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Reflection Ch. 35 •What happens when our wave hits a conductor? Ei E0 sin kx t •E-field vanishes in a conductor •Let’s say the conductor is at x = 0 Er E0 sin kx t •Add a reflected wave going other direction •In reality, all of this is occurring in three dimensions Incident Wave Reflected Wave Total 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 2 2 c k k ck x y Ei E0i sin k x x k y y t •What will reflected wave look like? E r E0 r sin k x x k y y t •Assume it is reflected at x = 0 c k x2 k y2 •It will have the same angular frequency •Otherwise it won’t match in time •It will have the same ky value k y k y •Otherwise it won’t match at boundary •kx must be negative • So it is going the other way c k x2 k y2 c k x2 k y2 k x k x k x k x 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 ki sini ki sini = kr sinr sini = sinr i = r ki = kr kr sinr i r Mirror y x Geometric Optics and the Ray Approximation i = r •The wave calculations we have done assume the mirror is infinitely large •If the wavelength is sufficiently tiny compared i r to objects, this might be a good approximation •For the next week, we will always make this approximation Mirror •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! Concept Question •This works for any angle •In 3D, you need three mirrors = 47 47 43 47 47 4343 Mirror 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 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 ½ ½ •The nodes will be separated by ½ •Then you get c from c = f •Biggest error comes from 133Cs 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 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 c 3.00 108 m/s •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 Indices of Refraction •This generally slows the wave down Air (STP) 1.0003 •n is called the index of refraction Water 1.333 •The amount of slowdown can depend Ethyl alcohol 1.361 on the frequency of the light Glycerin 1.473 c v n Fused Quartz 1.434 Glass 1.5 -ish Cubic zirconia 2.20 Diamond 2.419 Refraction: Snell’s Law ck n •The relationship between the angular frequency and the wave number k changes inside a medium •Now imagine light moving from one medium to another f c n •Some light will be reflected, but usually most is refracted •The reflected light again must obey the law of reflection 1 = r •Once again, the frequencies all match k1sin1 •Once again, the y-component of k must match k1 k2 c c n1 n2 n2 k1 n1k2 index n1 index n2 y k1 sin 1 k2 sin 2 n1n2 k1 sin 1 n2 n1k2 sin 2 n1 sin 1 n2 sin 2 Snell’s Law 1 r x 2 k2sin2 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 n5 = 1.5 6 n6 = 1 5 5 n3 = 2.4 n4 = 1.33 4 4 n1 = 1 n2 = 1.5 3 3 2 2 34 n1 sin 34 n2 sin 2 n3 sin 3 n4 sin 4 n5 sin 5 n6 sin 6 n1 sin 1 n2 sin 2 1 sin 34 1 sin 6 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 35-21) 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 35-16) 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 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 n1 = 2.4 60 2 2 n2 = 1 3 n3 = 2.4 n1 sin 60 n2 sin 2 sin 2 2.4 0.866 2.07 •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: n2 sin c •Set sin2 = 1 n 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) •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 P Q 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 i r X i Q’ Fermat’s Principle (2) P •What about refraction? n1 sin 1 n2 sin 2 •What’s the best path from P to Q? 1 s1 •Remember, light slows down in glass 1 L – x •Purple path is bad idea – it doesn’t avoid the d1 slow glass very much x d2 •Green path is bad too – it minimizes time 2 in glass, but makes path much longer 2 s2 •Red path – a compromise – is best Light always takes •To minimize, set derivative = 0 the quickest path Q n1s1 n2 s2 1 s1 s2 2 2 2 t n1 x d1 n2 L x d 22 c c v1 v2 c n2 L x 1 n1 sin 1 1 n1 x dt 0 2 2 2 2 c n2 sin 2 dx c x d1 L x d 2