2.Diffractdion
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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)
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Diffraction
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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
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DiffractionFor plane waves entering a single slit, the waves emerging from the slit start spreading out, diffracting.
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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
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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
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Types of Diffraction
1. Fresnel Diffraction
2. Fraunhofer Diffraction
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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.
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In laboratory Fraunhofer diffraction is realized by using converginglenses for conversion of spherical wavefront into plane wavefront andvice versa.
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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
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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.
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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
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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
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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,
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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.
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INTENSITY IN SINGLE – SLIT DIFFRACTION
sinx2
phase path length2difference difference
The phase relationships among the wavelets arriving from different
zones is given by
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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
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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.
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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
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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
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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,….
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[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
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[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
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[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.
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[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
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(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
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[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.
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[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
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(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
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[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
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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.
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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.
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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
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.
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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
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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
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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.
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[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
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[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
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[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
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[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
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DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
Fig. shows the double slit producing both interference and diffraction pattern
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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?
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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
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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:
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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.
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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.
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).........(2
cos 22
sin2
sin A
Also
)(
22
2sin2 1
or
where
EE
1
2/2
sindiagram, From
EE
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cos2
sin,),(
getweAeqninthisngSubstituti
sin)(2 ad
,sin
2aboveeqnofsidesbothto
aAdding
iswhichd sin
2
pathdiffPhasediff2
sin)(
2ad
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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
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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
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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)
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....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
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PROBLEMS
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[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
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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?
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[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
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[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.
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[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
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(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
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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?
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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.
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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
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Intensity pattern for (a) Two-slit diffraction (b) Five-slit diffraction(diffraction effect is neglected)
MULTIPLE SLITS
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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
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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.
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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
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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
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Effect of increasing the number of slits (N) in Fraunhoferdiffraction (Diffraction grating):
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PROBLEMS
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[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.
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(b) The angular width of the largest order maximum.
.)3.57)(cos2100)(10(
589cos
,max3)(
04 nmnm
Nd
imummtheForb
imumprincipalnarrowyexceedinglanisThisradx
max0030.0102.5 05
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[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
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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.
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[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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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GRATING
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[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
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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.
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[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.
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[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.
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[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
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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
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[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?
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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?
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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)?
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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?
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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?
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[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
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[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
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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.
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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
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• 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
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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
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Soft X-rays: possessing lower penetration power (low applied voltage, higherwavelengths) – medical imaging.
Soft X-rays and uses
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Typical medical X-ray tube
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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
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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
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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.
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X-RAY DIFFRACTION
NaCl crystal (a0 = 0.563nm)
A plane through a crystal of NaCl
NaCl unit cell
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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)
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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.
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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 = m
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PROBLEMS
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[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
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[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.
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[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
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TUTORIALExerciseE42-1, E42-16, E42-19, E42-26, E42-29
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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]
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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]
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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. [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]
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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