Reflection

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# Reflection - PowerPoint PPT Presentation

and Refraction. Reflection. Ch. 35. 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.

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Presentation Transcript

and Refraction

Reflection

Ch. 35

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 inthree dimensions

Incident WaveReflected WaveTotal 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
• 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 kyvalue
• Otherwise it won’t match atboundary
• kx must be negative
• So it is going the other way

Law of Reflection

ki=kr

• 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, kymust be the same before and after

kisini

krsinr

kisini=krsinr

ki

kr

sini=sinr

i=r

Incident

Reflected

i

r

Mirror

y

x

Geometric Optics and the Ray Approximation

i=r

• 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!

i

r

Mirror

Concept Question

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

= 47

47

43

47

Mirror

47

43

43

• This works for any angle
• In 3D, you need three mirrors

Mirror

Measuring the speed of light

½

½

• 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 frommeasuring the distance
• Since this is the best way tomeasure distance, we can use this to define the meter
• Speed of light is now defined as 2.99792458108 m/s

133Cs

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
• 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 dependon the frequency of the light

Indices of Refraction

Air (STP) 1.0003

Water 1.333

Ethyl alcohol 1.361

Glycerin 1.473

Fused Quartz 1.434

Glass 1.5 -ish

Cubic zirconia 2.20

Diamond 2.419

Refraction: Snell’s Law

k1sin1

1

r

2

k2sin2

• The relationship between the angular frequency  and the wave number k changes inside a medium
• 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

1=r

index n1

index n2

y

x

Snell’s Law

Snell’s Law: Illustration

34

2

n4 = 1.33

2

n5 = 1.5

3

3

n6 = 1

4

4

n2 = 1.5

n3 = 2.4

n1 = 1

5

5

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

6

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 canget 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

• 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 sin2 = 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)

P

Q

i

X

i

Q’

• 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

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)

1

s1

d1

1

L – x

x

d2

2

2

s2

P