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3/30/11 Converging lenses Announcements From last time… • HW set 9 due this week; covers Ch 23 and Ch 24.1-4 • Office hours: • My office hours Th 2 -3 pm • or make an appointment Thin Lenses n n Converging, diverging Magnification n Lens equation n M = • Come to class April 19 • course and instructor evaluation • 8 bonus HITT points h' q =! h p 1 1 1 + = p q f • Always check out http://www.phys.ufl.edu/courses/phy2054/spring11/ for more announcements n Diverging lenses Remember the sign conventions! (Table 23.3) Lens maker’s equation n QUESTIONS? PLEASE ASK! "1 1 1 %' = (n ! 1) $$ ! ' f # R1 R2 & Ray tracing n Wave Optics: Coherence and Interference http://www.black-holes.org/gwa4.html n n http://upload.wikimedia.org/wikipedia/commons/5/5d/ Michelson_Interferometer_Green_Laser_Interference.jpg Chapter 24 The wave nature of light à http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html n To produce interference – n Wave Optics n n Interference, diffraction, polarization sources must be coherent (maintain a constant phase w.r.t. each other) (waves should have identical wavelengths) Coherent sources n Old days: Single Slit + Double slit n Today: Lasers! n http://www.ligo.org/ sciencecity.oupchina.com.hk Young s double slit http://fuff.org/interference/two_sources_interference.gif 1 3/30/11 Interference Patterns Young s Double Slit Experiment n n n n n Constructive interference Light incident on a screen with a narrow slit, So Light emerging from So arrive at a second screen that contains two narrow, parallel slits, S1 and S2 The light from the two slits form a visible pattern on a screen The pattern consists of a series of bright and dark parallel bands called fringes n Constructive interference occurs where a bright fringe appears Destructive interference results in a dark fringe Question: What is the difference in the path length (distance) that the two waves travel of the above pictures ? DEMO Interference Equations Interference Equations δ = r2 – r1 = d sin θ n n n n n assumes the paths are parallel, a very good approximation since L >> d Bright fringe (constructive interference) à δ must be either zero or some integral multiple of the wavelength, λ δ = d sin θbright = m λ n n Constructive interference (again) Light waves emerging from S1 and S2 originate from the same wave front and therefore are always in phase n n Destructive interference m = 0, ±1, ±2, … m is called the order number Dark fringe (destructive interference) à δ must be an odd half wavelength δ = d sin θdark = (m + ½) λ n m = 0, ±1, ±2, … n n The positions of the fringes can be measured vertically from the zeroth order maximum y = L tan θ ≈ L sin θ n n n L>>d>>λ θ << 1 For bright fringes (constructive interference) sin θbright = n mλ d y bright = For dark fringes (destructive interference) sin θ dark = 1 ⎞ ⎜ m + ⎟ d ⎝ 2 ⎠ λ ⎛ !L m d y dark = m = 0, ± 1, ± 2 … !L ! 1$ ## m + && m = 0, ± 1, ± 2 … d " 2% 2 3/30/11 Problem 23.13, p 818 Phase Changes Due To Reflection n An electromagnetic wave undergoes a 180° phase change upon reflection from a medium of higher index of refraction than the one in which it was traveling n n Radio waves from a star, of wavelength 250 m, reach a radio telescope by two separate paths as shown in the figure. One is a direct path to the receiver, which is situated on the edge of a cliff. The second is by reflection off the water. The first minimum of destructive interference occurs when the star is is 25° above the horizon. Find the height of the cliff. (Assume no phase change on reflection.) n Analogous to a reflected pulse on a string There is no phase change when the wave is reflected from a boundary leading to a medium of lower index of refraction n Analogous to a pulse in a string reflecting from a free support Interference in Thin Films n n n n Interference is due to the interaction of the waves reflected from both surfaces of the film Ray 1 - phase change of 180° with respect to the incident ray Ray 2 - no phase change with respect to the incident wave Ray 2 travels an additional physical distance of 2t in the film n n n n The wavelength λ is reduced by n in the film à the optical path length is 2 n t 2 n t = (m + ½ ) λ n n n m = 0, 1, 2 … takes into account both the difference in optical path length for the two rays and the 180° phase change Destructive interference n 2nt=mλ m = 0, 1, 2 … Identify the thin film causing the interference Determine the indices of refraction in the film and the media on either side of it Determine the number of phase reversals: zero, one or two Interference is constructive if the path difference is an integral multiple of λ and destructive if the path difference is an odd half multiple of λ n Constructive interference n Handling thin films problems n DEMO NOTE: The conditions are reversed if one of the waves undergoes a phase change on reflection Equation 1 phase reversal 0 or 2 phase reversals 2nt = (m + ½) λ constructive destructive destructive constructive 2nt = m λ 3 3/30/11 Problem 24.26, p 819 n Anti-reflection coatings on solar cells A plano-convex lens with a radius of curvature R = 3.0 m is in contact with a flat plate of glass. A light source and the observer s eye are both close to normal, as shown below. The radius of the 50th bright Newton s ring is found to be 9.8 mm from the center. What is the wavelength of the light produced by the source? Answer to 23.13 Answer to 23.26 4