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 Incident Wave Reflected 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 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 k y value Otherwise it won’t match at boundary k x must be negative So it is going the other way

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, k y must be the same before and after Mirror Incident Reflected kiki krkr x y ii rr k i sin  i k r sin  r ki = krki = kr k i sin  i = k r sin  r sin  i = sin  r i = ri = r

Geometric Optics and the Ray Approximation The wave calculations we have done assume the mirror is infinitely large If the wavelength is sufficiently tiny compared to objects, this might be a good approximation For the next week, we will always make this 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i rr 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  Mirror 47  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 133 Cs atom This frequency is extremely stable Better than any other method of measuring time Defined to be frequency f = 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  10 8 m/s 133 Cs 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 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 Indices of Refraction Air (STP) Water1.333 Ethyl alcohol1.361 Glycerin1.473 Fused Quartz1.434 Glass1.5 -ish Cubic zirconia2.20 Diamond2.419

Refraction: Snell’s Law 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 index n 1 22 11 rr 1 = r1 = r index n 2 x y k 1 sin  1 k 2 sin  2 Snell’s Law

34  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 n 1 = 1 n 2 = 1.5n 3 = 2.4 n 4 = 1.33 n 5 = 1.5 n 6 = 1 66 22 22 33 33 44 44 55 55

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 60  22 22 33 n 1 = 2.4 n 3 = 2.4 n 2 = 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 sin  2 = 1

Optical Fibers Protective 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 glassHigh 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’ rr ii i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 the slow glass very much Green path is bad too – it minimizes time in glass, but makes path much longer Red path – a compromise – is best To minimize, set derivative = 0 P Q d1d1 x L – x d2d2 s1s1 s2s2 11 22 Light always takes the quickest path 11 22