14. Total Internal Reflection and Evanescent Waves

1 14. Total Internal Reflection and Evanescent WavesPhase shifts in reflectionTotal Internal Reflection and applicationsEvanescent wavesReminder: the Fresnel equations||cos( )cos( )cos( )cos( ) ittiittinnrnn ||2cos()cos( )cos( ) iiittintnn 2cos()cos( )cos( ) iiiittntnn cos( )cos( )cos( )cos( ) iittiittnnrnn s-polarized light:p-polarized light:And, for bothpolarizations: sin( )sin( ) iittnn plane of incidenceincident wavetransmitted waveinterfaceplane of incidenceincident wavetransmitted waveinterface || ititnnrrnn 00 Ifair to glass , at itinnr So there will be destructiveinterference between the incident and reflected beams near the surface, where they overlap in normal incidence, i= 0, we find:Phase Shift in ReflectionNote: for p-polarized light, the sign of r||changes for angles above Brewster s , ifni> nt(glass to air), r > 0, and there will be angle, iReflection coefficient, ||r0 30 60 90 Incidence angle, iReflection coefficient, ||r0 30 60 90 from air to glassBrewster s angleThe obvious answer is the front of the object, which sees the higher intensity constructive interference happens at the back surface between the incident light and the reflected you slowly turn up a laser intensity, where does damage happen first, the front or the back?

2 2(1 ) This yields an irradiance that is 44% higher just inside the back surface (for nglass= )!Phase shifts in Reflection (air to glass)ni< nt180 phase shift for all angles180 phase shift for angles below Brewster's angle; 0 for larger angles0 30 60 90 Incidence angle0 30 60 90 Incidence angle 0 0 ||The Reflection coefficients are real numbers, so the phase of the reflected wave is either 0 or (relative to the incident wave). Total Internal Reflectionnintik tk i tEiEtInterfaceSnell s Law:sinsin iittnn Solve for t :1 sinsin ititnn But is impossible to take the arcsinof a number larger than one!sin iis always 1, so if ni< ntthen this never becomes an : if ni> ntthen the argument of the sin-1can exceed one!When does this occur? As iincreases, talso increases. When treaches /2, the transmitted wave is grazing the interface. This occurs at a value of igiven by: 1 sin criticaltinn In this illustration, the light wave bends awayfrom the normal because nt< Internal Reflection occurs just as thetransmitted beam grazes the 0 15 20 30 42 45 70 60 ?

3 ? Total Internal Reflection is 100% the angle of incidence increases from 0 .. the refracted ray becomes the reflected ray becomes the angle of the refracted ray approaches 90 Transmitted intensity T (glass to air)Incidence angle0 30 60 90 0%100%Transmitted intensity ||Critical angleTransmitted power goes smoothly to zero as the critical angle is application of Total Internal reflectionCharles Kao (1965): first proposed that fiber could be used as a practical communication technology if the attenuation could be reduced below 20 dB/km. He showed that the loss was dominated by chemical impurities in the glass (1970): first commercial fiber for telecommunications. Attenuation = 17 dB/kmState of the art today (commercial fibers):Attenuation = dB/kmdata rate = 40 Gb/sec Wavelength division multiplexing (WDM) using multiple wavelengths, each carrying an independent data stream, on a single fiberCurrent record (2011): 370 WDM channels, 273 Gb/sec in each channelBeam steerersused to compressthe path insidebinocularsBeam steerersAnother application of Total Internal reflectionA thought experimentSuppose I have a glass prism, oriented as shown, with a laser beam undergoing Total Internal Reflection from the Internal Internal reflectionNow I bring a second identical prism close to the first : at what point do I see a transmitted beam?

4 Answer: when the prisms are close together, but not yet touching!Conclusion: something interesting must be happening on the low index (air) sideof the Waves 00expexpEjkxtEjjxt We know that E0ejkxe-j tis a solution to the wave if we allow the wave vector kto be a purely imaginary number? Let kbe replaced by j , where is a real number. 0expexpExjt This no longer oscillates as a function of position - it exponentially decays! But it still oscillates as a function of = 0xE(x)at t = 0xE(x)at t = xE(x)at t = 2 Real part: 0expcosExt Is this still a solution to the wave equation? 0,expexp ExtExjt Non-propagating Waves 2220222202expexp,expexp, dEExjtExtdxdEExjtExtdt So this isa solution to the wave equation! But it is a very different kind of solution from the ones we re used to does not propagate in space. It is localized. It is known as an Evanescent wave.

5 Such a wave can be found in a number of situations. In particular, Evanescent Waves are always present in the case of Total Internal Evanescent WaveThe Evanescent wave is the "transmitted wave" when Total Internal Reflection occurs. A very mystical quantity! So we'll do a mystical derivation:222cos( )1 sin ( )1sin ( )a negative number ittitnn 00cos( )cos( )cos( )cos( )iittrii it tnnErEnn Since sin( t) > 1, tdoesn t exist. So how can we compute r ?Use Snell s Law to eliminate tfrom the equation:Substitute this complex quantity into the formula for r , and redefine Ras: R= r r * = |r |2. Then we find that R= 1 for allangles above the critical , all of the power is reflected; the Evanescent wave contains no shifts in Reflection (glass to air)nt< ni0 phase shift for angles below the critical angle0 phase shift for angles below Brewster's angle; 180 for larger angles up to the critical angle0 30 60 90 Incidence angle0 30 60 90 Incidence angle 0- 0- ||In the case of Total Internal Reflection , the phase shift for the reflected wave is a complicated function of a real numberris complex and |r| = a real numberris complex and |r| = Evanescent -Wave k-vectorUsing Snell's Law, sin( t) = (ni/nt) sin( i), so ktxis a real quantity.

6 Txkkxjy Using this complex k-vector , we find:Et(x,y,t) = E0exp[ y] exp[ j( kt(ni/nt) sin( i) x t)]The Evanescent wave k-vector must have x and y components:Along surface: ktx= ktsin( t)Perpendicular to it: kty= ktcos( t)As before: ktcos( t) =kt[1 sin2( t)]1/2=kt[1 (ni/nt)2sin2( i)]1/2= a pure imaginary quantity; let s call it j = real The Evanescent wave decays exponentially away from the , kt= 2 nt/ as usual - a positive real numberThe quantity 1/ sets the length scale for the decay of the wave along and the Evanescent wave ievanescent wavereflected waveincident wavetotal wave:sum of incident and reflected wavesevanescent wave: decay length = 1/ The Evanescent wave: an exampleSo how far does the Evanescent wave extend away from the interface? Let s work an 1nglass= in= 45 crit= We saw earlier that: kt[(ni/nt)2sin2( i) 1 ]1/2where ktis the wave vector in the transmission medium (air):kt= 2 nair/ For our example, let s assume that the light is green: = 532 nmThen we find: = nm-11/ = 240 nmTypical result: somewhat less than the free-space wavelengthFrustrated Total Internal Reflection (TIR)By placing another surface in contact with a totally internallyreflecting one, Total Internal Reflection can be frustrated.

7 NnTotal Internal reflectionnnFrustratedtotal Internal reflectionWe can now calculate how close the prisms have to be before TIR is quantity 1/ tells us how far the Evanescent wave extends beyond the surface of the first prism, which tells us how close the second prism needs to be in order to frustrate the application of frustrated TIRThe ridges on a finger act as locations where TIR is frustrated, so less light comes from there. But between the ridges, there is still TIR so more light is Internal Reflection fluorescence (TIRF) microscopyOnly objects within ~100 nm of the interface are fluorescence microscope image is blurred due to fluorescence from out-of-focus sourcesTIRF image is sharperFiber optic sensorsrays propagating in a large-core fiberThe value of the critical angle depends on the ratio angleindex of transmitted mediumni= the value of ntincreases, then TIR is less likely to occur (a smaller range of angles experience TIR).

8 This effect can be used to sense small refractive index changes.