effect is neglected)

MULTIPLE SLITS

Width of the maxima: Central maximum 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 of number the is N whereN2

Corresponding path difference is,

N2L

Expression for the width of the central maxima

Also we know,

N2L

Nd

Ndsin

sindN

sindL

0

0

0

0

From the equation, for given and d

if we increase number of slits (N),

then the angular width of principal

maximum decreases. ie the principal

maximum becomes sharper.

Expression for the width of the other principal maxima

For the mth principal

maximum at by a

grating:

d sin = m .

For the first minimum at

+ after the mth

principal maximum

Nλmλθθsind

MINIMUM AT θ +θ

mth PRINCIPAL MAXIMUM AT θ

MINIMUM AT θ +θ

mth PRINCIPAL MAXIMUM AT θ

Nλmλθθsind

Nmd

sin coscos sin

1

Nmcos dsin d

Nmcos dm

cos d N

The principal maximum become sharper as number of slits (N) increases

ANGULAR HALF WIDTH OF mTH

PRINCIPAL MAXIMUM AT

Effect of increasing the number of slits (N) in Fraunhoferdiffraction (Diffraction grating):

PROBLEMS

[1] A certain grating has 104 slits with a spacing of d = 2100 nm. It

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

(a) the angular position of all principal maxima observed and

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

(a) Solution

d sin = m

sin = m / d

m=1 =16.30

m=2 =34.10

m=3 =57.30

For m=4 sin>1. Thus m=3 is the highest order observed, which gives

a total of 7 principal maxima. A central maxima and 3 on either side.

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

.)3.57)(cos2100)(10(

589cos

,max3)(

04 nmnm

Nd

imummtheForb

imumprincipalnarrowyexceedinglanisThisradx

max0030.0102.5 05

[2]Light of wavelength 600 nm is incident normally on a diffraction

grating. Two adjacent principal maxima occur at sin =0.20 and sin= 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).(a) SOLUTION

The principle maxima occur at points given by

d sinm = m

sinm = m / d

The difference of the sine of the angle between any two

adjacent orders is

sinm+1 - sinm = (m+1) / d - m / d= / d

sinm+1 - sinm =/ d

d= / (sinm+1 - sinm)

=600nm(0.30-0.20)

d=6μm

(b) If the fourth order maxima is missing it must be because the

diffraction pattern envelope has a minimum at that point.

dsinm = m sinm = m /d

sin4 = (4 x600nm) /6μm

sin4 = 0.4

We have diffraction minimum given by

a sinm = m a= m / sinm =m(600nm)/0.4

The minimum width is when m = 1 i.e ., a = 1.5 μm.

(c) The visible orders would be integer values of m except for when m

is a multiple of four.

[3] With light from a gaseous discharge tube incident normally on

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

green line appears with sharp maxima at measured transmission

angles = 17.6, 37.3, -37.1, 65.2 and -65.0. Compute

wavelength of the green line that best fits the data.

Solution:

We want to find a relationship between the angle and the order

number which is linear. We'll plot the data in this representation,

and then use a least squares fit to find the wavelength. The data to

be plotted is

We have

dsinѲm=mλ

sinѲm= (λ/d) m +c

Y= (slope).x+Intercept

Therefore

λ/d= Slope

λ/d=0.302

λ = (0.302)(1.73μ m)

λ = 522 nm

Y= 0.302X

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-4 -3 -2 -1 0 1 2 3 4

m θ Sin θ

-3 -65.0° -0.906

-2 -37.1° -0.603

-1 -17.6° -0.302

1 17.6° 0.302

2 37.3° 0.606

3 65.2° 0.908

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.

Diffraction grating spectrometer

Intensity versus sin for a diffraction grating

Any regular periodic structure can

be serve as a diffraction grating.

Diffraction grating spectrometer

set up is as shown in the figure.

The entire spectrum can be

viewed by rotating telescope

through various angles. In

general, gratings may produce

several images of spectral lines,

corresponding to

m = ±1, ±2, ±3…. in

dsinѲ = mλ

Sample spectra of visible light emitted by a gaseous source

m = 0 m = 1 m = 2 m = 3

DISPERSION AND RESOLVING POWER

linesspectral ofwavelengthbetweenDifferencelinesspectralbetweenseparationAngularDispersion

ΔλΔθD

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 (2) Resolving power

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.

Dispersion

Dispersion

d sin = m

Differentiating the above equation,

θcos dm

ΔλΔθ D

To achieve higher dispersion we must use a grating of smaller

grating spacing and work in higher order m .

ΔλΔθD

d cos = m

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 1 and 2, just

resolved by the grating, the resolving power is defined as

R21 2

21

RESOLVING POWER

θcos dm

ΔλΔθ D

cos dN

cosdmcosdN

mNR

Resolving power increases with increasing N

We have,

Putting second equation in first equation,

Resolving powerDISPERSION AND RESOLVING POWER

Intensity patterns of two close lines due to three gratings A, B, C.

(A)N = 5,000d = 10,000 nmR = 5,000D = 1.0x 10-4 rad/nm

(B)N = 5,000d = 5000nmR = 5,000D = 2.0 x 10-4 rad/nm

(C)N = 10,000d = 10,000 nmR = 10,000D = 1.0 x 10-4 rad/nm

DISPERSION AND RESOLVING POWER

Fig a: shows the maximum of one

line falls on the minimum of the

other using Grating A.

Fig b: Grating B has twice the

dispersion of A but same the

Resolving power.

Fig C: Grating C has twice the

Resolving power of A but same

Dispersion

The width of the grating is W=Nd

A=50mm B=25mm C=100mm(A)N = 5,000d = 10,000 nmR = 5,000

(B)N = 5,000d = 5000nmR = 5,000

(C)N = 10,000d = 10,000 nmR = 10,000

Intensity patterns of two closelines due to three gratings A, B, C.

cos dN

GRATING

[1] 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?

A) SOLUTION

Spacing b/n slits, d =W/N= (25x10-3m) /104

Wavelengths, λ1=589.0nm & λ2= 589.59nm,

Grating equation, d sin = mλ

= sin-1 (mλ1/d) Here m =1

=13.6 degrees

B. Dispersion relation, d = (mλ)/ d cos

=2.4 x 10-4 radians

or 0.014 degrees.

As the spectral separation increases with the order no. d value

increases with the order no.

[2] Given a grating with 400 rulings/mm, how many orders of the entire

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

SOLUTION:

:mm 10 2.5 mm 4001 d 3-

1-

3.57 m)10(700x m)10 x (2.5d m 9-

-6

To 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.

d sin = m

Using maximum angle 900

so there can be at most three orders of the entire spectrum.

[3] White light (400 nm < < 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.

Multiply wavelengths by the appropriate order

2(700nm) > 3(400nm)

Divide both side by d,

2(700nm)/d > 3(400nm)/d

We have d sin =m λ sin = =(m λ)/d

sin 2,λ=700 > sin 3,λ=400

Regardless of the value of d.

[4] 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 ( = 546nm)?

b) What is the resolving power of this grating in the fifth order?

nmcmNwd 3125

96003

06.31cos 31253

cos xnmdmDDispersion

nmnmradx /646.0/1013.1 03

SOLUTION (A)

011 6.313125

5463sinsin

nmnmx

dm

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

order?

Resolving power, R=Nm

R=(9600)(5)

R=4.80 x 104

Thus, near λ=546nm and in 5th order, a wavelength difference

given by

Δλ=λ/R

=546nm/4.80 x 104

Δλ=0.011nm can be resolved

[5] 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?

Nwd,separationslit)a( nm

xd 2083

102.11025

4

3

dmsinmsindMaxima 1

9

91

1 102083105891 )(sin

4253161 .:Answer

For the first order, m =1 and for the first of the wavelengths, 1 =589.0 nm

(a) At what angle does the first maximum occur for the first of these

wavelengths?

4253161 .:Answer

9

91

2 10208310595891 ).(sin&

4422162 .

)Answer(. 0169012

dmsinmsindMaxima,Again)b( 1

cosd)]([m

cosdm,elyAlternativ)b( 12

(b) What is the angular separation between these two lines (1st

order)?

mNR,powersolvingRe)c(

RR,Also

44 1021110211 .).(mNR,m,orderfirsttheFor

41021 .NmR,But

4

9

102.110589

But for sodium D lines, 2 – 1 = (589.59 – 589) nm = 0.59 nm. Therefore

the given grating can resolve the sodium D lines.

nm.04910

(c) How close in wavelength can two lines be (in first order) and still be

resolved by this grating?

Nm

mmNR

9

9

1059.010589

998: NAnswer

For just resolve sodium doublet lines in the first order,

As the grating has about 12 times [(1.2 104)/999 = 12] as many

rulings as this, it can easily resolve sodium lines.

(d) How many rulings can a grating have and just resolve the

sodium doublet line?

[6] The sodium doublet in the spectrum of sodium is s 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.

982nm) (589.0- nm) (589.6

nm) (589.0 R

bygiven is grating theofpower resolving required The:Solution

491(2)

(982)mR N

so maxima,order second at the looking are We982 R then ispower resolvingOur

[7] 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.

m 9.98 )sin(10.2

nm) (3)(589sin

m d 0m

328; (3)

(982)mR N

982 Rleast at be toneeds grating theofpower resolving The (b)

SOLUTION : (a)We have d sin = m d= m/ sin

The width of the grating W=dN=9.98 μm X 328

W=3.273 mm

DiffractionStructural coloring occurs in the

facial skin of mandrill baboons

(and the skin on some other

animals) because of diffraction

from parallel collagen fibers

beneath the skin surface. Light

penetrates the skin, diffracts from

the flbers, and then emerges from

the skin. The fiber separation

results in constructive interference

for wavelengths around 460 nm,

so the skin is blue.

1895 : Wihelm ConradRoentgen discovery of X-Rayswhile experimenting withdischarge tubes

X-unknown radiation or X – rays

When a beam of fast moving

electron is strikes on solid

target an invisible and high

penetrating radiation is

produced. These radiations are

called X – rays.

X-RAYS

• X-rays are electromagnetic radiations

of very short wavelength ranging from

0.1 Å (0.01 nm) to 100 Å (10 nm).

• Soft x-rays - possessing lower

penetration power (low applied voltage,

higher wavelengths) – medical

imaging.

• Hard x-rays - possessing higher

penetration power (high applied

voltage, lower wavelengths) – materials

crystal structure analysis

Hard X-rays: possessing higher penetration power (high applied voltage, lowerwavelengths) – materials crystal structure analysis. (X-ray diffraction experiment, X-ray scattering experiment, X-ray absorption experiment etc.)

Hard X-rays and uses

Soft X-rays: possessing lower penetration power (low applied voltage, higherwavelengths) – medical imaging.

Soft X-rays and uses

Typical medical X-ray tube

X-rays are electromagnetic radiation with wavelength ~1Å= 10-10m

(visible light ~5.5x10-7 m).

X-Ray Diffraction

X-ray generation

X-ray wavelengths too short to be

resolved by a standard optical

grating.

This is too close to the central

maximum to be practical. A grating

with d=λ is desirable, but, because x-

ray wavelengths are about equal to

atomic diameters, such gratings

cannot be constructed mechanically.

1 1 1 0.1 nmsin sin 0.0019

3000 nmmd

X-RAY DIFFRACTIONFor the observation of diffraction phenomenon by grating, the

grating space should have the dimension of the wavelength of

the wave diffracted. 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 producing Laue’s pattern X-ray tube

When a mono-energetic x-ray beam isincident on a sample of a single crystal,diffraction occurs resulting in a patternconsisting of an array of symmetricallyarranged diffraction spots, called Laue’sspots.

The single crystal acts like a grating witha grating constant comparable with thewavelength of x-rays, making thediffraction pattern distinctly visible.

Since the diffraction pattern is decidedby the crystal structure, the study of thediffraction pattern helps in the analysisof the crystal parameters.

A Laue pattern of asingle crystal. Eachdot represents apoint of constructiveinterference.

X-RAY DIFFRACTION

NaCl crystal (a0 = 0.563nm)

A plane through a crystal of NaCl

NaCl unit cell

Fig. 36-29

Interplanar spacing d is related to the unit cell dimension a0:

X-Ray Diffraction, cont’d

2 050 045 or 0.2236

20ad a d a

Not only can crystals be used to separate different x-ray wavelengths, but x-rays in turn can be used to study crystals, for example, to determine the type of crystal ordering and a0.

(36-22)

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.

Bragg’s Law

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.

X-RAY DIFFRACTION:BRAGG’S LAW

Glancing angle. ie angle

between the incident x-ray

beam and the reflecting

crystal planes.

For constructive

interference of diffracted x-

rays the path difference for

the rays from the adjacent

planes, (abc in the figure)

must be an integral number

of wavelength.

ie 2d sin = m

PROBLEMS

[1]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)

SOLUTION:

The inter planar d spacing for these planes is given by

5oad nmnm 252.0

5563.0

)252.0(2)110.0(sin

2sin 11

nmnmm

dm

2d sin = m

m=1 =12.60 m=2 =25.90

m=3 =40.90 m=4 =60.80

Higher order beams cant exist because they require sin >1

[2] 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.

SOLUTION: m sin 2d

09-

12-1- 2.68

m) 10 2(0.313m) 10 x (1)(29.3 sin

is angle minimum Then the

nm. 0.313 d and 1 m meansThat

We are looking for the smallest angle this will correspond to the

largest d and the smallest m.

[3] 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.

SOLUTION

m104.04x )sin(18.8 2

m)10 x (1)(0.261sin 2m d 10-

0

-9

d2 a 0

0.572nm

m) 10(4.04x 2 -10

02

022 a a (2d)

450

TUTORIALExerciseE42-1, E42-16, E42-19, E42-26, E42-29

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]

QUESTIONS – DIFFRACTION

Explain Rayleigh’s criterion for resolving images due to a circular apperture. [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]

QUESTIONS – DIFFRACTION

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 nmE42-11: 0.186°, 0.478 radian, 0.926E42-16: 36.2 mE42-19: 1400 kmE42-21: 15 mE42-26: (a) d =4a (b) Every 4th fringeE42-29: (a) 9 (b) 0.255E43-3: 523 nmE43-5: (a) 6 µm (b) 1.5 µm (c) m = 0, 1, 2, 3, 5, 6, 7, 9E43-9: 3E43-17: 491E43-21: (a) 9.98 µm (b) 3.27 nmE43-25: 2.68 degreeE43-33: 0.206 nm