# Refraction (pt 2) Snell’s Law Total Internal Reflection Fiber Optics

## Presentation on theme: "Refraction (pt 2) Snell’s Law Total Internal Reflection Fiber Optics"— Presentation transcript:

Refraction (pt 2) Snell’s Law Total Internal Reflection Fiber Optics
Dispersion Optical fibers are called light pipes. They work by a succession of total internal reflections

Snell’s Law When the wave fronts slow down at a boundary it can be seen that the angle that the fronts make with the boundary gets less. In other words the angle of incidence (1) between the incident ray and a line drawn perpendicular to the boundary (the normal) is greater than the angle of refraction (2). The angle of refraction is dependent on the angle of incidence and the difference in the type of medium from one side of the boundary to the other. For example, light and sound will travel at different speeds in different media. From the diagram it can be seen that if the incident wave is traveling at speed v1 and the refracted ray v2: sin 1 = L1 / A = v1 t / A and sin 2 = L2 / A = v2 / A Therefore sin 1 = v1 sin  v2 Refracted ray 2 2 1 A L1 L2 1 normal Incident ray Example An earthquake P wave passes across a boundary in rock where its velocity increases from 6.5 km/s to 8.0 km/s. If it strikes this boundary at 30O, what is the angle of refraction? 2 = sin-1( v2 sin 1 ) = sin-1 ( 8 km/s sin30o ) = sin-1 (0.615) = 380 v km/s Note: The wave speed ups when it passes across the boundary so will bend away from the normal.

But we now know that for light n = c / v or v = c / n and v = f 
So v1 = c n2 = n and v1 = f 1 = 2 v2 n1 c n v f  1 therefore v1 = 1 = n2 = sin 1 v  n sin 2 Refracted ray 2 Material of refractive index, n2 2 We can thus relate the relative refractive index of the two mediums across a boundary to the amount of bending of light. The amount of bending as you can see is directly related to how much the material “slows down” the light as it passes through (the inverse of the refractive index, 1 / n) and this in turn will influence the wavelength. For the example diagram to the right it can be seen that n2 must be bigger than n1 because 1 is bigger than 2. Thinking about it conceptually, this makes sense because we know that the refracted ray must be moving slower than the incident ray because it is closer to the normal. The refracted ray must consist of the same number of waves per second as the incident ray because you can’t “lose” waves but because the refracted waves move slower, their wavelength must be shorter than the incident waves. What this means is if 1 is big then 1 will be big compared with 2 and 2. This is shown in the final expression above. 1 A L1 L2 1 Incident ray Material of refractive index, n1 Example A jeweler wants to tell whether a stone he receives is diamond. He lets a light ray of wavelength 650 nm hit the stone at an angle of 350 to the normal. If the angle of refraction within the stone is what is a) the refractive index of the stone b) the wavelength of the light within the stone and c) the frequency of light within the stone. d) Is the stone diamond? 1 = 650 nm n1 = 1 = 2 = 23.10 a) n2 = n1 (sin 1 / sin 2) = (1.0) (sin350 / sin23.10) = 1.46 b) 2 = 1 (n1 / n2) = (650 nm) (1.0 / 1.46) = 445 nm c) f = v2 / 2 = v1 / 1 = (3 x 108 m/s) / (650 x 10-9m) = 4.6 x 1014 Hz d) Diamond has a refractive index of 2.41 so the stone is not diamond

Total Internal Reflection
What happens to a light ray when it goes from a more optically dense medium (bigger refractive index) to a less optically dense medium? It will speed up and bend away from the normal according to our theory of refraction. (Figure b in the diagram to the right shows that r is greater than I ). You can see that as the angle of incidence is increased (a through c), the angle of refraction gets closer to 900. At a particular angle of incidence (d) called the critical angle the angle of refraction is 900. Above this critical angle (e) most of the ray is not refracted but now reflected. Now sinI = nr = vi sinr = ni vr So when I = c then r = 900 and we can write sin c = nr sin ni But if the ray is being refracted into air (nr = 1) and sin900 = 1 so: sin c = 1 / ni or ni = 1 / sin c a b c Air (n = 1) r d Boundary / Interface c Material (water, glass, plastic) c e I* normal normal normal Why Diamond Sparkles Compute the critical angle for light incident onto air from water (nw =1.33), crown glass (ncg = 1.52) and diamond (nd = 2.41). How does diamond’s small critical angle relate to the appearance of a diamond gemstone? Water to air: sin c = 1 / nw therefore c = sin-1 (1 / 1.33) = 48.80 Glass to air: sin c = 1 / ncg therefore c = sin-1 (1 / 1.52) = 41.10 Diamond to air: sin c = 1 / nd therefore c = sin-1 (1 / 2.41) = 24.50 The unusually high refractive index of diamond leads to its small critical angle in air of Diamond gemstones are cut with many facets in such a way that much of the incident light undergoes multiple total internal reflections within the diamond before passing out again into the air. Many of the rays incident on the diamond from above are reflected back upward towards the faceted crown.

Optical Fibers Easily the most important application of internal reflection is fiberoptics. Techniques have evolved in recent times for efficiently conducting light and near IR (1300 nm) energy along thin, transparent, dielectric fibers. A typical fiber thickness might be in the range of say, 50m, just about the thickness of a human head-hair. An optical fiber is like a light pipe. Light entering it at the proper angle will zig zag its way through it as many as times paer meter without being lost through the walls of the fiber. The smooth surface of a single filament must be clean of contamination, or the boundary conditions change and the light leaks out at those spots. Accordingly, each fiber core is usually enshrouded in a transparent sheaf of lower-index material called cladding. The lower index of refraction of the cladding wrt the core ensures a small critical angle therefore allowing the light to undergo total internal reflection along the fiber Incidentally, just like an optical fiber, to keep a diamond ring sparkling you must keep the bottom of the diamond clean so that the sharp change in refractive index from diamond to air is maintained., and the critical angle is kept low. Dirt or moisture on the bottom of the diamond increases the critical angle, allowing much of the light incident on the diamond to escape through the bottom.

In the new era of optical communications, optical fibers are replacing metal wires, not for transmitting power, but information. The much higher frequencies of light allow for an incredible increase in data-handling capacity. For example, using some sophisticated transmitting techniques, a pair of copper telephone wires can be made to carry up to two dozen simultaneous conversations. To get a feel for how much information that is, consider the fact that a single ordinary TV transmission is equivalent to about 1300 simultaneous telephone conversations, which in turn, is roughly equal to sending more than 2500 typewritten pages each second! (See QWEST advertisement). So, at present it is quite impractical to send television over copper phone lines. Today, when you talk on the phone, your voice is converted into an electrical signal that goes to a tiny laser which transmits your voice as a series of flashes. Optical fibers can carry up to 20,000 conversations at once because of spaces in each person’s conversation. (multiplexing) This is only the beginning; achieved capacities to date don’t even begin to approach the theoretical limit. There are however some problems with this new medium of communication. Because the optical medium is not perfectly transparent, it is necessary to boost the signal every 50-km with repeater stations. This is still much less expensive than systems of metal wire, which require boosting every kilometer. Example The interior portion of an optical fiber has an index of refraction of If all the rays striking the interior walls of the pipe with incident angles greater than are subject to internal reflection, what is the index of refraction of the cladding? n2 = sin  therefore nclad = sin c and nclad = nfiber sin c = (1.60) sin (59.50) = 1.379 n sin  nfiber sin

Wavelength Relationship
Dispersion Dispersion involves the separation of light into colors according to their frequency.. Wavelength Relationship We have already seen from our discussions of Snell’s Law that the speed of light in a transparent medium depends on the frequency and therefore the wavelength of light. Since the natural or resonant frequency of most transparent materials is in the ultraviolet part of the spectrum, visible light of higher frequencies travels more slowly than light of lower frequencies. Violet light travels about 1% slower in ordinary glass than red light. The colors between red and red and violet travel at their own intermediate speed. Example Light of wavelength 400 nm is incident at an angle of 450 on acrylic and is refracted as it passes into the material. What wavelength of light incident on fused quartz at an angle of 450 would be refracted at exactly the same angle? The two materials must have the same index of refraction, n, so a wavelength of about 250 nm is needed. This conceptually makes sense because shorter wavelengths are refracted more by a certain material but Fused Quartz also acts as a less optically dense material than Acrylic. The shorter wavelength compensates for this. Normal Prism Since different frequencies of light travel at different speeds in transparent materials, they will refract differently and bend at different angles. When light is bent twice at non-parallel boundaries, as in a prism, the separation of light into colors of light is quite apparent. Dispersion is what enabled Sir Isaac Newton to produce a spectrum. In the diagram opposite, it can be seen that violet light ,(smallest wavelength) bends the most towards the normal line in the prism, and also bends the most away from it when it leaves. Normal

Rainbows Dispersion is most evident when seeing a rainbow. To see one you will need two things, water drops and sun light. Not only that, but to see one you will also need the sun behind you. The fact that rainbows are semicircular tells you that the light from one has to come from the water drops at the right angle angle*. When white light enters a water drop light is refracted and disperses into its component colors. Violet is bent the most and red the least. When the rays reach the opposite part of the drop they are partly reflected and leave the drop at an angle between 400 and 420 from their incident direction. (figure a) Even though each drop produces the spectrum of colors, an observer is in a position to only see one color from each drop. This is why red appears at the top of the rainbow. (figure c). Sometimes you will see a secondary rainbow above the primary rainbow with the colors inverted. This is a result of double reflection within the raindrop. (figure b). *Note: If you viewed a rainbow from an airplane you would see that it is circular. The horizon normally gets in the way making it look semicircular.

Diamonds Revisited Chromatic Aberration
Like a prism, the diamond is a dispersive medium (that is n varies somewhat with ), and so the various colors composing white light travel somewhat different paths and emerge in different directions. Hence as the diamond is turned, different colors can be seen, giving it its characteristic sparkling quality. Chromatic Aberration Chromatic aberration is the result of different speeds of light of various colors and hence the different refractions they undergo. In a simple lens red light and blue light bend by different amounts (as in a prism), so they do not come to focus in the same place.

Download ppt "Refraction (pt 2) Snell’s Law Total Internal Reflection Fiber Optics"

Similar presentations