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Diffraction
When a beam of light passes through a narrow slit, it spreads out to a certain extent
into the region of the geometrical shadow. This effect is one of the simplest
example of diffraction, i.e., of the failure of light to travel in straight lines.
This can be satisfactorily explained only by a assuming a wave character for light.
Classification of Diffraction
Diffraction phenomena are
conveniently divided into two
general classes:
(i) Fraunhofer diffraction
Fig 1. Diffraction of wave passing through a
(ii) Fresnel diffraction small aperture.
Distinction between Fraunhofer and Fresnel diffraction
1. Fraunhofer diffraction: If the source of light and the screen on which the
pattern is observed are effectively at infinite distances from the aperture
causing the diffraction is called Fraunhofer diffraction.
Fresnel diffraction: If either the source or screen or both, are at finite
distances from the aperture is called Fresnel diffraction.
2. Fraunhofer diffraction is much simpler to treat theoretically, Fresnel
diffraction is difficult to treat theoretically.
(3) In Fraunhofer diffraction lenses are required; on other hand no lenses are
necessary for Fresnel diffraction.
(4) In Fraunhofer diffraction wave fonts are plane and in Fresnel diffraction
wave fonts are divergent.
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Diffraction through single slit
A slit is a rectangular aperture of length large compared to its breadth.
Consider a slit ‘s’ to be setup with its long dimension perpendicular to the plane
of the plane, and to be illuminated by parallel monochromatic light from the
narrow slit ‘s’ at the principle focus of lens L1. The light focused by another
lens L2 on a screen or photographic plate P at its principle focus will form a
diffraction pattern.
Fig. 2. Experimental arrangement for obtaining the diffraction pattern of a single slit Fraunhofer diffraction.
The explanation of the single slit pattern lies in the interference of the Huygens secondary
wavelets which can be thought of as sent out from every point on the wave front. If we
assume wavelets to be uniform spherical waves, the emission of which stops abruptly at
the edges of the slit.
We consider a slit of width b, illuminated by parallel light from the left. Let ds be an
element of width of the wave front in the plane of the slit, at a distance s from the center
O, which we shall call the origin.
Fig. 3. Geometrical construction for investigating the intensity in the single slit Fraunhofer diffraction pattern.
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The parts of each secondary wave which travel normal to the plane at the slit will be
focused at P0, while those which travel at any angle will reach at P. Consider first the
wavelet emitted by the element ds situated at the origin, its amplitude will be directly
proportional to the length ds and inversely proportional to x.
At P it will produced an infinitesimal displacement which, for spherical wave, may be
expressed as ds
dy0 a sin(t kx)
x
As the position of ds is varied, the displacement it produces will vary in phase because of
the different path length to P. When it is at a distance s below the origin, the contribution
will be
We now wish to sum the effects of all elements from one edge of the slit to the
other. This can be done by integrating from s = - b/2 to b/2. The simplest way
is to integrate the contributions from pairs of elements symmetrically placed at s
and -s, each contribution being
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The resultant vibration will therefore be a simple harmonic one, amplitude of which varies
with the position of P, since this is determined by .
We may represent the amplitude as
sin
A A0
where,
ab
A0
x
1
kb sin
2
The intensity on the screen is then
2 sin2
I A2 A0
2
If the light (instead of being incident on the slit perpendicular to its plane) make an angle i,
then
b (sin i sin )
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Further investigation of the single slit diffraction pattern
1
kb sin
2
sin 2
2
sin 1
kb sin
A A0 I A0 2
2
1 2
b sin
ab 2
where A0
x b
sin
Fig. 4. Amplitude and intensity contour for Fraunhofer diffraction of single slit showing position of maximum
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and minimum.
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The maximum intensity of the strong central band comes at the point P0, where all
the secondary wavelets will arrive in phase because the path difference 0 .
For this point 0; then sin
Therefore sin
1
2
2 sin 2 2 sin 2
I A0 A0 A0 is the maximum intensity (at =0)
sin
From this principal maximum the intensity fall to zero at then passes through several
secondary maximum with equally spaced points at zeros intensity at
, , 2 , 3 , ........
Or in general m m 1, 2 ............... m
b sin
m
b sin m for minima.
The secondary maximum do not fall half way between these points, but are displaced
toward the center of the pattern by an amount which decreasing with increasing m.
The exact values of for these maxima can be found differentiating A A sin
0
with respect to and equating zero.
sin A
A0
d sin
0
d
cos sin
or , 0
2
or , cos sin
sin
or ,
cos
or , tan
The values of satisfying this relation are easily found graphically as the intersection of
the curve y tan
y tan
and straight line y . These points of intersection lie directly
below the corresponding secondary maxima . The intensity of the secondary maxima can be
y
calculated to a very close approximation by finding the values of sin 22
at the half way positions, where .
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The first secondary maximum is only 4.72 percent the intensity of
the central maximum, while the second and third secondary
maxima are only 1.65 and 0.83 percent respectively.
Figure 5. shows plots of the intensity of a single-slit diffraction pattern,
calculated with intensity equation for three slit widths: d =, d=5, and d= 10.
Note that as the slit width increases (relative to the wavelength), the width
of the central diffraction maximum (the central hill-like region of the graphs)
decreases; that is, the light undergoes less flaring by the slit. The secondary
maxima also decrease in width (and become weaker). In the limit of slit width a
being much greater than wavelength , the secondary maxima due to the slit
disappear; we then no longer have single-slit diffraction (but we still have
diffraction due to the edges of the wide slit, like that produced by the edges of the
razor blade.
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Fig. 5 .The relative intensity in single-slit diffraction for three values of the ratio a/. The wider the
slit is, the narrower is the central diffraction maximum.
Diffraction by double slit
Now, we have two equal slits of width b, separated by an opaque space of width
c, origin may be chosen at the centre of c, and the integration extended from
s=d/2-b/2 to s= d/2+b/2.
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Fraunhofer Diffraction by double slit
Now, we have two equal slits of width b, separated by an opaque space of width
c, origin may be chosen at the centre of c, and the integration extended from
s=d/2-b/2 to s= d/2+b/2.
Fig. 6
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This gives,
Therefore,
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sin 2
The factor in this equation is just that derived for the
2
2
single slit of width b. the second factor cos is characteristics
of the interference pattern produced by two beams of equal
intensity and phase difference .
Fig. 7. Path difference of parallel rays leaving a double slit.
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The resultant intensity will be zero when either of the two factors
is zero. For the first factor this will occur when
=, 2, 3………..
For the second factor this will occur when,
=/2, 3/2, 5/2………
The two variables and are not independent. The difference in
path from the two edges of a given slit is bsin. The
2
corresponding phase difference is b sin , which equals to
2.
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POSITIONS OF THE MAXIMA AND MINIMA, MISSING ORDERS
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Example 3: Monochromatic light of wavelength 441 nm is incident on a narrow
slit. On a screen 2.00 m away, the distance between the second diffraction
minimum and the central maximum is 1.50 cm. (a) Calculate the angle of
diffraction of the second minimum. (b) Find the width of the slit.
(a)
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Example 4: Light of wavelength 633 nm is incident on a narrow slit. The
angle between the first diffraction minimum on one side of the
central maximum and the first minimum on the other side is 1.20°. What is the
width of the slit?
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Example 5: Monochromatic light with wavelength 538 nm is incident on a slit
with width 0.025 mm. The distance from the slit to a screen is 3.5 m. Consider a
point on the screen 1.1 cm from the central maximum. Calculate (a) for that
point, (b) , and (c) the ratio of the intensity at that point to the intensity at the
central maximum.
(a)
(b)
(c)
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