Transcript Slide 1

5. Diffraction
1) What it is?
The bending of waves behind obstacles or apertures into the ”shadow region”,
that can be considered as interference of many waves.
2) Haw to observe?
Diffraction is most pronounced when the wavelength of the wave is similar to
the size of the obstacle or aperture.
For example, the diffraction of sound waves is commonly observed because
the wavelength of sound is similar to the size of doors, etc.
The waves
spread out from
the opening!
•Light will diffract around a single slit or obstacle.
•The resulting pattern of light and dark stripes on a screen is called
a diffraction pattern (fringes).
•This pattern arises because different points along a slit create
wavelets that interfere with each other just as a double slit would.
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3) Diffraction from a single slit
First minimum (dark):

r1
D/2
D/2
1
2

r2

y
D sin 1  12 
D sin 1  
Minima (dark fringes):
y x  tan
x
D sin  m  m
m D 
m  1,2,3,... mtot  2 D 
m
For small   tan  sin     m 
D
m
ym  x
D
2
3a) Diffraction from a single slit (intensity)
Minima (dark fringes):
D sin  m  m
m  1,2,3,...
Example: In order to obtain a good single slit
diffraction pattern, the slit width could be:
A. /100 ; B. /10; C. ; D. 10; E.100
Example: Light of wavelength 610 nm is incident on a slit 0.20 mm wide and
the diffraction pattern is produced on a screen that is 1.5 m from the slit.
What is the width of the central maximum?
  610nm
D  0.20m m
x  1.5m
2 y1  ?
ym  x
m
D

61010 m 
 2  1.5m 
 9.1510
0.2010 m
9
2 y1
3
3
m  9.2m m
Example: Light of wavelength 687 nm is incident on a single slit 0.75 mm wide.
At what distance from the slit should a screen be placed if the second dark fringe
in the diffraction pattern is to be 1.7 mm from the center of the screen?
  687nm
D  0.75m m
y 2  1 .7 m m
x?
y D
m
x m
D
m
1.7 103 m  0.75103 m 1.7  0.75
x

m  0.93m
9
2  68710 m
2  0.687
ym  x





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4) Multiple slits (Diffraction grating)
•A diffraction grating consists of a large number
of equally spaced narrow slits or lines.
•A transmission grating has slits, while a reflection
grating has lines that reflect light.
•The more lines or slits there are,
the narrower the peaks.
Two slits
Six slits
Parameters:

This is for : D  0
;
 
D d
;
L


Nd
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4a) Diffraction gratings and not monochromatic light
The maxima of the
diffraction pattern
are defined by:
d sin  m  m
m  0,1,2...
4b) The Spectrometer and Spectroscopy
•A spectrometer makes accurate measurements
of wavelengths using a diffraction grating or prism.
•The wavelength can be determined
to high accuracy by measuring the
angle at which the light is diffracted.
•Atoms and molecules can be identified
when they are in a thin gas through their
characteristic emission lines.
Example: A diffraction grating has 5000 lines per cm. The angle between the central
maximum and the fourth order maximum is 47.2°. What is the wavelength of the light?
d  L / N  N / L  
d sin  m  m
1
 1 / 5000cm  2  10 4 cm
  d sin  m / m
 4  47.2 
m4
  2  104 cmsin 47.2  / 4
 ?
  3.67  105 cm  367nm
Example: A diffraction grating has 6000 lines per centimeter ruled on it. What is the
angular separation between the second and the third orders on the same side of the
central order when the grating is illuminated with a beam of light of wavelength 550 nm?
1 / d  6000cm 1
d sin  m  m
 6  105 m 1
sin  m  m / d
  550nm
3  2  ?
sin  3  3  550 109 m  6  105 m 1  0.99
sin  2



 2  550 10 m  6  10 m   0.66
 3   2  40.6 
9
5
1
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5) Circular apertures. Limits of resolution
Thin slits:
Circular apertures:
sin 1   D
sin 1  1.22  D
Rayleigh criterion: two images are just resolvable when
the center of the diffraction disk of one image is directly
over the first minimum in the diffraction pattern of the other
 R  1  1.22   D
y
x
y
  R  1.22   D
x
Example: The Earth and Moon are separated by about 400x106 m.
When Mars is 5x1010 m from Earth, could a person standing on Mars
resolve the Earth and its Moon as two separate objects without a
telescope? Assume a pupil diameter of 5 mm and λ = 550 nm.
y  400106 m
x  5 1010 m
  550nm
D  5m m
 /R  ?
y 400106 m
3
 

8

10
x
5 1010 m
550109
3
 R  1.22   D  1.22

1
.
22

0
.
11

10
5 103
   R  yes!
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