PHYS 408, Optics. Problem Set 1 - Spring Posted: Fri, January 8, 2015 Due: Thu, January 21, 2015.

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1 PHYS 408, Optics Problem Set 1 - Spring 2016 Posted: Fri, January 8, 2015 Due: Thu, January 21, An electric field in vacuum has the wave equation, Let us consider the solution, 2 E 1 c 2 2 E = 0. (1) t2 E(r, t) = E 0 cos (kx ωt + φ 0 )ẑ, (2) where E 0 and φ 0 are some constants and ẑ the usual unit vector along the positive z-axis. (a) Show explicitly that this solution satisfies the wave equation if ω/k = c. (b) Show that the wave is spatially periodic with a period of λ = 2π/kand temporally periodic with period T = 2π/ω. (c) Find the corresponding solution for the magnetic field H(r, t). (Note: determine the integral constant with justification) 2. Consider a plane wave with its complex amplitude given by U(r) = A 0 e ik r, (3) where A 0 is a complex constant number and k is a real constant vector. (a) In which direction is this wave travelling?. (b) Calculate the optical intensity of this wave. Does it vary with time and space? (c) Write down the corresponding wave function. Determine the unit of A 0 and explain the concept of phase velocity. (d) Is U(r) and electric field? If not, how are they related? 1

2 3. Consider an infinitely large linear dielectric with homogeneous permittivity ɛ and permeability µ. (a) Assume no free charge or current density. Starting from Maxwell s equations, derive the wave equation. (b) What is the speed of light? How does it compare with the speed of light in vacuum? (c) Consider solutions E(r, t) = E 0 e i(k r ωt), B(r, t) = B 0 e i(k r ωt). (4) They must satisfy the all four Maxwell s equations in order to be solutions to the wave equation. i. What condition must the solutions meet to satisfy the divergence equations? What does that condition mean? Sketch the E and B fields along propagation direction in space. ii. Using the curl equations, show the relation between E and B. How do they compare in magnitude? iii. Calculate the time averaged Poynting Vector. 4. (a) Derive the Paraxial Helmholtz Equation from the Helmholtz Equation 2 TA( r) + i2k A( r) = 0 (5) 2 U( r) + k 2 U( r) = 0 (6) Justify each approximation and give its range of validity. Make sure to relate any approximations to the properties of the wave. Hint: recall that U( r) and A( r) are complex quantities, so you need to consider their real and imaginary parts in your derivation. (b) Show explicitly that the complex amplitude of a Gaussian beam (see equations 3.1-7, 3.1-8, 3.1-9, , in S&T) is a solution of the Paraxial Helmholtz Equation. 5. Consider the same situation with Problem 3 but assume the relative permeability is unity. Recall D = ɛ 0 E + P (7) (a) Show that P = 0. (Hint: ɛe = ɛ E + ɛ E) (b) Derive 2 (ɛ 0 µ 0 t )E = µ 0 P. (8) t2 2

3 (c) For a non-homogeneous wave equation, the non-homogeneous term is known as a source, or driving term. What causes the polarization? How does it work as a source? (d) Postulate that P takes the form: P = P 0 e i(kz ωt)ˆx, (9) Find E that safisfies the equation. (e) Based on your results of the above calculation, comment on the difference between the polarization density P, and the polarization of a plane wave solution of Maxwell s equations. 6. For a spherical wave with amplitude: U(r) = A 0 r eikr, (10) (a) Verify that this is a solution of the homogeneous Helmholtz equation everywhere except at the origin. (Hint: Use spherical coordinates.) (b) Why can t you extend your method to show that U(r) is a solution to the Helmholtz equation at the origin? Do you think it is or isn t a solution there? (c) The general non-homogeneous Helmholtz equation does not have such simple solutions and is usually difficult to solve. Green s function methods are very useful when dealing with non-homogeneous differential equations. Consider an non-homogeneous equation: Lu(x) = f(x), (11) where L is a linear differential operator with respect to x, u(x) is the function we want to solve for, and f(x) is some non-homogeneous term. The Green s function is given by G(x, x ) such that LG(x, x ) = δ(x x ), (12) where δ is the Dirac delta function. Show that once we know the form of G(x, x ), u(x) can be evaluated by: u(x) = G(x, x )f(x ) dx. (13) 3

4 Solutions: 1. (a) To start we observe that 2 E = 2 x [E 2 0 cos(kx ωt + φ 0 )ẑ] (14) = k 2 [E 0 cos(kx ωt + φ 0 )ẑ] (15) = k 2 E (16) 2 t E = 2 2 t [E 2 0 cos(kx ωt + φ 0 )ẑ] (17) = ω 2 E. (18) Therefore, if we plug this into the wave equation with c = ω/k, we find that 2 E 1 2 c 2 t E = 2 k2 E + ω2 E = 0. (19) c2 Hence, we have shown that it is a solution to the wave equation. (b) We first start with the a concise explanation before elaborating on our logic. In doing so, we note the simple cosine behaviour and that, at a fixed time we have that E cos(kx) (20) where the phase shift from φ 0 and the temporal component do not affect the oscillatory behaviour. From this, one finds that the period of these oscillations is λ = 2π k. (21) In examining the temporal period, we follow a similar procedure except we instead fix x. Then E cos(ωt) (22) and hence we find that T = 2π ω. (23) To elaborate, we first plot the electric field with some arbitrary test values. From Figure 1 one can better grasp the ideas of our argument. Notice that if we were to take a cut along the x or t-axis, shifting the position of the cut simply adds a different phase shift to the resulting wave as illustrated by Figure 2. (c) To find the corresponding H(r, t) we use Faraday s Law. t H = 1 µ E (24) = 1 µ 0 [E 0 cos(kx ωt + φ 0 )ẑ] (25) = k µ 0 E 0 sin(kx ωt + φ 0 )ŷ (26) 4

5 Figure 1: A plot of the electric field with k = 2π, ω = π, φ = 0 and E 0 = 1. Figure 2: A plot of the electric field along the x-axis for varying times. 5

6 Therefore, if we integrate both sides with respect to time we find that t t 0 dt t H(r, t ) = t t 0 dt k µ 0 E 0 sin(kx ωt + φ 0 )ŷ (27) = H(r, t) H(r, t 0 ) = k ωµ 0 E 0 ŷ [cos(kx ωt + φ 0 ) cos(kx ωt 0 + φ 0 )] (28) and hence H(r, t) = k ωµ 0 E 0 ŷ cos(kx ωt + φ 0 ) + C. (29) In general, as long as C is curl less and divergence less, any form of it can be added to the solution, and still satisfies Maxwell s equations. However, since C is time independent, or, static, it cannot generate or be generated by a harmonic field as given in this problem. To see this, a time derivative of C only gives zero, and a time derivative of a harmonic term still gives a harmonic term. Thus we have found that the magnetic counterpart to the given electric field is H(r, t) = E 0 cµ 0 ŷ cos(kx ωt + φ 0 ) (30) 2. (a) This plane wave travels along the direction of k. (b) First, we recall that the optical intensity for a monochromatic wave does not vary with time. Then we proceed by using the definition of optical intensity in terms of the complex amplitude I(r) = U(r) 2 (31) = A0 e jk r 2 (32) = A 0 2. (33) Hence, we see that the intensity doesn t vary in space either. (c) Recall that the wave function is defined by u(r, t) = Re {U(r, t)} (34) = Re { U(r)e jωt} (35) = Re { A 0 e jk r e jωt} (36) = a 0 cos(k r ωt + φ 0 ) (37) where A 0 = a 0 e jφ 0 for some real constants a 0 and φ 0. We know that the optical intensity has units of watts/cm 2 and hence a 0 has units of watts 1/2 m 1 from Equation 37. Therefore, A 0 has units of watts 1/2 m 1. 6

7 (d) The phase velocity of a wave is the velocity at which the wavefronts propagate. Consider wavefront with a constant phase φ 0 = kx ωt, take derivative with respect to time on both sides: 0 = k dx dt Then phase velocity is given by v p = dx/dt = ω/k ω. (38) (e) No, U(r) is not an electric field. We know this because it has the wrong units. U(r) has a unit of W 1/2 m 1 or kg 1/2 m 1/2 s 3/2, but the electric field should have a unit of kg m s 2 C 1. Further, U(r) is a scalar, while electric field is a vector. To get the electric field, one needs to attach to U(r) some constants to match units and a unit vector for the polarization. 3. (a) No free charge or current density corresponds to condition ρ = J = 0, and Maxwell equation is then given by D = 0 (39) B = 0 (40) E = B t (41) B = µɛ E t. (42) (Equation 39 can be reduced to E = 0 if ɛ is a constant. The proof is in Problem 5 part a).)take curl on both sides of equation 41, use the identity ( E) = ( E) 2 E and substitute equation 42 into it, we derive the wave equation: 2 E µɛ 2 E = 0. (43) t2 (b) The speed of light is simply given by 1/v 2 = µɛ and v = 1/ µɛ. Furthermore, µ0 ɛ 0 1 v = = c µɛ µ0 ɛ 0 n, (44) (c) where n = µɛ/µ 0 ɛ 0 is the refractive index. i. Substitute the given solutions into equations 39 and 40 respectively, we have (E 0 e i(k r ωt) ) = ik E 0 e i(k r ωt) = 0 (45) (B 0 e i(k r ωt) ) = ik B 0 e i(k r ωt) = 0, (46) which directly leads to: k E 0 = 0 (47) k B 0 = 0 (48) 7

8 Figure 3: A sketch of a transverse wave. Figure taken from wikipedia. What we can conclude is that E and B fields are both perpendicular to the direction of propagation. Such waves are called transverse waves. Furthermore, E and B should also be perpendicular to each other, which is guaranteed by the definition of cross product in the curl equations: by direct substitution of the solutions into equations 39 and 40 respectively, we have: which can be simplified to: ik E 0 e i(k r ωt) = iωb 0 e i(k r ωt) (49) ik B 0 e i(k r ωt) = iµɛωe 0 e i(k r ωt), (50) k E 0 = ωb 0 (51) k B 0 = µɛωe 0. (52) The sketch should be something like Figure 3. ii. Note that the magnitude of the wave vector k = ω/v = µɛω, the above equations can be further simplified by introducing the unit vector of k: n = k/k:. Therefore, E and B fields are related by: n E 0 = 1 µɛ B 0 (53) n B 0 = µɛe 0. (54) B = µɛn E 0 e i(k r ωt) (55) and B and µɛe should have the same dimension and magnitude: B / E = µɛ iii. Since the solutions are harmonic, the time-averaged Poynting vector is simply given by: S = 1 2 E H = 1 ɛ 2 µ E 0 2 n (56) 8

9 4. (a) First, note that U(r) = A(r)e ikz. (57) We then proceed to calculate 2 U. Starting with the x-derivative, we find that ( ) 2 U(r) 2 A(r) = e ikz. (58) x 2 x 2 Likewise, one finds that Before continuing we note that 2 U(r) y 2 = 2 ( 2 A(r) y 2 ) e ikz. (59) x x = 2 2 T (60) which we can use to simplify the x and y-derivatives. Lastly, we compute the z-derivative 2 U(r) = [( ) ] A(r) e ikz + ika(r)e ikz (61) 2 ( ) ( ) 2 A(r) A(r) = e ikz + 2ik e ikz k 2 A(r)e ikz. (62) 2 We can now plug these expression into the Helmholtz equation. ( ) 2 U( r) + k 2 U( r) = 2 T A + 2 A A + 2ik 2 k2 A e ikz + k 2 A(r)e ikz (63) ( ) = 2 T A + 2 A A + 2ik e ikz = 0 (64) 2 ( ) ( ) = 0 = 2 2 A A T A + + 2ik (65) 2 We see that this is almost the answer we re looking for, we just need to get rid of the z-derivative term. To do so, we note that if A(r) is slowly varying on the order of λ one can write { } { } A(r) A(r) Re << Re = Re {ka(r)}. (66) λ It therefore follows that { } 2 A(r) Re 2 { << Re k A(r) }. (67) An analogous argument applies for the imaginary components and hence, we can neglect this term. Therefore, we have found that 2 TA(r) + i2k A(r) 9 = 0 (68)

10 (b) To check that the complex amplitude satisfies the Paraxial Helmholtz equation, we recall that for a Gaussian beam, the complex envelope is given by where A(r) = A 1 q(z) ei We then proceed to calculate ( 2 2 T A(r) = k 2q(z) ρ2 (69) q(z) = z + iz 0 (70) ρ 2 = x 2 + y 2. (71) x y 2 ) A(r) (72) by first computing the x-derivative and noticing that the y-derivative is analogous. Hence 2 x A(r) = [ ] ikx 2 x q A(r) (73) = ik ( ) 2 ikx q A(r) + A(r) (74) q [ ) ] 2 x 2 = i k q ( k q A(r). (75) Therefore, [ 2 T A(r) = i 2k ( ) ] 2 k q ρ 2 A(r). (76) q We then proceed to calculate the z-derivative term. In order to make the algebra easier, we note that q(z) = (z + iz 0) = 1 (77) and therefore Therefore A(r) 2ik A(r) = q(z) A(r) q(z) = A(r) q (78) (79) = 1 q A(r) ik ( ρ q ) 2 A(r). (80) = [ i 2k q ( k q ) 2 ρ 2 ] and hence it satisfies the Paraxial Helmholtz equation. 10 A(r) (81)

11 5. (a) From Maxwell s equations, we know D = 0. Apply equation 7:. From another approach: D = ɛ 0 E + P = 0 (82) D = (ɛe) = ɛ E + ( ɛ) E = 0. (83). Since the material is uniform, i.e, ɛ is constant, ɛ = 0. Therefore, equation 83 gives E = 0. Substituting it into equation 82 gives: P = 0. (b) Maxwell s equations in medium are: D = 0 (84) B = 0 (85) E = B (86) t H = D t. (87) Take curl on both side of equation 86 and substitute equation 87 into it (note that B = µ 0 H here): 2 E + ( E) = 1 µ 0 2 D t E = µ 0 t (ɛ 0E + P) 2 2 (ɛ 0 µ 0 t )E = µ 0 t P 2 (Note: E = 0 has been proved in part a).) (c) The polarization is induced by the E field, and it is also the source that is generating the E field. The system is then self-consistent under the mutual generation of E and P. Only polarization and electric field that are self-consistent can be a solution to this equation, and that combination is a mode of the system. (d) Substitute the P expression into equation 8, and by inspection, the E field should have the form E = E 0 e i(kz ωt)ˆx to satisfy the equation. Substitute this E into equation 8: ( ɛ 0 µ 0 ω 2 + k 2 )E 0 = µ 0 ω 2 P 0 µ 0 ω 2 E 0 = ɛ 0 µ 0 ω 2 + k P 2 0 E 0 = E 0 = E 0 = 11 µ 0 ω 2 ɛ 0 µ 0 ω 2 + (nω/c) P ɛ 0 + ɛ P 0 1 ɛ 0 ( 1 + ɛ/ɛ 0 ) P 0

12 For linear material, ɛ = ɛ r ɛ 0. Define ɛ r = 1 + χ, we have: P 0 = ɛ 0 χe 0 (88) (e) The polarization density is the density of dipole moment in a material. Polarization of the wave is the direction along which the electric field oscillates, and has no units. Electric field in the wave drives and aligns the dipole by electric force. When ɛ is constant, the two should be along the same axis. 6. (a) Write the Laplacian in spherical coordinates, and the substitution should then be straight forward in the case where r 0. Since U(r) has no θ or φ dependence: Therefore, 2 U(r) = 1 r 2 r (r2 r U(r)) = A 0 r 2 r [r2 ( 1 r 2 eikr + i k r eikr )] = A 0 r 2 r [ eikr + ikre ikr ] = A 0 r 2 [ ikeikr + ikre ikr k 2 re ikr ] = k 2 A 0 r eikr = k 2 U(r) 2 U(r) k 2 U(r) = 0 (89) (b) The equation is diverging at r = 0, and one cannot do subtractions between two infinite quantities. No, it is not a solution there. The spherical wave satisfies the homogeneous (no source) equation, but there must be a source to generate the spherical wave. (c) Consider: f(x) = δ(x x )f(x ) dx = since L is a linear differential operator: f(x) = L[ G(x, x )f(x ) dx ]. LG(x, x )f(x ) dx, On the other hand, therefore, Lu(x) = L[ f(x) = Lu(x), G(x, x )f(x ) dx ], which indicates: u(x) = G(x, x )f(x ) dx 12