Phy2005 Applied Physics II Spring 2017 Announcements: HW Solns Chs. 26-28 posted Exam III April 10 in-class Reviews Today 4pm NPB 2205 (upstairs) Sunday April 10 4pm NPB 2205 Coverage: Chs. 26-28.4

Maxima or minima of intensity depend on difference in the Last time We know light is a wave because of interference phenomena: Interference patterns with 2 coherent sources Maxima or minima of intensity depend on difference in the path length from the 2 sources to the observation point

Last time s1 s2 • P r1 r2 |r1 – r2| = 0, l, 2l, 3l, …, ml (m: integer)  Constructive Int. |r1 – r2| = l/2, 3l/2, 5l/2, …, (2m+1)l/2  Destructive Int. l

Last time ml (0, ±l, ±2l, …) Constructive dsinq = (m+1)l/2 (±l/2, ±3l/2,…) Destructive d h q x 1 2 3 2nd-order bright fringe 1st order bright fringe

Interference in thin films Ds = Difference in two routes + = 2x (when qi << 1) Half –reflecting planes 2x = ml constructive = (2m+1)l/2 destructive x x q For an arbitrary angle q Ds = 2x/cosq

Ex. 1 The optical flat is a device used to measure the smoothness or flatness of surfaces. Two optical flats are illuminated by 580 nm light. A narrow, wedge-like gap between them gives rise to the fringes as shown. The distance between the fringes is 0.6 cm. Find the angle of the wedge assuming the gap is filled with (1) air and (2) water. q d x2 x1 L 2x1 = ml 2x2 = (m+1)l tanq = d/L = l/(2L) 2(x2 – x1) = l = 2d  d = l/2 tanq = 580 x 10-9/(2 x 0.006) tanq = 436 x 10-9/(2 x 0.006) For air, l = 580 x 10-9 m For water, l = 580 x 10-9/1.33 = 436 x 10-9 m q=0.0028o(air), 0.0020 (water)

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Q1: A thin film of polymer used in an antireflective coating has an index of refraction of 1.5. If light of wavelength% 545nm is incident on the film, but is found to be entirely transmitted and not reflected, what is a possible thickness x of the film (in m)? 9.1 x 10-8 1.8 x 10-7 1.3 x 10-7 (4) 0 (5) 3.6 x 10-7 x n

Dr = dsinq = ml  Constructive Diffraction Grating provides much clearer and sharper interference pattern and a practical device for resolving spectra. Dr = dsinq = ml  Constructive q d

Huygens’ principle construction

Q. A diffraction grating having 20,000 lines per inch is illuminated by parallel light of wavelength 589 nm. What are the angles at which the first- and second-order bright fringes occur? dsinq = ml d = 0.0254/20000 = 1.27 x 10-6 (m) First-order sinq1 = ml/d = 589 x 10-9/1.27 x 10-6 = 0.464 q1 = 27.6 Similarly, sinq2 = 2 x 0.464 = 0.928 q2 = 68.1