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Unformatted text preview: Ph 214, Spring 2009 Exam # 1 Question #1 SOLUTIONS 1. [20 pts] Interference in a thin film: How uneven is my microscope slide? a. Fringes arise from light reflecting from the top surface of the wedge interfering with light reflecting from the bottom surface. As in any interference problem, we must determine the phase difference between the two beams. This is composed of the phase delay due to the physical path length difference and, possibly, an additional term due to π-phase shift upon reflection. At the left edge of the wedge, the two beams have a phase difference of δ 1 = 2 π (2 d 1 ) λ g + π = 4 πn g d 1 λ air + π where the first term incorporates the physical path length and the second term is due to the fact that the Fresnel reflection coefficient (assuming normal incidence) is negative at the top surface r top ≈ 1- n g 1 + n g ! and positive at the bottom surface r bottom ≈ n g- 1 n g + 1 ! . Following the same logic, the two beams interfering at the right edge have a phase difference δ 2 = 4 πn g d 2 λ air + π . The net phase difference between the right and left edge is, therefore, given by Δ = δ 2- δ 1 = 4 πn g d 2 λ air + π- 4 πn g d 1 λ air + π = 4 πn g λ air ( d 2- d 1 ). For every integer multiple of 2 π , we get an additional bright fringe. In other words, 4 πn g λ air ( d 2- d 1 ) = 2 πf where f is the fractional number of additional fringes ( f ≥ 0). We have been told that there is a bright fringe at the left-edge so the total number of fringes is given by # = f + 1 = 2 n g λ air ( d 2- d 1 ) + 1 . [5 pts for correct determination of Fresnel coefficients; 5 pts for correct phase difference between two ends of wedge; 5 pts for final expression for # of fringes] b. Using the equation derived above, we may write 10 = 2 × 1 . 5 600nm ( d 2- d 1 )+1. Therefore, d 2- d 1 = 1800nm = 1 . 8 μ m . [ 5pts] 1 Ph 214, Spring 2009 Exam # 1 Question #2 SOLUTIONS 2. [30 pts] The interaction of electromagnetic plane waves and materials a. The electric field of an electromagnetic plane wave has a functional form of ~ E ( x,y,z,t ) = ~ E e i ( ωt- ~ k · ~ r ) . The wave is traveling in the +ˆ y direction which means that ~ k iI = k ˆ y and ~ k iI · ~ r = ky . The wave is polarized with the electric field oscillating in a plane tilted 30 ◦ CCW relative to the z- axis which means that ~ E = E 1 2 ˆ x + √ 3 2 ˆ z ! (see figure). Putting it all together, we obtain ~ E iI = E 1 2 ˆ x + √ 3 2 ˆ z ! e i ( ωt- ky )-or- ~ E iI = E 1 2 ˆ x + √ 3 2 ˆ z ! cos( ωt- ky ) . [3 pts for correctly decomposing ~ E (only 2 pts if you left your answer in terms of sin30 ◦ & cos30 ◦ ); 2 pts for correctly stating that ~ k · ~ r = + ky ] b. A λ/ 4 waveplate delays the component along the slow-axis ( z-axis) by π/ 2 in phase relative to the fast axis ( x-axis). In addition, the media in region I and II are identi- cal so the value of k will not change. This implies that ~ E iII = 1 2 E cos( ωt- ky ) ˆ x + √...
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This note was uploaded on 02/27/2012 for the course CHEMISTRY/ CH/ECE/PH/ taught by Professor Faculty during the Spring '08 term at Cooper Union.
- Spring '08