Wave Phenomena Physics 15c Lecture 15 Reflection and Refraction
What We (OK, Brian) Did Last Time Discussed EM waves in vacuum and in matter Maxwell s equations Wave equation Plane waves E t = c E B t = c B vacuum E and B are transverse Polarization Malus law of polarized EM waves EM waves in insulators propagate according to c w = c n = 1 Z = µ εµ ε µ 0 ε = Z 0 n Wave velocity Reflectivity
Goals For Today EM waves in conductors Skin depth Reflection and refraction of EM waves Light gets reflected/refracted as it enters different medium That s how lenses and prisms work Calculate angles and intensities Useful rules: Huygens principle, Fermat s principle Tackle Maxwell s equations for the complete answer Brewster s angle Reflected light becomes polarized We will discuss why
Conductors Electric field in conductor causes current σ = conductivity Now we need the full form of Maxwell s equations E = ρ ε B = 0 J = σe Free charge can t exist in conductor E = B t B = εµ E t + µj = µσ E Easy to derive wave equations E = εµ E t E + µσ t B = εµ B t + µσ B t New term is 1 st order derivative with time Similar to friction
Conductors We can still find plane waves Plug in to the wave equation E = E 0 e i(k x ωt ) B = B 0 e i(k x ωt ) E = εµ E t E + µσ t k = εµω + iµσω We ve seen complex k before Split real and imaginary parts k is complex { Re(k) + i Im(k) } = εµω + iµσω Waves moving toward +z look like E = E 0 e Im(k )z i(re(k )z ωt ) e B = B 0 e Im(k )z i(re(k )z ωt ) e Exponentiallyshrinking oscillation
Skin Depth EM waves in conductor shrinks exponentially Amplitude decreases by 1/e at d = 1/Im(k) d is the skin depth EM waves penetrate metal only a few times the skin depth Im(k )z e z Is the sign of Im(k) right? { Re(k) + i Im(k) } = εµω + iµσω Re(k)Im(k) = µσω > 0 Re(k) and Im(k) have the same sign e Im(k)z shrinks as ei(re(k)z ω t) moves forward
Skin Depth Solve for Re(k) and Im(k) explicitly Re(k) = ω εµ 1+ σ εω + 1 Im(k) = ω εµ Both Re(k) and Im(k) grow with frequency For large frequencies: For small frequencies: Skin depth gets shorter as the frequency increases High frequency waves can be stopped by thin metal 1/ Re(k) Im(k) ω εµ Re(k) Im(k) µσω 1+ σ εω 1 1/
Superconductors Perfect conductor has infinite conductivity { Re(k) + i Im(k) } = εµω + iµσω Skin depth becomes zero σ Im(k) EM waves cannot enter superconductors Generally, current in conductor flows in the skin depth For normal conductors, it s thinner for higher frequencies DC current (ω = 0) flows the entire thickness For superconductors, it s zero no matter what the frequency Only the surface matters. Bulk is useless Superconducting cables are made of very thin filaments
Reflection and Refraction We did reflection and transmission last time Found reflectivity R = Z Z 1 Z 1 + Z n n 1 n 1 + n,µ 1 ε,µ B i E i B t E t What if the light is at an angle? You know the answer E r B r Air Water?? Which way does the light go? Why is it? What is the intensity? And the polarization?
Huygens Principle Christiaan Huygens (169 1695) Draw circles to construct successive wavefronts Each circle has r = λ and centered on the previous wavefront Draw a common tangent of all circles New wavefront Easy examples: Plane waves Circular waves
Snell s Law Apply Huygens principle to the refraction problem Wavelength λ changes at the boundary λ 1 = c n 1 f λ = c n f Consider wavelength along the surface λ s θ 1 λ 1 λ s = λ 1 sinθ 1 = λ sinθ λ θ sinθ sinθ 1 = λ λ 1 = n 1 n Law of Refraction, or Snell s Law
Total Internal Reflection sinθ sinθ 1 = n 1 n Direction of refraction bend depends on n 1 /n Fast-to-slow n 1 < n θ 1 > θ bend down Slow-to-fast n 1 > n θ 1 < θ bend up For slow-to-fast transition, familiar pattern with air water sinθ 1 = n sinθ n < n < 1 θ 1 n 1 < arcsin n 1 n 1 θ critical = arcsin(1/ 1.5) = 4 Example: glass (n = 1.5) air Critical angle θ critical Total internal reflection
Optical Fibers Total internal reflection makes glass rod a light pipe Light gets trapped inside and bounce along Basic idea of optical fibers Real-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 Typical glass is not so transparent beyond a few meters Transoceanic fiber links extend several 1000 km Ultra-pure glass makes it possible Look up Charles K. Kao, 009 Nobel Laureate
Time Consider (again) light entering water from air Light goes from point A to point B How long does it take? Define x-y coordinates: A = (0,a) B = (b, a) Total time is C = (x,0) a a A b C B T = n 1 c AC + n c CB = 1 c n 1 a + x + n a + (b x) What is the fastest path? dt dx = 1 c n 1 x a + x { } n (b x) a + (b x) = 0
Fermat s Principle dt dx = 1 n 1 x n (b x) = 0 c a + x a + (b x) Angles θ 1 and θ are x b x sinθ 1 = sinθ a + x = a + (b x) a a A b C B Above equation becomes n 1 sinθ 1 = n sinθ Snell s law! Fermat s Principle of Least Time: The actual path between two points taken by a beam of light is the one which is traversed in the least time
Fermat s Principle Light chooses the fastest path to get from A to B Sounds good, but how does it know which path to take? Answer: it doesn t EM waves go all over the place All possible paths are in fact taken Each path gives a certain wave amplitude at point B 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 advanced classical mechanics
Boundary Conditions To calculate the intensity and the polarization, we need to solve the boundary-condition problem That 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? Components of E and B fields that are parallel/perpendicular to the boundary surface must satisfy: E 1 E E 1 E H 1 H H 1 H E 1 = E D 1 = D H 1 = H B 1 = B E 1 = ε E µ 1 H 1 = µ H
Polarizations Must consider two cases Let s do the vertical Horizontal will be left for your exercise Symmetry tells us that E R and E T must be vertical Define Es and Hs accordingly Write down all boundary conditions E 1 = E Vertical H I ( + E R )cosθ 1 = E T cosθ H I Horizontal H I H R E R E 1 = ε E ( E R )sinθ 1 = ε E T sinθ H 1 = H H I H R = H T H T E T µ 1 H 1 = µ H Nothing
Snell s Law Eliminate Hs using impedance Z 1 and Z E R = E T H I H R = H T Z 1 Z Combine with Recall: We ve found Snell s law! We are left with two equations Time to use brute force ( E R )sinθ 1 = ε E T sinθ Z 1 sinθ 1 = Z ε sinθ Zε = εµ = 1 c w = n c n 1 sinθ 1 = n sinθ ( + E R )cosθ 1 = E T cosθ ( E R )sinθ 1 = ε E T sinθ
Fresnel Coefficients ( + E R )cosθ 1 = E T cosθ ( E R )sinθ 1 = ε E T sinθ Solutions are E R = sinθ 1 cosθ ε cosθ 1 sinθ sinθ 1 cosθ + ε cosθ 1 sinθ E T = Simplify by introducing sinθ 1 cosθ 1 sinθ 1 cosθ + ε cosθ 1 sinθ α cosθ cosθ 1 and β ε sinθ sinθ 1 = ε n 1 n = ε µ 1 µ = Z 1 Z Note: α is a function of the angles, β is a constant E R = α β α + β E T = α + β Fresnel coefficients
Horizontal Polarization The other polarization can be solved similarly E 1 = E E R = E T H I E 1 = ε E H 1 = H Nothing (H I + H R )cosθ 1 = H T cosθ θ1 E R H R µ 1 H 1 = µ H This will be in the homework Solutions are E R = αβ 1 αβ + 1 E T = µ 1 (H I H R )sinθ 1 = µ H T sinθ αβ + 1 E T H T θ
Fresnel Coefficients α cosθ cosθ 1 β Z 1 Z Vertical polarization Horizontal polarization E R V V = α β α + β E T V V = α + β For air glass, β = 1.5 E R H H = αβ 1 αβ + 1 E T H H = αβ + 1 E T V V E T H H E R H H E R V V θ 1 θ 1 No reflection at this angle
Normal Incidence α cosθ cosθ 1 β Z 1 Z Going back to normal incidence θ 1 = θ = 0 This gives us α cosθ cosθ 1 = 1,µ 1 ε,µ B i E i B t E t E R V V = 1 β 1+ β = Z Z 1 Z + Z 1 E T V V = 1+ β = Z Z + Z 1 E r B r E R H H = β 1 β + 1 = Z 1 Z Z 1 + Z E T H H = β + 1 = Z Z 1 + Z Sign differs because of the way I defined the direction of E Agrees with what we found last time
Reflectivity and Transmittivity Intensity is S = E Reflectivity R is simply R = E R Transmittivity is a little more subtle Consider the power going through unit area on the boundary surface Z Acosθ 1 P I = E I Acosθ Z 1 1 P T = E T Acosθ Z A Acosθ T = P T P I = E T Z 1 Z cosθ cosθ 1 = E T αβ
Reflectivity and Transmittivity R V = α β α + β Vertical polarization Horizontal polarization R + T = 1 in both cases For air glass, β = 1.5 4αβ T V = R H = αβ 1 (α + β) αβ + 1 4αβ T H = (αβ + 1) T V T H R V No reflection around here R H θ 1 θ 1
Brewster s Angle At θ 1 = θ B, vertically-polarized light does not reflect Called Brewster s angle E R V = α β α + β = 0 α = cosθ cosθ B = β Use Snell s law to express θ in terms of θ B and solve tan θ B = 1 β ( ) 1 n 1 n = 1 µ1ε µ µ 1 ε µ 1 if µ 1 = µ tanθ B = n n 1 Light reflected at θ = θ B is horizontally polarized Polarized sunglasses cut glare from water Polarized photo filters remove reflection on windows
Brewster s Angle OK, so we know reflected light is polarized, so what? Is there anything deeper than solving equations? Look at the condition α = β again Assume µ 1 = µ α cosθ cosθ 1 β ε µ 1 µ = ε = n n 1 = sinθ 1 sinθ cosθ cosθ 1 = sinθ 1 sinθ A little trig gives us sin θ 1 = sin θ θ = θ or θ 1 1 + θ = π This isn t the case This must be
Brewster s Angle Why? At Brewster s angle, θ 1 + θ = π Reflected light and refracted light are perpendicular to each other Is this a coincidence? If not, what is the significance? We need to dig into the origin of refraction What makes matter dielectric? How does it cause light to slow down? We will do this in the next lecture θ 1 θ 1 θ
Summary EM waves in conductors are confined within the skin depth Higher frequencies thinner Studied reflection and refraction Derived Snell s law three times Using Huygens principle Using Fermat s principle Solving boundary-condition problem sinθ sinθ 1 = n 1 n Fresnel coefficients intensities of reflected/refracted light R V = α β α + β T V = 4αβ R H = αβ 1 (α + β) αβ + 1 T H = 4αβ (αβ + 1) Brewster s angle Reflection is polarized Next: Origin of refraction α cosθ cosθ 1 β Z 1 Z