10/13/2022
B.TECH FIRST YEAR
ACADEMIC YEAR: 2020-2021
SESSION OUTCOME
COURSE NAME: ENGINEERING PHYSICS
COURSE CODE
“UNDERSTAND THE BASIC
PRINCIPLES OF WAVE OPTICS”
: PY1001
LECTURE SERIES NO : 01 (ONE)
CREDITS
:
4
MODE OF DELIVERY : ONLINE (POWER POINT PRESENTATION)
FACULTY
:
DR. NILANJAN HALDER
EMAIL-ID
:
nilanjan.halder@jaipur.manipal.edu
DATE OF DELIVERY: 10 October 2021
1
2
Diffraction
Topics
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ASSIGNMENT
QUIZ
MID TERM EXAMINATION –I
END TERM EXAMINATION
ASSESSMENT CRITERIA’S
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3
4
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
10/13/2022
DIFFRACTION AND INTERFERENCE
Diffraction v/s Interference
Both involve superposition of coherent light waves.
Diffraction and interference are similar phenomena.
Interference is the effect of
superposition of 2 coherent waves.
Here Slit width a<<λ very small and
neglected. So, fringes are of equal
width and intensity on the screen is of
uniform distribution.
 Bending of light around the obstacle.
 The interfering beam originate from
continuous distribution of sources
(Huygens’ principle).
 The waves emerging from different
paths of the same wave front
superimpose with each other to
produce Diffraction pattern.
 The width of the diffraction fringes are
not equal.
 Minimum intensity point will not be
perfectly dark
 Bright fringes in the diffraction pattern
are not of same intensity.
Diffraction is the superposition
of many coherent waves. Slit
width ‘a’ is finite. Intensity on the
screen is non-uniform.
5
6
DIFFRACTION AND WAVE THEORY OF LIGHT
DIFFRACTION AND WAVE THEORY OF LIGHT
•
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 pattern of razor blade viewed in
monochromatic light
7
 Meeting of two waves.
 The interfering beam originate from
discrete number of sources.
 The superposition of waves coming
from two different wave front
originating from the same source,
produce Interference pattern.
 The width of the interference fringes
may/ may not be equal.
 Minimum intensity point will be
perfectly dark.
 Bright fringes in the interference
pattern are of uniform intensity.
8
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.
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Huygens’ Principle
DIFFRACTION AND WAVE THEORY OF LIGHT
Ꙫ A single slit placed between a distant light source and a
screen produces a diffraction pattern.
Ꙫ It will have a broad, intense central band called the
central maximum
Ꙫ The central band will be flanked by a series of narrower,
less intense secondary bands called side maxima or
secondary maxima
Ꙫ The central band will also be bordered by a series of
dark bands called minima.
Ꙫ The diffraction pattern consists of the central maximum
and a series of secondary maxima and minima.
Ꙫ The pattern is similar to an interference pattern as
shown in figure.
9
• Every point on a propagating
wavefront serves as the source
of spherical wavelets, such that
the wavelets at sometime later
is the envelope of these
wavelets.
• If a propagating wave has a
particular frequency and speed,
the secondary wavelets have
that same frequency and
speed.
The phenomenon of diffraction is caused by the interference of
innumerable secondary wavelets that are produced by
unobstructed portions of the same wave front or from the portions
of the wave front which are allowed to pass through a aperture.
10
Fresnel diffraction & Fraunhofer diffraction
Diffraction patterns are usually classified into two
categories depending on the source and screen are placed.
DIFFRACTION AND WAVE THEORY OF LIGHT
The pattern formed on the screen depends on the
separation between the screen C and the aperture B. Let us
consider the following three cases.
Case 1: Very small separation: when screen C is very close to B.
Fresnel diffraction: When either the source or the screen is
near the aperture or obstruction, the wavefronts are
spherical and the pattern is quite complex. (near-field)
Aperture
From
distant
source
Fraunhofer diffraction: When both the source and the screen
are at a great distance from the aperture or obstruction,
the incident light is in the form of plane wave and
the pattern is simpler to analyze. (far-field)
11
“Isotropic”
Screen
Geometrical
shadow of the
aperture
B
C
The waves travel only a short distance after leaving the
aperture, and the rays diverge very little. The effects
diffraction are negligible, and the pattern on the screen is
the geometric shadow of the aperture.
12
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Case 2: Very large separation: When screen C is far from
the aperture (Fraunhofer diffraction).
Plane wave front
Plane wave front
f
f
Very large separation
C
Very large separation
When the screen is so far from the aperture, then we can
regard the rays as parallel or wavefronts as planes.
In this case , we also assume the source to be far from the
aperture, so that the incident wavefronts are also planes.
This is one way of achieving Fraunhofer diffraction.
13
In the laboratory, two converging lenses are used to achieve
this condition.
The first lens converts the diverging light from the source in
to a plane wave, and the second lens focuses plane waves
leaving the aperture parallel to the point on screen.
14
Case 3: Intermediate separation: When screen C and
source (S) are at finite distances from the aperture (Fresnel
diffraction):
Plane wave front
Plane wave front
Spherical
wave front
Spherical
wave front
P
Source (S)
Very large separation
Finite distance
B
Finite distance
C
In Fresnel diffraction:
1. the incident and the diffracted wave fronts are
spherical.
2. The source and the screen are at finite distances from
the aperture/slit or obstacle causing diffraction.
3. No lenses/ mirrors are used.
15
Very large separation
C
In Fraunhofer type of diffraction:
1. Both the source and the screen are effectively at infinite
distances, from the aperture causing diffraction.
2. both the incident and emergent wavefronts are plane. That is,
both the incident and the diffracted beams are parallel.
3. can be realized in practice by using a pair of converging lenses
of suitable focal lengths (L1 and L2) and placing the source
and the screen at the foci of L1 and L2 respectively.
16
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SINGLE-SLIT DIFFRACTION
Fraunhofer diffraction is a special (limiting) case of
the more general Fresnel diffraction.
But, analysis of Fresnel diffraction is complicated
compared to Fraunhofer. That is, Fraunhofer
diffraction is easier to handle mathematically.
So, in our study on diffraction phenomenon, we
deal only with Fraunhofer diffraction.
17
All the diffracted rays arriving at P0 are in-phase.
Hence they interfere constructively and produce maximum
(central maximum) of intensity I0 at P0.
18
SINGLE-SLIT DIFFRACTION
SINGLE-SLIT DIFFRACTION
At point P1,
path difference between r1
and r2 is
(a/2) sin
⁕The finite width of slits is the basis for understanding Fraunhofer diffraction.
⁕According to Huygens’s principle, each portion of the slit acts as a source of light waves.
⁕Therefore, light from one portion of the slit can interfere with light from another portion.
⁕The diffraction pattern is actually an interference pattern.
⁕The different sources of light are different portions of the single slit.
19
So the condition for first minimum,
a

sin  
2
2
or
a sin   
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.
20
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SINGLE-SLIT DIFFRACTION
SINGLE-SLIT DIFFRACTION
Problem: 1
At point P2,
path difference between
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?
r1 and r2 is (a/4) sin
So the condition for second minimum,
a

sin  
or
a sin   2 
4
2
This is satisfied for every pair of rays, separated by a distance a/4.
In general, the condition for m TH minima,
a sin   m 
m   1,  2,  3, . . .
There is a secondary maximum approximately half way between
each adjacent pair of minima.
21
22
SINGLE-SLIT DIFFRACTION
SINGLE-SLIT DIFFRACTION
Problem: E42-5
Problem: 2
In P-1, 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?
23
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?
24
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INTENSITY IN SINGLE – SLIT DIFFRACTION
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.
Phasor showing
• The phase difference between two waves arriving at point P
a) Central maximum
from two points on the slit (with separation x) is,
 
b) A direction slightly shifted
2
 x sin 

from central maximum
c) First minimum
d) First maximum beyond the
central maximum
(corresponds to N = 18)
25
26
INTENSITY IN SINGLE – SLIT DIFFRACTION
From diagram,
E   2 R sin
 is the phase difference

2
Em
Also  
R
Combining,
E

E   m sin

2
2
sin 
Or , E   E m
where  
27

INTENSITY IN SINGLE – SLIT DIFFRACTION
between rays from the top
and bottom of the slit.
So we can write,
2
a sin 

 a
So,   
sin 
2



2
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10/13/2022
INTENSITY IN SINGLE – SLIT DIFFRACTION
The intensity    E 2
 sin  
E m2 

  

INTENSITY IN SINGLE – SLIT DIFFRACTION
2
2
 sin  
2
m 
 where  m  E m is the max. intensity
  
From the above eqn., for minima, sin   0
   m  where m   1,  2,  3,.....
 
or, a sin   m  where m   1,  2,  3,.....
The intensity distribution in
single-slit diffraction for three
different values of the ratio a/
29
30
Position of dark fringes in single-slit diffraction
sin  
Width of central maximum
•We can define the width of the central maximum to be the distance
between the m = +1 minimum and the m=-1 minimum:
m
a
y 
If, like the Young’s 2-slit treatment we assume small angles, sin ≈ tan  =ymin/D, then
ymin 
Dm
a
Positions of intensity
MINIMA of diffraction
pattern on screen,
measured from central
position.
the narrower the slit,
the more the diffraction
pattern “spreads out”
a sin θ = λ first minima
Very similar to expression derived for 2-slit experiment:
m
ym  D
d
31
D
D 2 D


a
a
a
Intensity
distribution
But remember, in this case ym are positions of MAXIMA
In interference pattern
image of diffraction
pattern
If we narrow the slit the
angle must get bigger more flaring
- what happens when a
= λ?
32
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INTENSITY IN SINGLE – SLIT DIFFRACTION
INTENSITY IN SINGLE – SLIT DIFFRACTION
Problem: SP42-3
Problem: SP42-4
Calculate, approximately, the relative intensities of the
maxima in the single slit Fraunhofer diffraction pattern.
33
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.
34
INTENSITY IN SINGLE – SLIT DIFFRACTION
DIFFRACTION AT A CIRCULAR APERTURE
Problem: E42-11
Monochromatic light with wavelength 538 nm falls on a slit
with width 25.2m. 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)  (c) ratio of the
intensity at this point to the intensity at the central
maximum.
DIFFRACTION PATTERN DUE TO A CIRCULAR APERTURE
35
36
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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 sin   1 . 22
d
where d is the diameter of aperture.
The equation for first minimum in single slit diffraction is

sin  
a
where a is the slit width
DIFFRACTION AT A CIRCULAR APERTURE
Raleigh’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.


R  sin11.22 
d

since R is very small, it can be appoximated as
R  1.22

d
separation for which we
can resolve the images of
two objects.
a. Not resolved
In case of circular aperture, the factor 1.22 arises when we divide the
aperture into elementary Huygens sources and integrate over the
aperture.
37
R is the smallest angular
b. Just resolved
c. Well resolved
38
DIFFRACTION AT A CIRCULAR APERTURE
Using a circular instrument (telescope, human eye),
when can we just resolve two distant objects?
Problem: SP42-5



d

 R is very small, since the sources are distant and closely spaced
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?
 R  sin 1 1.22
R is the smallest angular separation
for which we can resolve the images of two objects.
39
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DOUBLE-SLIT INTERFERENCE AND
DIFFRACTION
1.Young’s double-slit experiment is an idealized
situation in which slits were assumed to be very narrow
(a << λ).
2.This cannot occur with actual slits because the
condition a << λ cannot usually be met.
41
In case of double slit experiment:
Each of the two slit has a finite width and
the light is diffracted through it in a single slit diffraction pattern
Usually, the width of each of the slits (a) is quite a bit less than the
separation (d) between their centers
Two single slit diffraction pattern is to be superposed
to get the final intensity
The final result can be obtained by drawing a two slit interference
pattern inside the envelope of a single slit diffraction pattern
42
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
43
44
10/13/2022
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
Interference
I  , INT  I m,
INT
Diffraction
I  , DIF 
 m, DIF
cos
m
cos 2
β
 sin α 
 α 


Interference + Diffraction
I 
2
 sin α


 α 

2
2
45
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
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  =
/N between each of the N phasors where  is the phase
difference between1st phasor and Nth phasor.
46
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
From the figure ,
E   2E 1 sin
where
or
47


 
2
2
    (  )

 

 sin  
  cos 
 .........( A )
2
2 
2
 2 
 
and
 ( d  a ) sin 
2 
 a
Adding

sin  to both sides of above eqn , we get ,
2

 
 d sin  which is 
2

Also
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
48
sin
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DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
Substituti ng this in eqn( A ), we get,

sin  cos 
2
From sin gle  slit diffractio n, we have,
Problem: SP42- 6
Ina double slit experiment, the distance D of the screen
the electric amplitude at P due to one slit,
 sin  
E1  Em 

  

 E   2E1 sin
2
from the slits is 52cm, the wavelength is 480nm, slit
separation d is 0.12mm and the slit width a is 0.025mm.
 sin  
ie, E   ( 2Em )
 cos 
  
 sin  
     m (cos )2 

  
a) What is the spacing between adjacent fringes?
b) What is the distance from the central maximum to the
2
DOUBLE-SLIT
INTERFERENCE PATTERN
first minimum of the fringe envelope?
SINGLE-SLIT DIFFRACTION
PATTERN
49
50
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
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?
51
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53
54
MULTIPLE SLITS: Width of the Central maximum
Interference pattern from 3 slits
55
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For N slits, intensity of the primary maxima is N 2 times greater than that due to a
single slit.
57
58
MULTIPLE SLITS
Multiple slit arrangement
will be the interference
pattern multiplied by the
single slit diffraction
envelope. This assumes
that all the slits are
For N slits, intensity of the primary maxima is N 2 times greater than that due to a
single slit.
As the number of slits increases, the primary maxima increase in intensity and become
narrower, while the secondary maxima decrease in intensity
identical.
As number of slits increases, number of secondary maxima also increases. In fact,
the number of secondary maxima is always N - 2, where N is the number of slits.
59
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MULTIPLE SLITS
MULTIPLE SLITS
Condition for principal
maxima,
d sin  = m 
where d is the
separation between
Intensity pattern for
(a) Two-slit diffraction
(b) Five-slit diffraction
(diffraction effect is
neglected)
adjacent slits.
Location of principal
maxima is independent
of number of slits.
61
62
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,
2
 
where N is the number of slits
N
MULTIPLE SLITS
Width of the maxima: Central maximum

  
L  
 
2

N


 Also we know,
 L  d sin  0

 d sin  0
N

sin  0 
Nd

 0 
Nd
 Corresponding path difference is,

  
L  
 
N
 2 
63
64
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.
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MULTIPLE SLITS
MULTIPLE SLITS
Width of the maxima: Other principal maxima
Width of the maxima: Other principal maxima
λ
d sinθ  θ  mλ 
N



d sin  cos
  cos  sin
  m 


N

 
1

For the mth principal
maximum at  by a
grating: d sin = m .
For the first minimum
at  +  after the mth
principal maximum
d sinθ  θ 
d
sin
  d cos   

m   d cos    

 
N d cos 
λ
mλ 
N


ANGULAR HALF WIDTH OF mTH
PRINCIPAL MAXIMUM AT 
mth PRINCIPAL
MAXIMUM AT θ
MINIMUM AT θ
+θ
mth PRINCIPAL
MAXIMUM AT θ
66
MULTIPLE SLITS
MULTIPLE SLITS
Problem: SP43- 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.
67
m   N
The principal maximum become sharper as
number of slits (N) increases
MINIMUM AT θ
+θ
65
m   N
Problem: E43-5
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).
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Grating contains : greater
number of slits, or rulings,
as many as several 1000
per millimeter
69
70
Diffraction Gratings
DIFFRACTION GRATINGS
Grating contains : greater number of slits,
or rulings, as many as several 1000 per
millimeter
The diffraction grating, a useful device for analyzing light sources,
consists of a large number of equally spaced parallel slits.
Light passed through the grating forms
narrow interference fringes that can be
analyzed to determine the wavelength
 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.
As the number of rulings increases beyond 2
the intensity plot changes from that of a
double slit pattern to one with very narrow
maxima (called lines) surrounded by
relatively wide dark regions
71
 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.
72
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the condition for
interference maxima at a
specific angle () is
We can use this expression
to calculate the wavelength
if we know the grating
spacing d and the angle ().
If the incident radiation contains several
wavelengths, the mth-order maximum
for each wavelength occurs at
a specific angle.
All wavelengths are seen at  = 0,
corresponding to m = 0, the
zeroth-order maximum.
The first-order maximum (m = 1) is
observed at an angle that satisfies
the relationship sin  = λ/d.
The second-order maximum (m =2) is
observed at a larger angle , and so on.
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DIFFRACTION GRATINGS
Grating spectrometer
m=0
m=1
m=2
m=3
Sample spectra of visible light emitted by a gaseous source
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DIFFRACTION GRATINGS
Problem: SP43-2
A diffraction grating has 1.20 x 104 rulings uniformly
spaced over W= 2.50cm. 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? (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 lines?
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DIFFRACTION GRATINGS
DIFFRACTION GRATINGS
Problem: E43-9
Problem: E43-11
Given a grating with 400 rulings/mm, how many orders of
the entire visible spectrum (400-700nm) can be produced?
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.
A grating has 315 rulings / mm. For what wavelengths in the
visible spectrum can fifth-order diffraction be observed?
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DISPERSION AND RESOLVING POWER
The ability of a grating to produce spectra that permit precise
measurement of wavelengths is determined by two intrinsic
DISPERSION
A grating with high dispersive power must widely spread apart the
diffraction lines associated with nearly equal wavelengths.
Dispersion 
properties of the grating,
Angular separation between spectral lines
Difference between wavelength of spectral lines
(1) Dispersion: High dispersive powers refers to the wide separation
of the spectral lines
D 
(2) Resolving power: Ability of the instrument to show the
close spectral lines as the separate ones
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DISPERSION AND RESOLVING POWER
Dispersion
D 
Δθ
Δλ
d sin = m 
Differentiating the above equation,
d cos   = m 
D
Δθ
Δλ

m
d cos θ
To achieve higher dispersion we must use a grating of smaller
grating spacing and work in higher order m .
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Δθ
Δλ
RESOLVING POWER
Ability of grating to resolve two nearby spectral lines so that
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

    1   2   1   2
R
2

The limit of resolution is determined by
the Rayleigh criterion
the minimum wavelength separation we
can resolve min  2-1 occurs when
the maximum of 2 overlaps with the
first diffraction minimum of 1.
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DISPERSION AND RESOLVING POWER
DISPERSION AND RESOLVING POWER
Resolving power
We have,
Resolving power
D
Δθ
Δλ

m
d cos θ
 
N = 5,000
d = 10 m
R = 5,000
D = 1.0 x 10-4 rad/m

N d cos 
Putting second equation in first equation,





 N d cos  


N = 5,000
d = 5 m
R = 5,000
D = 2.0 x 10-4 rad/m
m
d cos 
N = 10,000
d = 10 m
R = 10,000
D = 1.0 x 10-4 rad/m

R
 Nm

Resolving power increases with increasing N
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DISPERSION AND RESOLVING POWER
DISPERSION AND RESOLVING POWER
Problem: SP43-3
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?
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Intensity patterns of two close
lines due to three gratings A, B, C.
Problem: SP43-4
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?
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DISPERSION AND RESOLVING POWER
DISPERSION AND RESOLVING POWER
Problem: E43-17
Problem: E43-21
The sodium doublet in the spectrum of sodium is s pair of
In a particular grating, the sodium doublet is viewed in
lines with wavelengths 589.0 and 589.6 nm. Calculate the
third order at 10.2 to the normal and is barely resolved.
minimum number of rulings in a grating needed to resolve
Find (a) the ruling spacing and (b) the total width of
this doublet in the second-order spectrum.
grating.
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X-RAY DIFFRACTION
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. Since the x-ray wavelength and the interplanar 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
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X-ray tube
 When a monoenergetic 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.
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A Laue pattern of a
single crystal.
Each dot
represents a
point of
constructive
interference.
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X-RAY DIFFRACTION
X-RAY DIFFRACTION
A plane through a crystal of NaCl
NaCl crystal (a0 = 0.563nm)
NaCl unit cell
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(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.
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X-RAY DIFFRACTION
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.
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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  = n 
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X-RAY DIFFRACTION
X-RAY DIFFRACTION
Problem: SP43-5
Problem: E43-25
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)
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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.
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QUESTIONS – DIFFRACTION
QUESTIONS – DIFFRACTION
Discuss the diffraction due to single-slit. Obtain the
locations of the minima and maxima qualitatively.
Explain Rayleigh’s criterion for resolving images due to a
circular apperture.
Obtain an expression for the intensity in single-slit
diffraction pattern, using phasor-diagram.
Obtain an expression for the intensity in double-slit
diffraction pattern, using phasor-diagram.
Calculate, approximately, the relative intensities of the first
three secondary maxima in the single-slit diffraction
pattern.
Discuss qualitatively the diffraction due to multiple slits
(eg, 5 slits).
Discuss qualitatively diffraction at a circular aperture.
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Obtain an expression for the width of the central
maximum in diffraction pattern due to multiple slits.
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QUESTIONS – DIFFRACTION
Obtain an expression for the width of a principal
maximum at an angle in diffraction pattern due to
multiple slits.
Obtain an expression for dispersion by a diffraction
grating.
Obtain an expression for resolving power of a diffraction
grating.
Discuss Bragg’s law for X-ray diffraction.
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