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Reflection• What happens when our wave hits a conductor?
• E-field vanishes in a conductor• Let’s say the conductor is at x = 0
• Add a reflected wave going other direction• In reality, all of this is occurring in
three dimensions
i 0 sin kx t E E
0 sinr kx t E E
Incident WaveReflected Wave
Total Wave
Ch. 35
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
0 sini i x yk x k y t E E2 2x yc k k ck
• What will reflected wave look like?• Assume it is reflected at x = 0
• It will have the same angular frequency• Otherwise it won’t match in time
• It will have the same ky value• Otherwise it won’t match at
boundary• kx must be negative
• So it is going the other way
0 sinr r x yk x k y t E E
y yk k
2 2x yc k k
2 2 2 2x y x yc k k c k k
x xk k x xk k
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
Mirror
IncidentRefl
ected
ki k r
x
y
i r
ki sini kr sinr
ki = kr
ki sini = kr sinr
sini = sinr i = r
Geometric Optics and the Ray Approximation
• The wave calculations we have done assumethe mirror is infinitely large
• If the wavelength is sufficiently tiny comparedto objects, this might be a good approximation
• For the next week, we will always makethis approximation• 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!
Mirror
i r
i = r
Concept Question
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 M
irro
r
47
4747
4343
43
= 47
• This works for any angle• In 3D, you need three mirrors
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• The nodes will be separated by ½ • Then you get c from c = f• Biggest error comes from
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 2.99792458108 m/s
133Cs
½ ½
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 83.00 10 m/sc
• 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• This generally slows the wave down
• n is called the index of refraction• The amount of slowdown can depend
on the frequency of the light
cv
n
Indices of RefractionAir (STP) 1.0003Water 1.333Ethyl alcohol 1.361Glycerin 1.473Fused Quartz 1.434Glass 1.5 -ishCubic zirconia 2.20Diamond 2.419
Refraction: Snell’s Law• The relationship between the angular frequency and the wave number k changes inside a medium
ck
n
• Now imagine light moving from one medium to another• Some light will be reflected, but usually most is refracted
• The reflected light again must obey the law of reflection• Once again, the frequencies all match• Once again, the y-component of k must match
index n1
2
1 r
1 = r
1 2
1 2
k kc c
n n
index n2
x
y1 1 2 2sin sink k
k1sin1
k2sin2
2 1 1 2n k n k
1 2 1 1 2 1 2 2sin sinn n k n n k
1 1 2 2sin sinn n
f c n
Snell’s Law
34
Snell’s Law: Illustration
1 1 2 2sin sinn n
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 34C) More than 34 D) This is too hard
n1 = 1 n2 = 1.5 n3 = 2.4
n4 = 1.33 n5 = 1.5
n6 = 1
6
2 2
3
3
4 4
5 5
1 2 2sin 34 sinn n 3 3sinn
4 4 5 5 6 6sin sin sinn n n
61 sin 34 1 sin 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 35-27) 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 35-22) 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 lightbend by different amounts in any givenmaterial
• For most materials, the index of refractionis 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 ReflectionA 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 60C) More than 60 D) None of the above
60
22
3
1 2 2sin 60 sinn n 2sin 2.4 0.866 2.07
n1 = 2.4
n3 = 2.4
n2 = 1
• 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:
• Set sin2 = 12
1
sin c
n
n
Optical FibersProtective
Jacket
• 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
Low n glass High n glass
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’
P
Q
X
Q’
ri
i
i = r
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
Fermat’s Principle (2)• What about refraction?
• What’s the best path from P to Q?• Remember, light slows down in glass
• Purple path is bad idea – it doesn’t avoid theslow glass very much
• Green path is bad too – it minimizes timein glass, but makes path much longer
• Red path – a compromise – is best• To minimize, set derivative = 0
P
Q
d1
x
L – x
d2
s1
s2
1 2
1 2
s st
v v 1 1 2 2n s n s
c c 22 2 2
1 1 2 2
1n x d n L x d
c
0dt
dx
21
2 2 2 21 2
1 n L xn x
c x d L x d
1 1
2 2
sin1
sin
n
nc
1
2
1 1 2 2sin sinn n
Light always takes the quickest path
1
2