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solution 2.19 - mm from the surface(if we disregard for a...

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Solution 2.19 a) At what distance is the amplitude down by a factor of exp(-1)? exp(-alpha*distance) = exp(-1) alpha * distance = 1 distance = 1/alpha, where alpha = (2*pi/lambda) * sqrt[sin(pi/4)^2 - n^2], lambda = 500 nm, n = n2/n1 = 1/1.5 alpha = 0.00296192 nm^-1 distance = 1/alpha = 337.6 nm b) By what factor is the intensity down at a distance of 1 mm? d = 1 mm = 10^6 nm I(d)/I(0) = [amplitude(d)/amplitude(0)]^2 = exp(-alpha*d)^2 = exp(-2*alpha*d) = exp(-5923.84) = 10^- 2572.69 = 2 * 10^-2573 Note that this is an awfully small number, actually zero for all practical purposes. Consider that the total power output of the sun is “only” about 4*10^26 Watts, and the combined power output of all the stars in all the galaxies in the observable universe is of the order of 10^49 Watts. If the entire power of the universe were incident on our prism, only 10^49 * 10-2573 = 10^-2524 Watts would be left at 1
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Unformatted text preview: mm from the surface (if we disregard for a moment that the prism would be instantly vaporized). This corresponds to one photon every 10^2506 seconds or one photon in 10^2499 years. Considering that the age of the universe is only about 10^10 years, we would still have to wait for 10^2489 “universe-ages” until we would see a single photon at 1 mm from the surface! Even our outrageous astronomical number crunching didn’t help to get a grip on this number, since it didn’t bring down the value of the exponent much at all: 10^2489 is still unimaginably large. This fast decay of the evanescent wave is the reason why a relatively thin cladding layer is enough to completely block the evanescent field from leaking energy out of optical fibers....
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