And Refraction
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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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Reflection

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


Reflection

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


Reflection

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


Reflection

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


Reflection

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


Reflection

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


Reflection

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

Water1.333

Ethyl alcohol1.361

Glycerin1.473

Fused Quartz1.434

Glass1.5 -ish

Cubic zirconia2.20

Diamond2.419


Reflection

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


Reflection

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


Reflection

Solve on Board


Reflection

Solve on Board


Reflection

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


Reflection

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


Reflection

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


Reflection

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


Reflection

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


Reflection

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


Reflection

Fermat’s Principle (2)

1

s1

d1

1

L – x

x

d2

2

2

s2

P

  • 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

Light always takes the quickest path

Q


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