Diffraction and Interference
6.1 Diffraction
Diffraction grating
Single slit diffraction
Circular diffraction
Diffraction grating
• Diffraction and interference are similar
phenomena.
• Interference is the effect of superposition
of 2 coherent waves.
• Diffraction is the superposition of many
coherent waves.
Diffraction grating
λ
m=2
• Consists of a flat barrier which contains many parallel
slits separated by a short distance d.
• A parallel monochromatic light beam passing through the
grating is diffracted by an angle θ
m=1
d
m=0
two slit interference
m=-1
m=-2
dsinθ=mλ
dsinθ=mλ
similar to two slit interference.
m=2
m=1
However, the intensity of the diffracted light is higher and
the peaks are much narrower.
m=0
m=-1
multi-slit interference
Diffraction grating
Use of a diffraction grating
in a spectrometer
m=-2
Question
A grating in a spectrometer has a length of 2 cm and has
contains 104 lines. Find the first order diffraction angle
for light with a wavelength of 500 nm.
first order m=1
d=
L
N
=
2x10 −2 m
= 2x10 −6 m
10 4
dsinθ=mλ
sin θ =
Dispersion of light of different wavelengths
L=2.0 cm
θ
N
=104
slits
mλ 500x10 −9
=
= 0.25
d
2x10 −6
θ=14.5o
1
Optical compact disc
Single slit diffraction
For single slit diffraction
the complex behavior of the
wave as a function of the
aperture a is explained by
interference between multiple
coherent point sources in the slit.
a
10-6 m
λ
The closely spaced dots act like a diffraction grating.
barrier
Two Limiting Cases
Barrier
Ray picture
point
sources
Laser
screen
a>>λ
Light moves in a straight line
intensity
slit open
a
Wave picture
a<< λ
What happens to the laser beam
as the slit decreases?
Light spreads out when passed
through small aperture.
What happens to the
light for the general case?
The pattern spreads out as the slit becomes narrower
Single slit diffraction pattern
screen intensity
slit narrow
How do we
account for the
minima and
maxima?
What happens to the laser beam
as the slit decreases?
Single slit diffraction pattern
Analyze assuming Fraunhofer diffraction conditions.
a) Screen far from the slit or
b) Converging lens with screen at the focal length
2
Single slit diffraction
Single slit diffraction
Huygens principle – Each point in the wave in the slit
acts as a source of spherical waves.
Find the angle at the first minimum
amplitude = 0
sum the waves with different phases
θdark
θ
Divide the slit into 2 halves
Slit width
sum
a
amplitude(θ)
θ
Light from every point in the top
half interferes destructively with
a light from a point in the bottom
half when
a
λ
sin θdark =
2
2
Condition for a minimum
asinθdark=λ
Circular diffraction
asinθdark =mλ
m=+ 1, + 2, .........
θdark
Waves passing through a circular hole forms a
a circular diffraction pattern.
Circular diffraction limits the minimum size do the
spot that can be formed by a lens.
Circular diffraction.
Circular hole
According to ray optics.
The first minimum
occurs at
diameter
D
θmin = 1.22
λ
D
parallel rays from
a point at infinity
Can focus
on a point
at the focal point
of the lens
Circular diffraction pattern
3
Circular diffraction limits the minimum size do the
spot that can be formed by a lens.
A camera lens with an f- number (f/D) equal to 1.4 is used
to focus light from a distant source. What is the
diffraction limited diameter of the spot that can be formed
for 500 nm light?
According to wave optics.
θmin
diameter D
Example
θmin = 1.22
parallel rays from
a point at infinity
λ
D
=
has a diffraction pattern
with a width of θmin
d = 2r = 2(1.22
f
= 1.4
D
r
f
θmin
D
r
f
λf = 2(1.22)(500x10 −9 )(1.4) = 1.7x10 −6 m
)
D
about 3 x the
wavelength of the
light
Optical CD
Resolution of two images by a lens
resolved
focused laser
beam
just
resolved
10-6 m
not
resolved
The amount of information that can be encoded is limited
by the diameter of the diffraction-limited spot.
The resolution of images is limited by the
diffraction pattern.
Resolution limit of the eye
Rayleigh criterion
For resolution of two object by a circular lens of
diameter D the diffraction limit of resolution occurs when
the image of the second object is at position of the first
minimum of the diffraction pattern of the first object.
θmin
D
λ
= 1.22
D
θmin
Two light sources (λ= 500nm) are separated vertically by 2.0
mm. How far away can these objects be resolved by the eye
Assume a diameter of the pupil of 2.0 mm. nwater =1.33
L
D=2.0 mm
y
y
(for small angles)
L
y
λ
= 1.22
L
Dnwater
θmin =
L=
yDnwater
1.22λ
=
λ water
D
λ
= 1.22
Dnwater
θmin = 1.22
(2x10 −3 )(2x10 −3 )(1.33)
= 8.7m
1.22(500x10 −9 )
4
Diffraction limit
• Diffraction limits the resolution of objects
viewed through an optical system.
– atoms cannot be seen with a light microscope
(shorter wavelengths are required)
– Satellite cameras have a limited resolution.
• To attain higher resolution.
– Larger diameter lenses
– Shorter wavelengths.
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