Download Phys-272 Lecture 22

Phys-272 Lecture 22
Wave Propagation
Reflection
Refraction
Phys-272 Lecture 22
Reflection
Refraction
Huygen’s Principle
Dispersion
Light Scattering
Polarization
θi = θr
n1 sin θ1 = n2 sin θ 2
Index of Refraction (n)
• Speed of light, c, in vacuum is 3x108 m/s (2.99792458x108 m/s)
• Speed of light, v, in a medium can be lower i.e. v < c.
• index of refraction, n = c/v
•frequency, f, is unmodified as v = f l,
• wavelength, l, depends on medium, λ = vT = v/f = c/nf = λ0/n
• In some media, n, depends on f, this is called dispersion.
Waves, wavefronts, rays
Plane waves moving in +x direction
r
E ( x, y, z , t ) = E0 yˆ cos(kx − ωt )
r
B( x, y, z , t ) = B0 zˆ cos(kx − ωt )
Wavefront is a surface
of constant phase
E field wavefront (y-z plane surface of constant phase)
y
Ray propagation
z
x
Huygen’s Principle
• Huygen’s principle; a wave front can be a source of secondary
wavelets that spread out in all directions at the speed of propagation
in the medium. The envelope of leading edges forms a wave front.
• This principle was stated by Huygen in 1678, it can be derived
from Maxwell’s eqns. It is a geometrical description of ray
propagation.
vt
Plane wave example;
Secondary wavelets create
another wave front (plane)
Reflection from Huygen’s Principle
Consider wave fronts, separated by vt, the incident wave fronts in
contact with the surface will create a wavelets according to Huygen’s
Principle and leds to another “reflected” wave front. Result is qi = qr
reflected wave front
Incident wave front
vt
qr
vt
vt
vt
qi
qi = qr
Reflection Law
ray diagram
qi
qr
The same effect can be seen with water
waves in a ripple tank
Refraction from Huygen’s Principle
Now the speed changes, from medium a to medium b, so the
Speed may change and the wavefront spacing differs.
L sinqa = vat = ct/na
vat
L sinqb = vbt = ct/nb
vat
Medium a, va=c/na
Medium b,
vb=c/nb
qb
vbt
L
vbt
L
qa
e
v
a
W
t
n
fro
t
n
o
r
ef
v
a
W
na sinqa = nb sinqb
same thing happens to
marching soldiers
Snell’s Law
What happens to the frequency ? Nothing
ide
Inc
normal
Snell’s Law (law of refraction)
y
ra
nt
e
av
W
qa
Medium a, va=c/na
qa
qb
qb
a
ed r
t
n
o
r
ef
v
a
W
d
e
t
ac
r
f
e
R
ract
Ref
Medium b, vb=c/nb
t
n
fro
y
na sinqa = nb sinqb
Snell’s Law
Angles are
defined
relative to the
normal to plane
Clicker question
2) A ray of light passes from air into
water with an angle of incidence of
30o. Which of the following quantities
does not change as the light enters the
water. .
a) Wavelength
b) frequency
c) speed of propagation
d) direction of propagation.
Material
Index
Vacuum
1.00000
Air at STP
1.00029
Ice
1.31
Water at 20 C
1.33
Acetone
1.36
Ethyl alcohol
1.36
Sugar solution(30%)
1.38
Fluorite
Fused quartz
Glycerine
Sugar solution (80%)
Typical crown glass
Crown glasses
Spectacle crown, C-1
Sodium chloride
1.433
1.46
1.473
1.49
1.52
1.521.62
1.523
1.54
Indices of Refraction
Polystyrene
Carbon disulfide
Flint glasses
Heavy flint glass
Extra dense flint, EDF3
Methylene iodide
Sapphire
1.551.59
1.63
1.571.75
1.65
1.7200
1.74
Arsenic trisulfide glass
1.77
1.71.84
1.821.98
2.04
Diamond
2.417
Rare earth flint
Lanthanum flint
High Index n Lots of TIR
Clicker question
• Which of the following ray diagrams could represent the passage of light from air
through glass and back to air?
(a)
(nair=1 and nglass=1.5)
(b)
(c)
air
air
air
glass
glass
glass
air
air
air
Clicker question
• Which of the following ray diagrams could represent the passage of light from air
through glass and back to air? (nair=1 and nglass=1.5)
(a)
air
glass
air
q1
(b)
q2
(c)
air
air
glass
glass
air
air
• The behavior of these rays is determined from Snell’s Law:
n1 sin θ1 = n2 sin θ 2
• Since n(glass) > n(air), sinq (glass) < sinq (air) .
• Therefore, moving from air to glass, ray will bend toward normal.
• this eliminates (a).
• Moving from glass to air, ray will bend away from normal.
• this eliminates (c).
• As a matter of fact, the final angle in air must be equal to the initial
angle in air!!
EXAMPLE, from air into glass;
Suppose we have light in air (n=1)
incidence on glass (n=1.55) at an
angle qa=45 deg. What is the angle
of the refracted light, qb?
nair sin θ a = n glass sin θ b
(1) sin (45°) = (1.55) sin θ b
sin θ b =
( )
(1) sin (45°)
θ b = 27°
1.55
1
=
1.55 2
qa
qb
EXAMPLE, from glass into air;
Suppose we have light in glass (n=1.55)
incident into air at an angle qa=30 deg.
What is the angle of the refracted light, qb?
n glass sin θ a = nair sin θ b
(1.55) sin (30°) = (1) sin θ b
1.55
sin θ b = (1.55) sin (30°) =
2
θ b = 51°
qa
qb
Suppose you are stranded on a tropical island with no food.
You see a fish in the water. Where should you aim your spear
to hit the fish?
ANSWER; do not aim directly at the apparent position of
the fish. (Your spear will miss). Aim at the inside of the fish.
Suppose in the previous question instead of a spear you
had a high power laser to simultaneously kill and cook the
fish (in the water). Where should you aim the laser??
ANSWER; aim directly at apparent fish position as the laser
beam will refract to the correct fish position.
Total Internal Reflection
– Consider light moving from glass (n1=1.5) to air
(n2=1.0)
incident
reflected
sin θ 2 n1
ray
ray
=
>1
θ 2 > θ1
q1 qr
n1
sin θ1 n2
GLASS
n2
q2
refracted
ray
AIR
I.e., light is bent away from the normal.
as q1 gets bigger, q2 gets bigger, but q2
can never get bigger than 90°!!
In general, if sin q1 > (n2 / n1), we have NO refracted ray;
we have TOTAL INTERNAL REFLECTION.
For example, light in water which is incident on an air surface
with angle q1 > qc = sin-1(1.0/1.33) = 48.8° will be totally reflected.
This property is the basis for the optical fiber communication.
Total Internal Reflection of a Laser Beam
Material: PMMA
Endoscope used
by physicians to
look inside body.
Clicker question
I) The path of light is bent as it passes from medium
1 to medium 2. Compare the indexes of refraction in
the two mediums.
a) n1 > n2
b) n1 = n2
Snell’s Law: n1sinq1 = n2sinq2
Here, q2 >q1 implies n2 < n1
c) n1 < n2
II) A light ray travels in a medium with n1 and
completely reflects from the surface of a
medium with n2. The critical angle depends on:
a) n1 only
b) n2 only
c) n1 and n2
Critical angle occurs when q2 = 90o
Therefore, sinqcritical = n2/n1
Clicker question on Critical Angle…
An optical fiber is
surrounded by another
dielectric. In case I this is
water, with an index of
refraction of 1.33, while in
case II this is air with an
index of refraction of 1.00.
Compare the critical angles
for total internal reflection
in these two cases
a) qcI>qcII
b) qcI=qcII
c) qcI<qcII
water n =1.33
Case I
qc
glass n =1.5
water n =1.33
Case II
air n =1.00
qc
glass n =1.5
air n =1.00
Clicker question on Critical Angle…
An optical fiber is
surrounded by another
dielectric. In case I this is
water, with an index of
refraction of 1.33, while in
case II this is air with an
index of refraction of 1.00.
qc
glass n =1.5
water n =1.33
Case II
Compare the critical angles
for total internal reflection
in these two cases
a) qcI>qcII
water n =1.33
Case I
air n =1.00
qc
glass n =1.5
air n =1.00
n2
b) qcI=qcII
n1
c) qcI<qcII
Since n1>n2 TIR will occur for q > critical angle.
Snell’s law says sinqc=n2/n1.
n1>n2
If n2=1.0, then qc is as small as it can be.
So qcI >qcII .
Total Internal Reflection
Total internal reflection occurs when q>qc and provides 100%
reflection. This has better efficiency than silvered mirror.
Examples of devices using Critical Angle
• Prism Binoculars
• Fiber Optics
Fiber optics is extremely important for high speed
Internet and digital data transfer at long distances.
Many companies (Lucent, Oceanic Cable…) have
laid fiber over long distances to provide internet service.
Dispersion: n = n(w)
The index of refraction depends on
frequency, due to the presence of
resonant transition lines.
For example, ultraviolet absorption
bands in glass cause a rising index
of refraction in the visible, i.e.,
n(higher w) > n(lower w):
Index of refraction
nred = 1.52
nblue = 1.53
1.54
ultraviolet
absorption
bands
1.52
white light
prism
1.50
frequency
Split into Colors
Rainbows (how they
form)
Rainbow
s
Hence we also see a faint
secondary rainbow
A meter stick lies at the bottom of a
rectangular water tank of height 50cm.
You look into the tank at an angle of
45o relative to vertical along a line that
skims the top edge of the tank.
45o
nwater = 1.33
50 cm
What is the smallest number on the
ruler that you can see?
0
Conceptual Analysis:
- Light is refracted at the surface of the water
Strategy:
- Figure out the angle of refraction in the water
and extrapolate this to the bottom of the tank.
20
40
60
80
100
A meter stick lies at the bottom of a
rectangular water tank of height 50cm.
You look into the tank at an angle of
45o relative to vertical along a line that
skims the top edge of the tank.
45o
nwater = 1.33
qR
50 cm
0
20
If you shine a laser into the tank at an angle of 45o, what is the
refracted angle qR in the water ?
A) qR = 28.3o
B) qR = 32.1o
C) qR = 38.7o
nairsin(45) = nwatersin(qR)
sin(qR) = nairsin(45)/nwater = 0.532
qR = sin-1(0.532) = 32.1o
40
60
80
100
A meter stick lies at the bottom of a
rectangular water tank of height 50cm.
You look into the tank at an angle of
45o relative to vertical along a line that
skims the top edge of the tank.
45o
nwater = 1.33
qR
50 cm
0
20
40
60
80
100
qR = 32.1o
What number on the ruler does the laser beam hit ?
A) 31.4 cm
B) 37.6 cm
C) 44.1 cm
A meter stick lies at the bottom of a
rectangular water tank of height 50cm.
You look into the tank at an angle of
45o relative to vertical along a line that
skims the top edge of the tank.
45o
nwater = 1.33
qR
50 cm
0
20
d
What number on the ruler does the laser beam hit ?
A) 31.4 cm
B) 37.6 cm
C) 44.1 cm
tan(qR) = d/50
d = tan(32.1) x 50cm = 31.4cm
40
60
80
100
qR = 32.1o
A meter stick lies at the bottom of a
rectangular water tank of height 50cm.
You look into the tank at an angle of
45o relative to vertical along a line that
skims the top edge of the tank.
45o
nwater = 1.33
50 cm
What is the smallest number on the
ruler that you can see?
0
20
40
If the tank were half full of water, would the laser hit a bigger or a
smaller number than it did when the tank was full of water ?
A) bigger
B) smaller
C) The same number
60
80
100
45o
45o
qR
45o
50 cm
50 cm
d = 31.4 cm
0
20
40
d = 50 cm
60
80
100
0
20
45o
50 cm
qR
0
20
d = 40.7 cm
40
25 cm + (31.4 / 2) cm
60
80
100
40
60
80
100