Transcript Diffraction
Diffraction
A Phenomenon of bending of light or spreading of light by an obstacle is known as diffraction. Italian scientist Grimaldi
A’ A S B Slit B’ Screen
Difference between interference and diffraction
Interference Diffraction
1) All maxima have same intensity.
2) Interference fringes have almost same width.
3) The region of minimum intensity is usually almost dark.
4) This phenomenon is the result of interaction taking place between two separate wave fronts obtaining from two coherent sources.
1) All maxima have varying intensity.
2) The diffraction fringes don’t have same width.
3) The region of minimum intensity is not perfectly dark.
4) This phenomenon is the result of interaction of light between the secondary wavelets originating from different pints of the same wave front.
Type of diffraction:
Fraunhofer Diffraction Fresnel Diffraction
Fraunhofer Diffraction Fresnel Diffraction
1) The source and the screen are at infinite distance from the diffracting aperture.
2) Diffracted light is collected by a lens as in a telescope.
3) The centre of the diffraction pattern is always bright for all paths parallel to the axis of the lens.
4) The wave fronts are plane which is realized by using convex lens.
5) For this, single, double slit or plane diffraction gratings are used.
1) The source and the screen are at finite distance from the diffracting aperture.
2) No mirror and lenses are used for observation.
3) The centre of the diffraction pattern may be bright or dark depending upon the number of Fresnel Zone.
4) The wave fronts are divergent either spherical or cylindrical.
5) For obtaining Fresnel zone plates are used.
diffraction
Fresnel’s Zone: A wave front can be divided into a large number of elements or zones called Fresnel’s zone.
Fraunhofer’s diffraction at a single slit:
S Monochromatic Source A θ e e e B θ e θ N e N Slit Convex lens P Y Screen X O
The path difference (Δ) = BN.
BN
AB
sin
So that the path difference = e sin
θ
Therefore,
phase difference
= 2 e sin
Let the width AB of the slit be divided into n equal parts. So that n is the no.
of vibrating particles and let d is the constant phase difference between those particles.
d
1
n
phase difference
d
2
e
sin
n
This is the constant phase difference of each vibrating particle
To calculate the resultant amplitude (like that: the resultant amplitude for a simple harmonic vibration having the common phase difference).
Draw a polygon, to determine the resultant amplitude
R
and resultant phase
δ
.
R 3d 2d δ a d
Resolved part of a system of vibration in the horizontal direction:
R
cos
a
cos 0
a
cos
d
a
cos 2
d
a
cos 3
d
a
cos
n
1
d
and for vertical direction:
R
sin
a
sin 0
a
sin
d
a
sin 2
d
a
sin 3
d
a
sin
n
1
d
By solving these two equations
R
a
sin sin
nd
2
or
R
a
sin sin
e
sin
e
sin
n
Let us put
e
sin then
R
a
sin sin
n
n is the no. of vibrating particles, when n is very large then
should be very small, so that:
sin
n
n n R
a
sin
n
R
na
sin Let
na
A
, then
R
A
sin
Therefore, the resultant intensity at point
P
is being proportional to the square of the amplitude.
I
R
2
I
A
2 sin 2
(1)
2
Positions of Minima:
It is clear from above equation that the intensity should be minimum (zero) when, sin 0 sin 0 , (but 0 )
e
sin
m
,
Finally, we have tan This equation is solved graphically by plotting the curves
y
and
y
tan y 45 ◦
π/
2
min
1.43
π (max)
3
π/
2
min
2.46
π (max) 5 π/
2
α
Form the graph the point of intersection gives the values of α , these values are approximately given
0 , 3 2 , 5 2 , 7 2 ,
or more exactly as
0 , 1 .
43 , 2 .
46 , 3 .
47 ,
Put the approximately values of α in equation (1), we get the intensities of various maxima. Thus the intensities of the central maxima is
I
A
2 sin 2 (i) For principal maxima 0 then
I
0
A
2
(ii) Intensity for the first secondary maxima 3 2
I
4 9 2
I
0
I
0 22 Now intensity for the second secondary maxima 5 2
I
4 25 2
I
0
I
0 61
So that the ratios of successive maxims are:
or
The minima lie at 1 : 4 9 2 : 4 25 2 : 4 49 2 : 1 1 : 22 : 1 61 1 : 121 :
m
where
m
= 1, 2, 3,…………….. , 2 , 3 , 4 ,
It is clear that most of the incident light is concentrated in the principal maxima, which occur when
0
I -4π -3 π -2 π π 0 + π +2 π +3 π +4π
α
Fraunhofer Diffraction at N-slit (Diffraction grating):
e d A
K
θ θ B θθ X P Y O
By the theory of Fraunhofer diffraction at a single slit, the secondary waves from all points of a slit in the direction wave of amplitude
θ are equivalent to a single
A o
A
sin
A o
is the resultant amplitude due to a single slit, in the normal direction.
A = na, where A is the amplitude of n number of vibrating particles in a slit, α is the phase difference between the wavelets.
e
sin
If N is the total number of slits in grating, the diffracted rays from all the slits act as N number of parallel rays.
Draw an equal lengths (like vector) MP 1,
P
1
P
2 , P 2
P
3 ------------ , centre of polygon, then from geometry: the closing side MP N of the polygon represents the resultant amplitude (R). If O is the
P N MP MP N
1 2
OM
2
OM
sin sin
N
O
2
β β
2
N β M N P 1
2
β P 2
4
β
6
β MP N
MP
1 sin sin
N
Resultant amplitude (R)
A
o or amplitude of each vibration
Resultant amplitude (R)
R
A
sin sin sin
N
Then the resultant intensity (
I
) is:
I
A
2 sin 2 2 sin 2 sin 2
N
The first factor
A
2 sin 2 2
gives the diffraction pattern due to single slit; while the second factor
sin 2
N
sin 2
gives the intensity distribution pattern due to interference pattern for N slits.
Let us consider the intensity distribution due to the second factor.
Principal Maxima:
i. e.
When sin 0
n
where
n
= 0, 1, 2, 3, ---------- Then sin
N
0 and thus sin sin
N
0 0 Let us find the value of usual method of differentiating the numerator and the denominator. By using ‘
L
’ Hospital rule: lim
n
sin sin
N
N
So that the intensity from equation (1), we have,
I
A
2 sin 2 2
N
2
This is the maximum intensity.
e
d
sin
n
n
where
n
= 0, 1, 2, 3, ---------- For
n
= 0, we get the zero order maxima. For
n
=1 this is
e
d
sin For
n
= ±1, ±2, ±3, -------------, we obtain the first, second, third order principal maxima respectively.
For Minima:
When sin
N
0 sin sin
N
but 0 sin 0 Hence, from equation (1) we get
I
= 0 This is minimum intensity. These minima are obtained in the direction given by or
N
m
N
e
d
sin
m
Where
m
value of
m
takes all integral values except 0,
N
, 2
N
, 3
N
, ------- make sin 0 which gives the principal maxima.
nN
, because these It is clear from above that
m
= 0 gives a principal maxima and
m
= 1, 2, 3, -------- (
N
-1) give minima.Then
m
=
N
gives again a principal maxima. Thus, there are (
N
-1) minima between two consecutive principal maxima.
Secondary Maxima:
As there are (
N
-1) minima between to consecutive principal maxima, there must be (
N
-2) secondary maxima between two principal maxima.
Principal Maxima Secondary Maxima Minima
The position of the secondary maxima are obtained by differentiate equation (1) with respect to
β
and equating it to zero.
dI
0
d
N
tan tan
N
1
N
2 tan 2
N
tan
Nβ
1 sin 2
N
sin 2 1
N N
2 1 sin 2 This equation shows that the intensity of the secondary maxima is proportional to 1
N
2 1 sin 2 where as the intensity of the principal maxima is proportional to
N
2 therefore
Intensity of Intensity of the Secondary
max
ima the principal
max
ima
1
N
2 1 1 sin 2 Hence, greater value of
N
(Number of slits), the weaker are secondary maxima, In actual grating,
N
is very large. Hence these secondary maxima are not visible in the grating spectrum.
Formation of Multiple Spectra by the grating: When a beam of wavelength
fall normally on the grating, the principal maxima are formed in the direction given by
e
d
sin
n
n
= 0, 1, 2, 3, -------------------
R
2
V
2
II nd order R
1
I st order V
1
Zero order R
1
V
1
I st order R
2
V
2
II nd order
Condition for absent spectra:
e
d e
n m
This is the condition for the spectrum of the order n to be absent. Maximum number of orders:
n
max
e
d
Dispersive power of a grating:
The dispersive power of a grating is defined as the angle of diffraction with the wavelength of “the rate of change of
d
light”. It is expressed as
d
The angle of diffraction
, for the principal maxima is related to the corresponding wavelength by grating equation
e
d
sin
n
Differentiating the above equation with respect to we get
d
d
e
d n
cos
The above equation shows that the dispersive power is 1.
2.
3.
directly proportional to the order
n
.
inversely proportional to the grating element inversely proportional to cos
It means larger value of
smaller value of
cos
then the value of dispersive power is higher. Example-
The angle of diffraction for red is greater than the violet in a given order spectrum, the dispersion in the red region is greater then the violet region.
Rayleigh’s Criterion for resolution:
A B Resultant intensity A C B λ λ' Widely separated λ λ' Very close
In this figure, the resultant intensity shows a maxima C and spectral lines appear to be one.
two
λ θ A Dip B θ
'
λ
'
In this case, the resultant intensity shows a dip and two wavelengths are said to be just resolved.
Resolving Power of a Diffraction grating:
It defined as the ratio of the wavelength of light between two wavelengths to be resolved. If
λ
and
λ
to the difference
λ+dλ
wavelengths to be resolved then the resolving power is given as
d
Grating Ist minima of the wavelength λ overlapping the maxima of wavelength λ
+
dλ .
dθ n θ n Resolving power of a diffraction grating using Rayleigh’s criterion of resolution
Let the angle of diffraction for the n th be θn and that of wavelength
d
be θn+dθn.
Using Rayleigh Criterion
d
nN
A grating with larger numbers of slits will have larger resolving power
Relation between Resolving Power and Dispersive Power: The resolving power of a grating is equal to the product of the order of the spectrum and total number of lines on the grating.
d
nN
We know that the dispersive power of the grating is
d
d
e
d n
cos
so that from the above equation, we have
d
Re
solving Power
Total Apperature
Dispersive Power N
e
d
cos
is total aperture width.