Midterm Exam 2. Nov. 17, Optics I: Theory CPHY72250

Transcription

1 Midterm Exam. Nov. 17, 015 Optics I: Theory CPHY750 - time: 1hr 15 mins. - show all work clearly - indicate reasoning where appropriate name 1. (a) Show that two cascaded quarter-wave retarder plates with parallel fast axes are equivalent to a half-wave retarder. The T-matrix for a quarter-wave retarder is ( ) 1 0 T = 0 e i π and for two of these, we have ( ) ( T tot = 0 e i π 0 e i π ) = ( e iπ ) = ( ) which is the T-matrix for a half-wave plate. 1

2 (b) What if the fast axes are orthogonal? If the two fast axes are orthogonal, then we have ( ) ( ) ( ) 1 0 e i π 0 i 0 T tot = 0 e i π = i which is a half-wave retarder (not a plate; there is a half-wave retardation for all polarizations.)

3 . Explain, quantitatively, why metals reflect light very effi ciently. In the simplest approximation, the relative dielectric constant in isotropic materials is given by ε r = 1 + ω p ω o ω iβ ω In metals, the electrons are free, their spring constant is zero, and hence ω 0 = 0. Ignoring dissipation, we have ε r = 1 ω p ω Since the plasma frequency is in the UV, ( ω p Hz), for visible light, ω < ω p, and ε r < 0. Since the refractive index n is given by n = ε it follows that n is imaginary; that is, n = i n. The reflection coeffi cient r σ is given, for normal incidence, by r σ = 1 n 1 + n = 1 i n 1 + i n The reflection coeffi ceint for the intensity is R = rr = 1 i n 1 + i n 1 + i n 1 i n = 1 and all the incident light is therefore reflected. 3

4 3. Light is incident on an interface as shown; n 1 > n and total internal reflection takes place. y 1 ik iθ rθ rk x The components of the wave vectors are where (a) What are their magnitudes? k i = (n 1 k o sin θ i, n 1 k o cos θ i ) k r = (n 1 k o sin θ i, n 1 k o cos θ i ) k o = π λ o The magnitudes of the wave vectors are k i = k r = k i k i = n 1 k 0 = πn 1 (b) What is the magnitude of the wave vector of the transmitted wave k t? The magnitude of the transmitted wave is 4 k t = πn

5 (c) Give the components of k t, given that n 1 > n. Wave vector components must be continuous across the interface, so we must have k tx = k ix = πn 1 sin θ i Since we have that and k tx + k ty = k t k ty = k t k tx = ( πn ) ( πn 1 sin θ i ) k ty = πn 1 n 1 sin θ n i This indicates that k ty is imaginary if sin θ i > n /n 1, and we have an evanescent wave decaying in the -y direction 5

6 4. You are standing outside on a clear afternoon, facing south. The sun is on your right, about 45 o above the horizon. You are looking at the sky straight ahead, just above the horizon. What is the polarization of light coming from the sky? Light from the sun is polarized perpendicular to its propagaition direction, which is SE at the point you are looking at. Imagine a plane perpendicular to this direction. Light reaching your eyes from the southern sky is polarized perpendicular to its propagation direction, which is N. Imagine a plane perpendicular to this direction. Since light originates in the sun, the light reaching your eyes must be polarized so that it is in both planes; it must therefore be along the intersection of the two planes, that is, it must be vertically polarized. 6

7 5. Plane polarized light is normally incident on a nematic cell with orientation as shown. The refractive indices are n o = 1.3, and n e = 1.5 the optic axis is at 45 o from the horizontal. E k 45 o (a) What is the direction of wave propagation inside the cell? Since the tangential components of the wave vector k are continuous accross the interface, the direction of the wave vector does not change on normal incidence. The direction of wave propagation inside the cell is normal to the cell walls. (b) What is the speed of wave propagation inside the cell? Since E is in the plane of ˆn and k, the wave is the extraordinary eigenmode. The refractive index is n = n e n o n o sin θ+n e cos θ = where θ is the angle betweeen ˆn and k. The speed is v = c n = = m/s 7

8 (c) What is the direction of the Poynting vector inside the cell? The Poynting vector is in the plane of ˆn and k, making an agle β with the k vector, where β is given by and β = 8.09 tan β = (n e n o) sin θ cos θ n o sin θ+n e cos θ = 0.14 (d) What is the direction of the electric field inside the cell? The E field is normal to the index elllipsoid; it makes the angle β with the D vector The Poynting vector and the electric field direction inside the cell are shown below E k E 45 P o β 8

9 Useful Information c = m/s q = C m e = kg ε o = F/m µ o = 4π 10 7 H/m Z = µ ε T = S = E H [ ] Ex J = E y [ e iγ ] n 1 sin θ i = n sin θ t r π = n cos θ i n 1 cos θ t n cos θ i + n 1 cos θ t r σ = n 1 cos θ i n cos θ t n 1 cos θ i + n cos θ t k ω 0 = m k = kx + ky + kz n = n e n o n e cos θ + n o sin θ tan β = (n e n o) sin θ cos θ n e cos θ + n o sin θ ε r = 1 + ω p ω o ω iβ ω 9