Download Single Slit Diffraction and Resolution

Diffraction around an edge and
through an aperture
Huygens’ principle is consistent with
diffraction:
The larger the wavelength with respect to the
aperture size, the more significant the
diffraction
Young’s Double Slit
Experiment
If light is a wave, interference effects will be seen, where one part
of wavefront can interact with another part.
One way to study this is to do a double-slit experiment:
Two-Source Interference
Pattern
Nodal points – blue dots
Anti-nodal points – red dots
Young’s Double Slit
Experiment
If light is a wave, there should be an interference pattern
Path Difference
1) there is no difference
path difference have the waves
2) half a wavelength
from each slit traveled to give a
3) one wavelength
minimum at the indicated
4) three wavelengths
position?
5) more than three wavelengths
Intensity
In a double-slit experiment, what
Path Difference
1) there is no difference
In a double-slit experiment, what
2) half a wavelength
path difference have the waves
from each slit traveled to give a
3) one wavelength
minimum at the indicated
4) three wavelengths
position?
5) more than three wavelengths
Intensity
3
2

For Destructive Interference
7/2
5/2
3/2
= (m + 1/2) 
/2
 = 1/2 , 3/2 , 5/2 , 7/2 , …
Diffraction by a Single Slit or
Disk
bright
dark
Note: There are
bright and dark
fringes beyond the
shadow. These
resemble
interference fringes
of a double slit.
Diffraction by a Single Slit or
Disk
Wavelets from the center and edge of the slit
Dark Band
Y
D
D
D
Path difference = m/2
Wave front going through the
slit is made up of many
wavelets.
X
If D is small we can use…..
similar triangles, for small angles
sin D = tan D = Y / X
so…..
or….
Y/X
=
m/D
Y =mX/D
m = 1,2,3…….
(Interference of
wavelets)
Contrary to expectation, an interference
pattern is observed because the wave
front approaching slit, D, is itself a
source of wavelets. It is these individual
wavelets that will interfere destructively
at the part of the screen where the dark
band is observed.
Now from the diagram the angle to the
first dark band is given by:
sinD = (/2) / (D/2) =  / D
Therefore for any dark band
sin D = m  / D
m = 1,2,3…….
Diffraction by a Single Slit or
Disk
If we look at the intensity of the bright
bands we find that they are not the
same. The brightest band is in the
center and the bands seem to get
dimmer as the angle increases. The
diagram shows how the intensity
changes with angle. You can see that
as sin approaches  / D the intensity
falls to zero as expected from our
theory. Also note that the second band
is much less intense than the central
band and is half as wide.
Practice Problem
Light of wavelength 750 nm passes through a slit 1.0 x 10-3 mm
wide. How wide is the central maximum a) in degrees and b) in
centimeters on a screen 20 cm away?
 = 750 x 10-9 m
D = 1.0 x 10-6 m
X = 0.2 m away
m=1
Width of central maximum is 2 or 2Y
a) Width = 2 = 2 sin-1 (m  / D)
b) Width = 2Y = 2 X tan D
= 2 sin-1 (1) (750 x 10-9m) / (1.0 x 10-6 m )
= 2 (0.2) tan 48.590
= 0.45 m
= 97.20
Check Your Understanding
The diffraction pattern below arises
from a single slit. If we would like
to sharpen the pattern, i.e., make
the central bright spot narrower,
what should we do to the slit width?
1) narrow the slit
2) widen the slit
3) enlarge the screen
4) close off the slit
Check Your Understanding
The diffraction pattern below arises
from a single slit. If we would like
to sharpen the pattern, i.e., make
the central bright spot narrower,
what should we do to the slit width?
1) narrow the slit
2) widen the slit
3) enlarge the screen
4) close off the slit
The angle at which one finds the first
minimum is:
sin  =  / d
The central bright spot can be
narrowed by having a smaller angle.
This in turn is accomplished by
widening the slit.
d


Check Your Understanding
Blue light of wavelength  passes
through a single slit of width d and
forms a diffraction pattern on a screen.
If the blue light is replaced by red light
of wavelength 2, the original diffraction
pattern can be reproduced if the slit
width is changed to:
1) d/4
2) d/2
3) no change needed
4) 2 d
5) 4 d
Check Your Understanding
Blue light of wavelength  passes
through a single slit of width d and
forms a diffraction pattern on a screen.
If the blue light is replaced by red light
of wavelength 2, the original diffraction
pattern can be reproduced if the slit
width is changed to:
1) d/4
2) d/2
3) no change needed
4) 2 d
5) 4 d
d sin = m (minima)
If   2 then we must have d 
2d for sin  to remain unchanged
(and thus give the same diffraction
pattern).
d


When violet light of wavelength 415 nm falls on a single slit,
it creates a central diffraction peak that is 9.20 cm wide
on a screen that is 2.55 m away. How wide is the slit?
 = 415 x 10-9 m
Y = 4.6 x 10-2 m
X = 2.55 m away
m=1
D=?
Because X is much bigger than Y, D is small, so we can
use…..
Y =mX/D
D = mX/Y
m = 1,2,3…….
= (1) (415 x 10-9 m) (2.55 m) / (4.66 x 10-2 m)
D = 2.30 x 10-5 m = 2.30 x 10-2 mm
Rayleigh Criterion for Resolution of Two
Diffracted Images
Two images are said to be just resolved when the central maximum of one image
falls on the first minimum of the diffraction pattern of the other image.
From single slit diffraction
sinD =  / D
If  << d, sinD ≈ D
min =  / D
Where min is in radians
The apertures of microscopes and
telescopes are circular. Analysis
shows…
min = 1.22  / D
Check Your Understanding
An optical telescope has a 21 cm
mirror diameter. To give the same
angular resolution as an optical
telescope the effective diameter of a
radio telescope would have to be….
1) bigger
2) smaller
3) the same
4) doesn’t matter
Check Your Understanding
An optical telescope has a 21 cm
mirror diameter. To give the same
angular resolution as an optical
telescope the effective diameter of a
radio telescope would have to be….
The minimum angle of resolution is
given by:
min = 1.22  / d
Radio waves have a much longer
wavelength, so to give the same
minimum angle of resolution the
dish diameter has to be much larger
1) bigger
2) smaller
3) the same
4) doesn’t matter
Check Your Understanding
A scientist wants to observe finer
1) Make the objective bigger
detail on a specimen under a
2) Use smaller wavelengths
microscope. They could….
3) Use oil between the
specimen and the objective
4) All of the above
Check Your Understanding
A scientist wants to observe finer
1) Make the objective bigger
detail on a specimen under a
2) Use smaller wavelengths
microscope. They could….
3) Use oil between the
specimen and the objective
4) All of the above
The minimum angle of resolution is
given by:
min = 1.22  / d
To make min smaller you would have
to make the objective lens bigger or
use a smaller wavelength. Oil has a
higher n than air so n = air / n.
Spherical and chromatic
aberration however limit
the size of the objective,
d. Also UV is absorbed
by glass.
Electrons can be used
because of their wave
properties (E = h f)
Two of Jupiter’s largest moons (it has 16 or more) are separated
by a distance of 3.0 x 106 km at a time when the planet is 3.3 x 108 km
from Earth. a) Would a person be able to resolve both moons with the
unaided eye, assuming night-time pupil diameter of 7.5 mm? b) What
minimum diameter of mirror would be needed in a radio telescope?
light = 550 nm, radio = 21 cm and s = r 
Use:
 = 550 x 10-9 m
s = 3.0 x 109 m
r = 3.3 x 1011 m
d = 7.5 x 10-3 m dmin = ?
=s/r
= (3.0 x 109 m / (3.3 x 1011 m) = 0.009 rad
a) min = 1.22  / d = 1.22 (550 x 10-9 m / (7.5 x 10-3 m) = 0.000089 rad
Yes  > min , so a person would be able to resolve both moons
b) dmin = 1.22  /  = 1.22 (21 x 10-2 m) / 0.009 = 28 m
Now you can see why big dishes are needed for radio telescopes
A camera on a spy satellite orbiting at 200 km has a diameter of 35 cm.
What is the smallest distance this camera can resolve on the
surface of the earth?
Use:
 = 550 x 10-9 m
s=?
light = 550 nm, and s = r 
r = 200 x 103 m
d = 35 x 10-2 m
min = 1.22  / d = 1.22 (550 x 10-9 m / (35 x 10-2 m)
s = r min = (200 x 103 m)(1.92 x 10-6 )
= 1.92 x 10-6 rad
= 0.38 m (38 cm)
This size of lens would not be good enough to read license plate
numbers but good enough to resolve individual people.