Transcript Diffraction by edges

Diffraction vs. Interference
• Both involve superposition of coherent light
• Custom
– Interference – only a few waves
– Diffraction – large number of waves
• Fresnel correction to Huygens Principle:
– Every unobstructed point of of a wavefront, at a given
instant, serves as a source of spherical secondary
wavelets (with the same frequency of that of the
primary wave). The amplitude of the optical field is the
superposition of all these wavelets (considering their
amplitudes and relative phases)
Fraunhofer vs. Fresnel
• Fresnel – Diffraction is occurring near the
aperture. a.k.a “near field diffraction”
– Initially the fringe pattern looks like the aperture but
then the pattern changes as the distance from the
aperture increases.
• Fraunhofer – Diffraction occurs far from the
aperture. a.k.a “far field diffraction”
– In this region the fringe pattern remains constant,
changing only in size as distance from the aperture
increases.
Diffraction by edges
Fig 38-2, p.1207
Single-Slit Diffraction
• The finite width of slits is
the basis for
understanding Fraunhofer
diffraction
• According to Huygens’s
principle, each portion of
the slit acts as a source of
light waves
• Therefore, light from one
portion of the slit can
interfere with light from
another portion
Diffraction by a single slit
Minima:
a sin   m
m = 1,2,3,…
Diffraction Pattern, Single Slit
• The diffraction pattern
consists of the central
maximum and a series
of secondary maxima
and minima
• The pattern is similar
to an interference
pattern
Intensity
• The light intensity at a point on the screen is
proportional to the square of ER:
 sin  β 2 
I  I max 

β
2


2
2π
β
a sin θ
λ
• Imax is the intensity at θ = 0
– This is the central maximum
Combination of interference and
diffraction for 2 slits
Diffraction
Interference
Final Exam Problem 48
Light of wavelength 632 nm is incident on a single slit.
The distance from the slit to a screen is 3 m. If the
distance from the first minimum on one side of the
center of the diffraction pattern to the first minimum
on the other side is 8 mm, the width of the slit is
closest to
A. 0.22 mm
screen
B. 0.31 mm
C. 0.47 mm
D. 0.59 mm
E. 0.66 mm
3m
The Square Aperture
y
x
5
0
-5
1
0.8
0.6
0.4
Y
0.2
0
-5
0
5
X
Circular Aperture
I
I  0
I
I  0
1
1
Airy Pattern
0.8
0.8
0.6
0.4
0.6
0.2
-7.5
0.4
-5
-2.5
2.5
5
7.5
 2

J1 
a sin    0
 

0.2
4
2
0
0
-4
-2
2

a sin   3.83
-2
0
2
4
-4
ka sin
sin   1.22

2a
 1.22

D
Circular apertures
1.22

D
Rayleigh resolution criteria
Rayleigh Criteria for Resolving Two Objects
• Overlapping images from two apertures are just resolved
when the center of one Airy disk falls on the first minimum
of the other.
I
I  0 1
0.8
0.6
0.4
0.2
-7.5
-5
-2.5
2.5
l
5
7.5
ka sin
Rayleigh resolution criteria
slit


a
circ
1.22

D
Resolution, Example
• Pluto and its moon, Charon
• Left: Earth-based telescope is blurred
• Right: Hubble Space Telescope clearly resolves
the two objects
Final Exam Problem 49
A boat has lights on a mast that are 1 m apart. The
dominant wavelength in the lights is 600 nm. The
pupil in a person’s eye has an opening of 1 mm.
For simplicity, we assume that the eye has a
refractive index of 1. If the boat is closer, the
person sees two lights on the mast. If the boat is
farther away, the person sees only one light on the
mast. The best value for the distance from the
person to the boat is
A. 1.4 km
1m
B. 1.2 km
C. 2.0 km
D. 1.6 km
E. 1.8 km
Diffraction Grating
Two slits
Grating
Maxima:
dsin   m
Final Exam Problem 50
A beam of light is incident on a diffraction grating that has
600 lines/mm. The second order maximum occurs at a
distance 0.7 m from the center of a screen that is 1.0 m
from the grating. The wavelength of light is closest to:
a. 478 nm
b. 613 nm
c. 574 nm
grating
0.7 m
d. 589 nm
e. 542 nm
1.0 m
Polarization
“Linear” or “plane” polarization
Vertically polarized
Processes for
accomplishing
polarization:
•selective
absorption
•reflection
•double
refraction
•scattering
Unpolarized?
Horizontally polarized
Polarization by Selective
Absorption
• The most common technique for polarizing light
• Uses a material that transmits waves whose electric field
vectors lie in the plane parallel to a certain direction and
absorbs waves whose electric field vectors are
perpendicular to that direction
Polarizing Sheets – Selective absorption
Io
I
2
Law of Malus:
I  I o cos 2 
Polarization by reflection
Brewster’s Angle:
n2
tan p 
n1
Polarization by Double Refraction
• Unpolarized light
splits into two planepolarized rays
• The two rays are in
mutual perpendicular
directions
– Indicated by the dots
and arrows
Polarization by Scattering, Rayleigh Scattering
• The horizontal part of the electric
field vector in the incident wave
causes the charges to vibrate
horizontally
• The vertical part of the vector
simultaneously causes them to
vibrate vertically
• If the observer looks straight up, he
sees light that is completely
polarized in the horizontal direction
Iscattered
1
4
Final Exam Problem 37
When unpolarized light is passed through two polarizing
filters in succession, its intensity is decreased by 80%.
The angle, , between the transmission axis of the
filters is
A. 78.5 degrees
B. 63.4 degrees

C. 26.6 degrees
D. 36.9 degrees
I=0.2Io
E. 50.8 degrees
Polaroids
3.
A screen is placed 50.0 cm from a single slit, which is
illuminated with 690-nm light. If the distance between the first
and third minima in the diffraction pattern is 3.00 mm, what is the
width of the slit?
6.
Light of wavelength 587.5 nm illuminates a single slit
0.750 mm in width. (a) At what distance from the slit should a
screen be located if the first minimum in the diffraction pattern is
to be 0.850 mm from the center of the principal maximum? (b)
What is the width of the central maximum?
18.
A binary star system in the constellation Orion has an
angular interstellar separation of 1.00 × 10–5 rad. If λ = 500 nm,
what is the smallest diameter the telescope can have to just
resolve the two stars?
41.
Plane-polarized light is incident on a single polarizing
disk with the direction of E0 parallel to the direction of the
transmission axis. Through what angle should the disk be
rotated so that the intensity in the transmitted beam is
reduced by a factor of (a) 3.00, (b) 5.00, (c) 10.0?
45.
The critical angle for total internal reflection for
sapphire surrounded by air is 34.4°. Calculate the polarizing
angle for sapphire.