Quantum Physics
Many physical phenomena of great practical interest to engineers
chemists, biologists, physicists, etc. were not in Gen. Phys. I & II
Quantum Physics:
The development of experimental equiment and techniques
modern physics can go inside the microscopic world (atoms,
electrons, nucleus, etc.)
New principles, new laws for the microscopic (subatomic)
world were discoverved
GENERAL PHYSICS III
Optics
&
Quantum Physics
What does we learn in Gen. Phys. III?
Many physical phenomena of great practical interest to engineers,
chemists, biologists, physicists, etc. were not in Gen. Phys. I & II
Wave phenomena of light:
Interference: what happens when two or more waves overlap?
Interference!
(Light passing through two slits give such kind of picture)
• Diffraction: The image of an object is not
exact in fine details.
For example, the image of a
circular disk is diffused
Interference & diffraction can be
analyzed if we regard light as a wave
Quantum Physics:
The development of experimental equiment and techniques
modern physics can go inside the microscopic world (atoms,
electrons, nucleus, etc.)
New principles, new laws for the microscopic (subatomic)
world were discoverved
The basis concept: wave particle duality
Examples: light wave photon
electron electron wave etc…
• Subatomic objects obey new mechanics: quantum mechanics
On the basis of quatum mechanics we study structure and
properties of atoms, nucleus, solids, laser rays, etc…
Chapter XVII
Interference of light
§1. Interference of coherent sources of light
§2. Interference in thin films
§3. Interferometer
§1. Interference of light from coherent sources:
1.1 Coherent sources of light:
We consider an overlap of light that comes from two sources.
A remarkable phenomen takes place, if two sources satisfy some
following conditions:
The sources are monochromatic. It means that they emit light
of a single color.
A monochromatic light corresponds to a sinusoidal electromagnetic
wave with a single frequency f and wave length
Two sources have the same frequency f (the same wave length )
Two souces are permanently in phase, or , at least, have any definite
constant phase difference
Then, two sources are called coherent sources.
Notes:
Recall the formula for a sinusoidal e-m wave:
2
y , t A cos A cos ft A cos t
x x vt kx 2 kx
v speed A amplitude
2
wavelength k wavenumber or wavevector
f frequency 2f angular frequency
Common sources of light do not emit monochromatic light
(single-frequency light)
→ However one can produce approximately monochromatic light:
• by using filters which block all but a very narrow range
of wave length
• by using light from a laser
1.2 Interference of light through narrow slits:
Young’s experiment on double-slit interference
double-
(Thomas Young performed in 1800)
Monochromatic
light source at a
great distance, Observation
or a laser.
Slit pattern screen
Light (wavelength is incident on a two-slit (two narrow,
rectangular openings) apparatus:
I1
If either one of the slits is closed, a
diffuse image of the other slit will appear S1
on the screen. (The image will be
“diffuse” due to diffraction. We will
discuss this effect in more detail
later.) S2 Diffraction
profile
Monochromatic light
screen
If both slits are now open, we see (wavelength )
interference “fringes” (light and dark
bands), corresponding to constructive S1
and destructive interference of the
electric-field amplitudes from both
slits.
S2
Light fringes
Dark fringes I
Observer
r1
S1
Light r2
d
S2
Important quantity: path difference r2 - r1
=
The light density at the location of observer depends on the
path difference
A path difference corresponds to a phase difference
of
two waves at the observer’s point
One has a simple formula for the path difference, ,
when the observer is far from sources.
sources.
(Assume 2 sources radiating in phase)
Observer
d r Normal to d
When observer distance >> slit spacing
(r >> d) : = dsin
d
The corresponding phase difference
at the observer’s point:
sin
d
At observer’s points that satisfy the condition
= 0, ±1, ±2,... → two waves
m
m
are in phase and reinforce each other, we say
/d
that there is constructive interference
at these points y
r /d
Constructive
= dsin= m Interference
d
0
If m + 1)→ two waves
I
cancell each other → there is -/d
destructive interference
Destructive L
= dsin= (m + 1/ )
2 Interference
m=2 Usually we care about the linear
“lines” of
(as opposed to angular)
constructive m=1 displacement y of the pattern
interference: (because our screens are often
m=0
= sin-1(md
flat):
m=-1 y = L tan
m=-2
The slit-spacing d is often large compared to so that small.
, is
Then we can use the small angle approximations to simplify our results:
For small angles: ( Example:
S1
A laser of wavelength 633 nm is y
incident on two slits separated by
0.125 mm.
S2
I
1. What is the spacing between fringe maxima on a screen 2m away?
y
a. 1 m b. 1 mm c. 1 cm
2. If we increase the spacing between the slits, what will happen to y?
a. decrease b. stay the same c. increase
3. If we instead use a green laser (smaller will?
), y
a. decrease b. stay the same c. increase
Example (continue)
S1
A laser of wavelength 633 nm is y
incident on two slits separated by
0.125 mm.
S2
1. What is the spacing between fringe maxima on a screen 2m away?
y
a. 1 m b. 1 mm c. 1 cm
First question: can we use the small angle approximation?
d = 125 m; = 0.633
m d >> is small
d sin = m i ~ d
i i mi (
i d)
y L( – L(2 – 1) ( = L = (2 m)(0.663
2 1) d)
/d m)/125 = 0.01 m
m
2. If we increase the spacing between the slits, what will happen to y?
a. decrease b. stay the same c. increase
3. If we instead use a green laser (smaller will?
), y
a. decrease b. stay the same c. increase
Example (continue)
S1
y
A laser of wavelength 633 nm is
incident on two slits separated by
0.125 mm.
S2
1. What is the spacing between fringe maxima on a screen 2m away?
y
a. 1 m b. 1 mm c. 1 cm
2. If we increase the spacing between the slits, what will happen to y?
a. decrease b. stay the same c. increase
Since ~ 1/d, the spacing decreases. Note: This is a general phenomenon
y
– the “far-field” interference pattern varies inversely with slit dimensions.
3. If we instead use a green laser (smaller will?
), y
a. decrease b. stay the same c. increase
Since ~ the spacing decreases.
y ,
Note:
If a monochromatic source is replaced by a white light one →
how is the interference picture?
We have known that the location of (light or dark) fringers depends
on the wavelength
therfore
• At y = 0 (the center light fringer) the maxima for all wavelengths
coincide → there is a white light fringer
• The nearby fringers have spectrum colors (like in rainbow)
• Far fringers are not visible (are diffused)
§2. Interference in thin films:
This kind of interference takes place when light reflects from a
thin film (for example, a soap bubble, a thin layer of oil floating
on water)
In this case there is an overlap
eye
of light waves: one is reflected
from the upper, the other from
the lower surface.
2.1 Calculation of path difference of two light waves:
First we consider the case that the incident light is monochromatic
with the wavelength
b: the thickness of the film
n: the index of refraction
path difference between
the
the reflected waves 1 & 2
By elementary geometry it is
not difficult to obtain
Using the formula one can eliminate i1 or i2
We must take into account one more effect: the phase shift of
a wave after reflection
We use the follwing theoretical results from Maxwell’s theory of
electromagetism:
• If na > nb → the phase shift of
reflected wave relative to the incident na
wave is zero
nb
• If na < nb → the phase shift of
reflected wave relative to the incident
wave is radian ( a half cycle)
We can take into account this phase shift by introducing a
complementary term in the formula of path difference
Remark:
The path difference of two reflected waves depends on b, i1 (or i2)
At the points that satisfy
m (m = 0, ±1, ±2,...) → we have
cnstructive interference
the points that satisfy = (m+1/2) destructive interference
At
→
