Diffraction Modified Aug2011
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Transcript of 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.
DIFFRACTION AND WAVE THEORY OF LIGHT
MIT- MANIPAL
8
DIFFRACTION AND WAVE THEORY OF LIGHT
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
9BE-PHYSICS-DIFFRACTION-2010-11
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.
DIFFRACTION AND WAVE THEORY OF LIGHT
MIT- MANIPAL
SINGLE-SLIT DIFFRACTION
10BE-PHYSICS-DIFFRACTION-2010-11
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
SINGLE-SLIT DIFFRACTION
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
SINGLE-SLIT DIFFRACTION
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
SINGLE-SLIT DIFFRACTION
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
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 14
SINGLE-SLIT DIFFRACTION
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.
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 15
SINGLE-SLIT DIFFRACTION
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
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 16
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
INTENSITY IN SINGLE – SLIT DIFFRACTION
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)
BE-PHYSICS-DIFFRACTION-2010-11MIT- MANIPAL
Some Phasor representations
19
INTENSITY IN SINGLE – SLIT DIFFRACTION
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
INTENSITY IN SINGLE – SLIT DIFFRACTION
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
INTENSITY IN SINGLE – SLIT DIFFRACTION
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
==
=
BE-PHYSICS-DIFFRACTION-2010-11MIT- MANIPAL
Since α = mπ =
∴∴∴∴ a sin θθθθ = mλλλλ
θλπ
sina
22
INTENSITY IN SINGLE – SLIT DIFFRACTION
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
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 23
Problem: SP42-3
Calculate, approximately, the relative intensities of the
maxima in the single slit Fraunhofer diffraction pattern.
INTENSITY IN SINGLE – SLIT DIFFRACTION
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:
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 24
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.
INTENSITY IN SINGLE – SLIT DIFFRACTION
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
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 25
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.
INTENSITY IN SINGLE – SLIT DIFFRACTION
Ans: θ = 0.00325 = 0.1860 , α = 0.478 rad, Iθ/Im = 0.926
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 26
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
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. 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?
DIFFRACTION AT A CIRCULAR APERTURE
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)
DIFFRACTION AT A CIRCULAR APERTURE
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
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
MIT- MANIPAL
Where ββββ = ππππdsinθθθθ/λλλλ
Where αααα = ππππasinθθθθ/λλλλ
39
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 ∆φ = φ/N between each of the N phasors
where φ is the phase difference between 1st phasor and Nth phasor.
MIT- MANIPAL
40
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
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
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
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
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
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
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 54
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
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 58
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
DIFFRACTION GRATINGS
Sample spectra of visible light emitted by a gaseous source
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 62
DIFFRACTION GRATINGS
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
DISPERSION AND RESOLVING POWER
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
DISPERSION AND 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
DISPERSION AND RESOLVING POWER
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
DISPERSION AND RESOLVING POWER
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
DISPERSION AND RESOLVING POWER
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
DISPERSION AND RESOLVING POWER
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
DISPERSION AND RESOLVING POWER
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
��� 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 =
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 88
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
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 89
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
QUESTIONS – DIFFRACTION
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
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]
MIT- MANIPAL BE-PHYSICS-DIFFRACTION-2010-11 93
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