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DIFFRACTION BY SINGLE SLIT
Diffraction : Phenomena where light or waves being spread to a wider region
after its passes through narrow slit or narrow opening.
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Diffraction of waves
As the width of slit increase, the dispersion of wave decrease.
As the wave length decrease, the dispersion of wave decrease.
The diffraction is appreciable when the width of slit is comparable to the wavelength of the
waves and very small when the width is large compare to wavelength.
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FRAUNHOFER DIFFRACTION When parallel light rays incident on a narrow rectangular slit ( width 0.01mm). Light will
be disperse to a region bigger than the width of the slit. This phenomena is called
Fraunhofers Diffraction.
O
When parallel light rays reaches the slit,
each point on the wave front acts as a
secondary coherent source, sending out
new waves beyond the slit. From principle
of superposition, overlapping of coherent
wave front produce diffraction pattern in
front of the slit.
By putting a screen in front, the diffraction pattern consists of dark
and bright fringes:
(a) A bright fringes is obtain at point O.
(b) Other bright fringes with decreasing intensity.
(c) Dark fringes located in between 2 bright fringes. m = 1, 2
m=2
m=1
m=-1
m=-2
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DIFFRACTION AT SINGLE SLIT
(A) Central bright band:
With O is the central of the diffraction image, a bright band is obtain at O. Assuming the slit is divided into 2 equal section, each point source
pairing with its corresponding point source, acting as 2 coherent point
sources, producing waves overlap at O with zero path different.
All waves arrive in phase at O, producing a center bright band with maximum
intensity.
The point O is directly in front of the slit, = 0.
o
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(B) First minimum: (Dark fringe)
The slit is divided into 2 equal halves. The top point source of the upper half and
the top point source of the lower half send out
waves to Q, with path difference /2, therefore
the waves overlap out of phase at Q.
All other pairs of corresponding points in
the 2 halves also have path difference /2 and
overlap out of phase at Q. Therefore, the resultant amplitude at Q is
zero, having zero intensity, (dark fringe).
Point Q also know as the edge of central
bright band.
DIFFRACTION AT SINGLE SLIT
/2
/2
1d1
If 1is the direction of first minimum:Then , d sin 1=
For small angle, ( in radian):
1= /d
Q
Q
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SECONDARY FRINGES
Other than the central bright band and first minimum, other secondary bright
and dark fringes also obtained on the screen beyond Q.
For secondary bright fringes:
Assuming the slit is divided into 3, 5,7 equal
section.
Each consecutive parts have path difference of
, give rise to zero resultant.
The last remaining section give rise to bright
fringes of less intensity. (Calculation show that the Firstmaximum is only about 5% of the central bright band.)
Equation for the position of secondary bright
fringes:
d sin n= (n + )
For secondary dark fringes: (zero intensity)
Assuming the slit is divided into 2,4,5 equal
section.
Each consecutive parts have part difference of
give rise to zero resultant.
Therefore, the final resultant is also zero.
Equation for position of dark fringes:
d sin n= nwhere n = 1,2,3,4,.
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Half angular width of center bright band
For first minimum: d sin 1= , [1= /d if 1small in radian,
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VARIATION IN DIFFRACTING PATTERN
1. When width of the slit increase, the dispersion of light decrease and the width of the
diffraction pattern decrease but intensity of center bright band increase. [sin 1/d] 2. When width of slit decrease, the dispersion of light increase, the width of the diffraction
pattern increase but intensity of center bright band decrease.
3. When the slit become too wide, the dark and bright fringes become too close to be seen
by naked eye. So dispersion of light (diffraction of light) disappear.
Sin-1/d Sin-12/d Sin-12/dSin-1/dSin-1/d Sin-1/dSin-12/d Sin-12/d
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Intensity
Intensity
Graph of variation in intensity for diffraction pattern
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Example A rectangular slit of width 0.02 mm is placed in front of a parallel light of wavelength 600nm. A
screen is placed a distance 60 cm from the slit.
Find (a) half angular width of the diffraction pattern form on the screen.
(b) the width of the centre bright fringes.
(c) the half angular width if the width is reduced to 0.3 mm
'o
2-
3
9-
431
10x3
1002.0
600x10sin
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x
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'o
b
rad
x
3-
3-
3
9-
10x2
10x2
103.0600x10sin
,sindFrom(c)
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Resolving power of Optical Instrument
When light entering the objective lens of an optical instrument, (e.g Telescope)
Diffraction of light occur, produce diffracting pattern for each of the image form.
Angular separation of object: Angle between both object subtended at the optical
center of the objective lens.
Image, 1
Image, 2
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x
L
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Rayleigh Criteria: 2 images is perfectly resolved when the center maximum of the one fall
onto the first minimum of the other image.
Resolving power of Optical Instrument
Perfectly resolved Totally resolvednot resolved
With, D sin = 1.22 , [ factor 1.22 because of round opening]sin = 1.22 / D , [ D- diameter of objective lens] NEXT
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CONDITION FOR PERFECTLY RESOLVED
From Rayleigh criteria: Two images is perfectly resolved when maximum of
one image fall on the first minimum of the other image.
Angle subtended by images at objective lens:
D sin = 1.22 [ 1.22 is a factor to correct the round opening of the lens]
Perfectly resolved
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Image, 1
Image, 2
With tan = x/L, L- distance of object from lens.
x - distance separation of 2 object.
L
x
And, sin = 1.22 /D, angle of perfectly resolved.
When > , then the 2 image form is totally resolved.When < , then the 2 images is not resolved.When = , the 2 images is perfectly resolved.Therefore:
The resolving power of optical instrument directly proportional to 1/.The bigger the value of D, the higher the resolving power.
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Resolving Power of Optical Instrument
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Example
2 object with separation of 2 mm is located 50 m from the objective lens of a
telescope. If the lens have diameter D, determine whether the image form can be
resolved or not, when
(i) D = 20 cm
(ii) D = 40 cm
rad
x
5-
5
10x4
10450
002.0
L
xtan
50 m
2 mm
resolved.notisimages2the,Since
.10x3.66
10x3.6610x20
10x600x1.22sin
6-
5-2-
-9
rad
resolved.totallyisimages2the,,
.10x1.83
10x1.8310x40
10x600x1.22sin
6-
5-
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-9
Since
rad
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Example
A Youngs double slit, each slit have width 0.04mm and separation 2mm. Monochromatic light of
wave length 580 nm fall on the 2 slits produce an interference pattern on screen placed 90 cm in
front of the slits.
(i) Determine the angular separation between 2 successive bright fringes.
(ii) If one of the slit is covered with an opaque material, find the half angular width of the center
bright fringes form on screen.
.10x2.9
109.2
102
10x580sin
:sinaFrom(i)
4-
4
3
9-
rad
x
x
49'
1045.11004.0
10x580sin
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3
9-
xx
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EXERCISE
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EXERCISE
1. A single slit is located at infinity in front of a lens of focal length 1.0 m and is illuminated normally
with light of wavelength 600 nm. The first minima on either side of the central maximum of the
diffraction pattern observed in the focal plane of the lens are separated by 4 mm. What is the
width of the slit ? [a = 0.3mm]
2. A parallel beam of blue light (420nm) incident on a small aperture. After passing through the
aperture, the beam is no longer parallel but diverges at 10to the incident direction. What is the
diameter of the aperture ? [12m]
3. The headlights of a distant automobile are 1.4 m apart. If the diameter of the pupil of the eye is
3mm, What is the maximum distance at which the headlight can be resolved ? [7.0 km]
4. Two light sources are viewed by the eye at a distance L = 2500m. The entrance opening of the
viewers has a diameter of 3mm. If the eye were perfect, the limiting factor for resolution of the
two sources would be diffraction. How large could the separation of the two light sources and
still have the sources seen as separate entities. [ 0.5m]
5. Monochromatic light from a distant source is incident on a slit 0.800mm wide. On a screen 3.00m
away, the distance from the central maximum of the diffraction pattern to the first minimum is
measure to be 1.80 mm. Calculate the wavelength of the light. [480nm]
6. Parallel rays of green mercury light with a wavelength of 546 nm pass through a slit covering a lens
with a focal length of 40.0cm.In the focal plane of the lens the distance from the central
maximum to the first minimum is 12.0 mm. What is the width of the slit.
7. Red light with a wavelength of 633 nm from a helium-neon laser passes through a slit o.300 mm
wide. The diffraction pattern is observed on a screen 4.0 m away. Define the width of a bright
fringe as the distance between the minima on either side.. (a) what is the width of the central
bight fringes ? (b) What is width of the first bright fringe on either side of the central one.
[16.9mm , 8.44mm]
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8. Light with a wavelength of 589nm from a distant source is incident on a slit 0.850 mm wide, and the resulting
diffraction pattern is observed on a screen 2.00 m away. What is the distance between the two dark fringes on
either side of the central bright fringes ?
9. A slit 0.200 mm wide is illuminated by parallel rays of light that has a wavelength of 500 nm. The diffraction pattern is
observed on a screen that is 4.00 m from the slit. What is the distance on the screen from the center of the central
maximum to the first minimum ? [1.00 cm]
10 A diffraction pattern is formed by passing parallel rays of light 500 nm through a slit of 0.250 mm wide. What is thephase difference between wavelets from the top and bottom of the slit at (a) the center of the central maximum, (b)
the third minimum out from the central maximum ?
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