Download Document

Reflection
Ch. 35
•What happens when our wave hits a conductor?
Ei  E0 sin  kx  t 
•E-field vanishes in a conductor
•Let’s say the conductor is at x = 0
Er  E0 sin  kx  t 
•Add a reflected wave going other direction
•In reality, all of this is occurring in
three dimensions
Incident Wave
Reflected Wave
Total Wave
Waves going at angles
•Up to now, we’ve only considered waves going in the x- or y-direction
•We can easily have waves going at angles as well
2
2

c
k

k
  ck
x
y
Ei  E0i sin  k x x  k y y  t 

•What will reflected wave look like? E r  E0 r sin k x x  k y y   t
•Assume it is reflected at x = 0
   c k x2  k y2
•It will have the same angular frequency
  
•Otherwise it won’t match in time
•It will have the same ky value
k y  k y
•Otherwise it won’t match at
boundary
•kx must be negative
• So it is going the other way
c k x2  k y2  c k x2  k y2
k x  k x
k x  k x

Law of Reflection
•Since the frequency of all waves are the same, the total k
for the incident and reflected wave must be the same.
•To match the wave at the boundary, ky
must be the same before and after
ki sini
ki sini = kr sinr
sini = sinr
i = r
ki = kr
kr sinr
i r
Mirror
y
x
Geometric Optics and the Ray
Approximation
i = r
•The wave calculations we have done assume
the mirror is infinitely large
•If the wavelength is sufficiently tiny compared
i r
to objects, this might be a good approximation
•For the next week, we will always make
this approximation
Mirror
•It’s called geometric optics
•In geometric optics, light waves are represented by rays
•You can think of light as if it is made of little particles
•In fact, waves and particles act very similarly
•First hint of quantum mechanics!
Concept Question
•This works for any angle
•In 3D, you need three mirrors
 = 47
47
43
47
47
4343
Mirror
Mirror
A light ray starts from a wall at an angle
of 47 compared to the wall. It then
strikes two mirrors at right angles
compared to each other. At what angle
 does it hit the wall again?
A) 43 B) 45 C) 47 D) 49 E) 51
Measuring the speed of light
•Take a source which produces EM waves with a known frequency
•Hyperfine emission from 133Cs atom
•This frequency is extremely stable
•Better than any other method of measuring time
•Defined to be frequency f = 9.19263177 GHz
•Reflect waves off of mirror
½
½
•The nodes will be separated by ½
•Then you get c from c = f
•Biggest error comes from
133Cs
measuring the distance
•Since this is the best way to
measure distance, we can use this to define the meter
•Speed of light is now defined as 2.99792458108 m/s
The Speed of Light in Materials
•The speed of light in vacuum c is the same for all wavelengths of
light, no matter the source or other nature of light c  3.00 108 m/s
•Inside materials, however, the speed of light can be different
•Materials contain atoms, made of nuclei and electrons
•The electric field from EM waves push on the electrons
•The electrons must move in response
Indices of Refraction
•This generally slows the wave down
Air (STP)
1.0003
•n is called the index of refraction
Water
1.333
•The amount of slowdown can depend
Ethyl alcohol 1.361
on the frequency of the light
Glycerin
1.473
c
v
n
Fused Quartz 1.434
Glass
1.5 -ish
Cubic zirconia 2.20
Diamond
2.419
Refraction: Snell’s Law
ck

n
•The relationship between the angular frequency 
and the wave number k changes inside a medium
•Now imagine light moving from one medium to another  f  c n
•Some light will be reflected, but usually most is refracted
•The reflected light again must obey the law of reflection 1 = r
•Once again, the frequencies all match
k1sin1
•Once again, the y-component of k must match
k1
k2
c c
n1
n2
n2 k1  n1k2
index n1
index n2
y
k1 sin 1  k2 sin 2
n1n2 k1 sin 1  n2 n1k2 sin 2
n1 sin 1  n2 sin  2
Snell’s
Law
1 r
x
2
k2sin2
Snell’s Law: Illustration
A light ray in air enters a region at an
angle of 34. After going through a layer
of glass, diamond, water, and glass, and
back to air, what angle will it be at?
A) 34
B) Less than 34
C) More than 34
D) This is too hard
n5 = 1.5
6
n6 = 1
5
5
n3 = 2.4
n4 = 1.33
4 4
n1 = 1 n2 = 1.5
3
3
2
2
34
n1 sin 34  n2 sin 2 n3 sin 3
 n4 sin 4  n5 sin 5  n6 sin 6 n1 sin 1  n2 sin  2
1 sin 34  1 sin 6 6  34
Ex - (From MCAT practice book). If a ray is refracted from air into a
medium with n = 1.47 at an angle of incidence of 50, the angle of refraction
is
A. 0.059
B. 0.087
C. 0.128
D. 0.243
CT-1- A fish swims below the surface of the water at P. An observer at O
sees the fish at
A. a greater depth than it really is.
B. the same depth.
C. 3. a smaller depth than it really is.
CT –2 A fish swims below the surface of the water. Suppose an observer is
looking at the fish from point O'—straight above the fish. The observer sees
the fish at
A. a greater depth than it really is.
B. the same depth.
C. a smaller depth than it really is.
Ex- (Serway 35-21) An opaque cylindrical tank with an open top has a
diameter of 3.00 m and is completely filled with water. When the setting Sun
reaches an angle of 28.0° above the horizon, sunlight ceases to illuminate
any part of the bottom of the tank. How deep is the tank?
Solve on Board
Ex- (Serway 35-16) When the light illustrated below passes through the
glass block, it is shifted laterally by the distance d. If n = 1.50, what is the
value of d?
Solve on Board
Dispersion
•The speed of light in a material can depend on frequency
•Index of refraction n depends on frequency
•Confusingly, its dependence is often given as a function of
wavelength in vacuum
•Called dispersion
•This means that different types of light
bend by different amounts in any given
material
•For most materials, the index of refraction
is higher for short wavelength
Red Refracts Rotten
Blue Bends Best
Prisms
•Put a combination of many wavelengths (white light) into a triangular
dispersive medium (like glass)
•Prisms are rarely used in research
•Diffraction gratings work better
•Lenses are a lot like prisms
•They focus colors unevenly
•Blurring called chromatic dispersion
•High quality cameras use a combination of
lenses to cancel this effect
Rainbows
•A similar phenomenon occurs when light bounces off of the inside of a
spherical rain drop
•This causes rainbows
•If it bounces twice, you can
get a double rainbow
Total Internal Reflection
A trick question:
A light ray in diamond enters an air gap at an
angle of 60, then returns to diamond. What
angle will it be going at when it leaves out the
bottom?
A) 60
B) Less than 60
C) More than 60
D) None of the above
n1 = 2.4
60
2
2 n2 = 1
3
n3 = 2.4
n1 sin 60  n2 sin 2
sin 2  2.4  0.866   2.07
•This is impossible!
•Light never makes it into region 2!
•It is totally reflected inside region 1
•This can only happen if you go from a high index to a low
•Critical angle such that this occurs:
n2
sin  c 
•Set sin2 = 1
n
1
Optical Fibers
Protective
Jacket
Low n glass
High n glass
•Light enters the high index of refraction glass
•It totally internally reflects – repeatedly
•Power can stay largely undiminished for many kilometers
•Used for many applications
•Especially high-speed communications – up to 40 Gb/s
Fermat’s Principle (1)
•Light normally goes in straight lines. Why?
•What’s the quickest path between two points P and Q?
•How about with mirrors? Go from P to Q but touch the mirror.
•How do we make PX + XQ as short as possible?
•Draw point Q’, reflected across from Q
•XQ = XQ’, so PX + XQ = PX + XQ’
• To minimize PX + XQ’, take a straight line from P to Q’
i = r
P
Q
We can get: (1) light moves in
straight lines, and (2) the law
of reflection if we assume light
always takes the quickest path
between two poins
i r
X
i
Q’
Fermat’s Principle (2)
P
•What about refraction?
n1 sin 1  n2 sin  2
•What’s the best path from P to Q?
1
s1
•Remember, light slows down in glass
1 L – x
•Purple path is bad idea – it doesn’t avoid the d1
slow glass very much
x
d2
•Green path is bad too – it minimizes time
2
in glass, but makes path much longer
2
s2
•Red path – a compromise – is best
Light always takes
•To minimize, set derivative = 0
the quickest path
Q
n1s1 n2 s2 1 
s1 s2
2
2
2


t 

n1 x  d1  n2  L  x   d 22 

c
c
v1 v2
c 


n2  L  x   1  n1 sin 1  
1  n1 x
dt
 


0

2
2
2
2


c  n2 sin  2 
dx c
x  d1
L

x

d


2 
