Physics 1402: Lecture 34 Today

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Physics 1402 Lecture 34Todays Agenda

• Announcements
• Midterm 2 graded soon
• Homework 09 Friday December 4
• Optics
• Interference
• Diffraction
• Introduction to diffraction
• Diffraction from narrow slits
• Intensity of single-slit and two-slits
  diffraction patterns
• The diffraction grating

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Interference
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A wave through two slits
In Phase, i.e. Maxima when DP d sinq nl
Out of Phase, i.e. Minima when DP d sinq
(n1/2)l
q
Screen
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A wave through two slits
In Phase, i.e. Maxima when DP d sinq nl
Out of Phase, i.e. Minima when DP d sinq
(n1/2)l
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The Intensity

• What is the intensity at P?

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The Intensity
We can rewrite intensity at point P in terms of
distance y
Using this relation, we can rewrite expression
for the intensity at point P as function of y
Constructive interference occurs at
where m/-1, /-2
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Phasor Addition of Waves
Consider a sinusoidal wave whose electric field
component is
Consider second sinusoidal wave
The projection of sum of two phasors EP is equal
to
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Phasor Diagrams for TwoCoherent Sources
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SUMMARY 2 slits interference pattern (Youngs
experiment)
How would pattern be changed if we add one or
more slits ? (assuming the same slit separation )
3 slits, 4 slits, 5 slits, etc.
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Phasor 1 vector represents 1 traveling wave
2 wave interference
single traveling wave
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N-slits Interference Patterns
F0
F90
F180
F270
F360
N2
N4
N3
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Change of Phase Due to Reflection
The reflected ray (red) can be considered as an
original from the image source at point I. Thus
we can think of an arrangement S and I as a
double-slit source separated by the distance
between points S and I.
Lloyds mirror
P2
S
P1
An interference pattern for this
experimental setting is really observed .. but
dark and bright fringes are reversed in order
L
Mirror
I
This mean that the sources S and I are different
in phase by 1800
An electromagnetic wave undergoes a phase change
by 1800 upon reflecting from the medium that has
a higher index of refraction than that one in
which the wave is traveling.
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Change of Phase Due to Reflection
n1
n2
n1
n2
1800 phase change
no phase change
n1ltn2
n1gtn2
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Interference in Thin Films
1800 phase change
no phase change
A wave traveling from air toward film undergoes
1800 phase change upon reflection. The
wavelength of light ln in the medium with
refraction index n is
The ray 1 is 1800 out of phase with ray 2 which
is equivalent to a path difference ln/2. The ray
2 also travels extra distance 2t.
Constructive interference
Destructive interference
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Chapter 34 Act 1
Estimate minimum thickness of a soap-bubble film
(n1.33) that results in constructive
interference in the reflected light if the film
is Illuminated by light with l600nm.
B) 250nm
A) 113nm
C) 339nm
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Problem
Consider the double-slit arrangement shown in
Figure below, where the slit separation is d and
the slit to screen distance is L. A sheet of
transparent plastic having an index of refraction
n and thickness t is placed over the upper slit.
As a result, the central maximum of the
interference pattern moves upward a distance y.
Find y
where will the central maximum be now ?
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Solution
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Phase Change upon Reflection from a
Surface/Interface
Reflection from Optically Denser Medium (larger n)
Reflection from Optically Lighter Medium (smaller
n)
by analogy to reflection of traveling wave in
mechanics
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constructive 2t (m 1/2) ln destructive
2t m ln
Examples
constructive 2t m ln destructive 2t
(m 1/2) ln
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Application Reducing Reflection in Optical
Instruments
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Diffraction
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Experimental Observations (pattern produced by a
single slit ?)
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How do we understand this pattern ?
First Destructive Interference
(a/2) sin Q l/2
sin Q l/a
Second Destructive Interference
(a/4) sin Q l/2
sin Q 2 l/a
mth Destructive Interference
sin Q m l/a m1, 2,
See Huygens Principle
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So we can calculate where the minima will be !
sin Q m l/a m1, 2,
So, when the slit becomes smaller the central
maximum becomes ?
Why is the central maximum so much stronger than
the others ?
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Phasor Description of Diffraction
Lets define phase difference (b) between first
and last ray (phasor)
b S (Db) N Db
(a/l? sin Q 1 1st min.
Db / 2p Dy sin (Q) / l

• N Db
• N 2p Dy sin (Q) / l
• 2p a sin (Q) / l

Can we calculate the intensity anywhere on
diffraction pattern ?
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Yes, using Phasors !
Let take some arbitrary point on the diffraction
pattern This point can be defined by angle Q or
by phase difference between first and last ray
(phasor) b
The resultant electric field magnitude ER is
given (from the figure) by
sin (b/2) ER / 2R
The arc length Eo is given by Eo R b
ER 2R sin (b/2) 2 (Eo/ b) sin (b/2)
Eo sin (b/2) / (b/2)
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Other Examples
Light from a small source passes by the edge of
an opaque object and continues on to a screen. A
diffraction pattern consisting of bright and dark
fringes appears on the screen in the region above
the edge of the object.

• What type of an object would create a diffraction
  pattern shown on the left, when positioned midway
  between screen and light source ?
• A penny,
• Note the bright spot at the center.

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Fraunhofer Diffraction(or far-field)
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Fresnel Diffraction(or near-field)
Lens
P
Incoming wave
Screen
(more complicated not covered in this course)
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Resolution (single-slit aperture)

• Rayleighs criterion
• two images are just resolved WHEN
• When central maximum of one image falls on
• the first minimum of another image

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Resolution (circular aperture)
Diffraction patterns of two point sources for
various angular separation of the sources
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EXAMPLE
A ruby laser beam (l 694.3 nm) is sent outwards
from a 2.7-m diameter telescope to the moon, 384
000 km away. What is the radius of the big red
spot on the moon?
a. 500 m b. 250 m c. 120 m d. 1.0 km e.
2.7 km
Qmin 1.22 ( l / a) R / 3.84 108 1.22 6.943
10-7 / 2.7 R 120 m !
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Two-Slit Interference Pattern with a Finite Slit
Size
Itot Iinter . Idiff
smaller separation between slits gt ?
The combined effects of two-slit and single-slit
interference. This is the pattern produced when
650-nm light waves pass through two 3.0- mm slits
that are 18 mm apart.
smaller slit size gt ?
Animation
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Example
The centers of two slits of width a are a
distance d apart. Is it possible that the first
minimum of the interference pattern occurs at the
location of the first minimum of the diffraction
pattern for light of wavelength l ?
1st minimum interference d sin Q l /2
1st minimum diffraction a sin Q l
No!
The same place (same Q) l /2d l /a a /d 2
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Application X-ray Diffraction by crystals
Can we determine the atomic structure of the
crystals, like proteins, by analyzing X-ray
diffraction patters like one shown ?
A Laue pattern of the enzyme Rubisco, produced
with a wide-band x-ray spectrum. This enzyme is
present in plants and takes part in the process
of photosynthesis.
Yes in principle this is like the problem of
determining the slit separation (d) and slit
size (a) from the observed pattern, but much much
more complicated !
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Determining the atomic structure of crystals With
X-ray Diffraction (basic principle)
Crystals are made of regular arrays of atoms that
effectively scatter X-ray
Scattering (or interference) of two X-rays from
the crystal planes made-up of atoms
Braggs Law
Crystalline structure of sodium chloride (NaCl).
length of the cube edge is a 0.562 nm.