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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 in three dimensions.

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Reflection

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  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. Solve on Board

  11. Solve on Board

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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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