Wave Phenomena Physics 5c Lecture 6 eflection and efraction (&L Sections. 3) Administravia Laser safety video at the end of this lecture ecause you ll be doing take-home lab using a laser pointer Lecture will be 0 minutes shorter than usual erm paper due in days April 5 th, 7:00 know it s not tax day in Massachusetts
What We Did Last ime Discussed M waves in vacuum and in matter Maxwell s equations " Wave equation " Plane waves t c t c and are transverse " Polarization M waves in insulators propagate according to c w M waves in conductors limited by the skin depth hinner for higher frequencies ero for perfect (super) conductors vacuum c n εµ µ µ 0 0 ε ε n Wave velocity eflectivity Goals For oday eflection and refraction of M waves Light gets reflected/refracted as it enters different medium hat s how lenses and prisms work Calculate angles and intensities Useful rules: uygens principle, Fermat s principle ackle Maxwell s equations for the complete answer rewster s angle eflected light becomes somehow polarized We ll discuss how later
eflection and efraction We did reflection and transmission last time Found reflectivity n n + n+ n What if the light is at an angle? You know the answer? Air Water? ε, µ ε, µ i r i r t t # Which way does the light go? # Why is it? # What is the intensity? # And the polarization? uygens Principle Christiaan uygens (69 695) Draw circles to construct successive wavefronts ach circle has r λ and centered on the previous wavefront Draw a common tangent of all circles " New wavefront asy examples: Plane waves Circular waves
Snell s Law Apply uygens principle to the refraction problem Wavelength λ changes at the boundary c c λ λ nf n f Consider wavelength along the surface λ θ λ λ λs sinθ sinθ sinθ λ n Law of efraction, or Snell s Law sinθ λ n λ s θ λ otal nternal eflection sinθ n sinθ n Direction of refraction bend depends on n /n Fast " slow n < n θ > θ " bend down Slow " fast n < n θ θ For slow " fast transition, > familiar pattern with air " water n n sinθ sinθ < < θ < arcsin n Critical angle n n n xample: glass (n.5) " air θ critical arcsin(/.5) 4 θ critical otal internal reflection
Optical Fibers otal internal reflection makes glass rod a light pipe Light gets trapped inside and bounce along asic idea of optical fibers eal-world optical fibers made of two types of glass core: large n cladding: small n Light loss due to impurity in glass must be very small ypical glass is not so transparent beyond a few meters ransoceanic fiber links extend several 000 km ime Consider (again) light entering water from air Light goes from point A to point A ow long does it take? a C Define x-y coordinates: a A (0, a) (, b a) C ( x,0) otal time is b n n AC+ C { n a + x + n c c c a + ( b x) } What is the fastest path? d nx n( b x) 0 dx c a + x a + ( b x) Where are we going?
Fermat s Principle d nx n( b x) 0 dx c a + x a + ( b x) Angles θ and θ are x sinθ sinθ a + x Above equation becomes ( b x) a + ( b x) Fermat s Principle of Least ime: n sinθ n sinθ Snell s law he actual path between two points taken by a beam of light is the one which is traversed in the least time a a A b C Fermat s Principle Light chooses the fastest path to get from A to Sounds good, but how does it know which path to take? Answer: it doesn t M waves go all over the place All possible paths are in fact taken ach path gives a certain wave amplitude at point When integrated, contributions from all but the fastest path cancel out Similar principles govern broader range of physics You will see them in QM and classical mechanics
oundary Conditions o calculate the intensity and the polarization, we need to solve the boundary-condition problem hat is, we must find incoming/reflected/refracted waves that satisfy the continuity conditions at the boundary We must do this for two possible polarizations Q: What are the boundary conditions, anyway? D D ε ε Polarizations Must consider two cases Let s do the vertical orizontal will be left for your exercise Symmetry tells us that and must be vertical Define s and s accordingly ertical Write down all boundary conditions And try to solve them together orizontal
oundary Conditions ε ε ( + )cosθ cosθ ε ( )sinθ ε sinθ Nothing What else do we know? Connection between and Got 6 equations in total " Seems enough to me θ θ Snell s Law liminate s using Combine with ε ( )sinθ ε sinθ ut We are left with two equations ( )cosθ cosθ ε ( )sinθ ε sinθ ime to use brute force ε sinθ ε sinθ ε εµ c n c w n sinθ n sinθ Snell s Law +
Fresnel Coefficients ( + )cosθ cosθ ε( )sinθ ε sinθ Solutions are εsinθcosθ εcosθsinθ ε sinθ cosθ + ε cosθ sinθ Simplify by introducing cosθ α cosθ α β α + β εsinθcosθ ε sinθ cosθ + ε cosθ sinθ εsinθ εn ε µ and β ε sinθ ε n ε µ α + β Fresnel coefficients orizontal Polarization he other polarization can be solved similarly ε ε Solutions are αβ αβ + Nothing ( + )cosθ cosθ ( )sinθ sinθ αβ + Problem Set θ θ
Fresnel Coefficients cosθ a cosθ β ertical polarization α β α + β α + β For air " glass, β.5 orizontal polarization αβ αβ + αβ + No reflection at this angle Normal ncidence Going back to normal incidence θ θ 0 α his gives us β + β + β β + + cosθ cosθ + β + β + + Sign differs because of the way defined the direction of ε, µ ε, µ i i t t r r β Agree with what we found last time
eflectivity and ransmittivity ntensity is S eflectivity is simply ransmittivity is a little more subtle Consider the power going through unit area on the boundary surface P Acosθ P Acosθ A Acos θ Acos θ P cosθ αβ P cosθ eflectivity and ransmittivity ertical polarization α β α + β 4αβ ( α + β ) + for both cases orizontal polarization αβ αβ + 4αβ ( αβ + ) For air " glass, β.5 No reflection around here
rewster s Angle At θ θ, vertically-polarized light does not reflect Called rewster s angle α β cosθ 0 α β α + β cosθ Use Snell s law to express θ in terms of θ and solve µ ε β ε µ n tan θ εµ if µ n µ tanθ εµ n ( n ) Light reflected at θ θ is horizontally polarized Polarized sunglasses cut glare from water Polarized photo filters remove reflection on windows Summary Studied reflection and refraction Derived Snell s law three times Using uygens principle Using Fermat s principle Light chooses the fastest path between two points Solving boundary-condition problem Fresnel coefficients " intensities of reflected/refracted light cosθ α β 4αβ αβ 4αβ a cosθ α + β ( α + β ) αβ + ( αβ + ) rewster s angle " eflection is polarized β Next week: Origin of refraction sinθ n sinθ n