Physics 1202: Lecture 25 Today’s Agenda Announcements: –Midterm 2: Friday Nov. 6… –Chap. 18,...
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Transcript of Physics 1202: Lecture 25 Today’s Agenda Announcements: –Midterm 2: Friday Nov. 6… –Chap. 18,...
Physics 1202: Lecture 25Today’s Agenda
• Announcements:
– Midterm 2: Friday Nov. 6…
– Chap. 18, 19, 20, and 21
• Homework #7:Homework #7:– Due today at 11:59 PM !Due today at 11:59 PM !
– No HW for next week (midterm)No HW for next week (midterm)
• Optics – interference
Interference
A wave through two slits
Screen
P=d sin
d
In Phase, i.e. Maxima when P = d sin = nOut of Phase, i.e. Minima when P = d sin = (n+1/2)
A wave through two slitsIn Phase, i.e. Maxima when P = d sin = n
Out of Phase, i.e. Minima when P = d sin = (n+1/2)
+
+
The IntensityWhat is the intensity at P?
The only term with a t dependence is sin2( ).That term averages to ½ .
If we had only had one slit, the intensity would have been,
So we can rewrite the total intensity as,
with
The Intensity
We can rewrite intensity at point Pin 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 …
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
E0E1(t) t
E2(t)E0
EP(t)
ER
/2
E0
tE1(t)
t+E0
E2(t)
Phasor Diagrams for TwoCoherent Sources
ER=2E0
E0 E0 E0
E0
ER
450
E0
E0ER
900
ER=0
E0 E0
E0
E0ER
2700 ER=2E0
E0 E0
SUMMARY2 slits interference pattern (Young’s 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.
Phasor: 1 vector represents 1 traveling wave
single traveling wave 2 wave interference
N=2 N=4N=3
N-slits Interference Patterns
Change of Phase Due to Reflection
Lloyd’s mirror P2
P1
S
I
LMirror
The reflected ray (red) can be considered asan 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.
An interference pattern for this experimentalsetting is really observed …..but dark and bright fringes are reversed in order
This mean that the sources S and I are different in phase by 1800
An electromagnetic wave undergoes a phase change by 1800 uponreflecting from the medium that has a higher index of refraction thanthat one in which the wave is traveling.
Change of Phase Due to Reflection
1800 phase change
n1 n2
n1<n2
n1 n2
no phase change
n1>n2
Interference in Thin Films
Air
Air
Film t
12
1800 phase change no phase
changeA wave traveling from air toward filmundergoes 1800 phase change upon reflection.The wavelength of light n 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 n/2.The ray 2 also travels extra distance 2t.
Constructive interference
Destructive interference
Chapter 25 – Act 1
Estimate minimum thickness of a soap-bubble film (n=1.33) thatresults in constructive interference in the reflected light if the film isIlluminated by light with =600nm.
A) 113nm B) 250nm C) 339nm
ProblemConsider 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 thecentral
maximum be now ?
Solution
Corresponding path length difference:
Phase difference for going though plastic sheet:
Angle of central max is approx: Thus the distance y’ is:
gives
Phase Change upon Reflection from a Surface/Interface
Reflection fromOptically Denser Medium (larger n)
Reflection fromOptically Lighter Medium (smaller n)
by analogy to reflection of traveling wave in mechanics
180o Phase Change No Phase Change
Examples :constructive: 2t = (m +1/2) n
destructive: 2t = m n
constructive: 2t = m n
destructive: 2t = (m +1/2) n
ApplicationReducing Reflection in Optical Instruments
Diffractio
n
Experimental Observations:(pattern produced by a single slit ?)
First Destructive Interference:
(a/2) sin = ± /2
sin = ± /a
mth Destructive Interference:
(a/4) sin = ± /2
sin = ± 2/a
Second Destructive Interference:
sin = ± m /a m=±1, ±2, …
How do we understand this pattern ?
See Huygen’s Principle
So we can calculate where the minima will be !
sin = ± m /a m=±1, ±2, …
Why is the central maximum so much stronger than the others ?
So, when the slit becomes smaller the central maximum becomes ?
Phasor Description of Diffraction
Can we calculate the intensity anywhere on diffraction pattern
?
= = N
/ 2 = y sin () /
= N = N 2 y sin () / = 2 a sin () /
Let’s define phase difference () between first and last ray (phasor)
1st min.
2nd max.
central max.
(a/ sin = 1: 1st min.
Yes, using Phasors !Let take some arbitrary point on the diffraction pattern
This point can be defined by angle or by phase difference between first and last ray (phasor)
The arc length Eo is given by : Eo = R
sin (/2) = ER / 2R
The resultant electric field magnitude ER is given (from the figure) by :
ER = 2R sin (/2) = 2 (Eo/ ) sin (/2) = Eo [ sin (/2) / (/2) ]
I = Imax [ sin (/2) / (/2) ]2
So, the intensity anywhere on the pattern :
= 2 a sin () /
Other Examples
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.
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.
Fraunhofer Diffraction(or far-field)
Incoming wave
Lens
Screen
Fresnel Diffraction(or near-field)
Incoming wave
Lens
Screen
P
(more complicated: not covered in this course)
Resolution(single-slit aperture)
Rayleigh’s criterion:• two images are just resolved WHEN:
When central maximum of one image falls on the first minimum of another image
sin = / a
min ~ / a
Diffraction patterns of two point sources for various angular separation of the sources
Resolution(circular aperture)
min = 1.22 ( / a)
Rayleigh’s criterionfor
circular aperture:
EXAMPLE
A ruby laser beam ( = 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 mb. 250 mc. 120 md. 1.0 kme. 2.7 km min = 1.22 ( / a)
R / 3.84 108 = 1.22 [ 6.943 10-7 / 2.7 ]
R = 120 m !
EarthMoon
Two-Slit Interference Pattern with a Finite Slit Size
Idiff = Imax [ sin (/2) / (/2) ]2
Diffraction (“envelope” function):
= 2 a sin () /
Itot = Iinter . Idiff
Interference (interference fringes):
Iinter = Imax [cos (d sin / ]2
smaller separation between slits => ?
smaller slit size => ?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. Animation
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 ?
da
a
1st minimum interference:d sin = /2
1st minimum diffraction:a sin =
The same place (same ) : /2d = /a
a /d =
No!
ApplicationX-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 !
Determining the atomic structure of crystalsWith X-ray Diffraction (basic principle)
2 d sin = m m = 1, 2, ..Crystalline structure of sodium chloride (NaCl). length of the cube edge is a = 0.562 nm.
Crystals are made of regular arrays of atoms
that effectively scatter X-ray
Bragg’s Law
Scattering (or interference) of two X-rays from the
crystal planes made-up of atoms