Chapter 82 Example and Supplementary Problems

Transcription

1 Chapter 82 Example and Supplementary Problems Nature of Polarized Light: 1) A partially polarized beam is composed of 2.5W/m 2 of polarized and 4.0W/m 2 of unpolarized light. Determine the degree of polarization of the beam. Here we make use of the definition of the degree of polarization: V! I p 2.5 = = 0.38 or 38% I p + I u ) What must be the relative orientation of two perfect linear polarizers if a beam of unpolarized light is reduced to 1/4 its intensity upon transmission through the two polarizers? Here the trick is remember that the first polarizer reduces an unpolarized beam by half. We make use of Malus' law for another 50% reduction: I(0) 2 = I(0)cos2! and so: cos! =1/ 2 and so! = 45 o Birefringence and Wave Plates: 3) Given incident 590-nm light, compute the minimum thickness a quartz retarder must have if it is to be a quarter-wave plate. The indices of refractions are n o = and n e = For a phase shift of! /2 between the e and o waves we require:!" = 2! ( # n o " n e )d where d is the thickness. Equivalently:! 4 = ( n " n o e)d and solving for d gives: 590 #10 "9 d = 4(1.544 "1.553) =1.639 #10"5 meters

2 Chapter 3 Example and Supplementary Problems Interference: 4) Two waves of light of wavelength 600 nm (in air) are initially in phase before they encounter two thin plastic films. If the index of refraction of the two layers is n 1 = 1.60 and n 2 = 1.40 and their thicknesses are L1 = 4.00 mm and L 2 = 3.50 mm, what is the phase difference between the two waves after they emerge from the plastic films? The phase difference is related to the optical path length difference:!! = 2"!OPL (0.004m)(1.60)" (0.0035m)(1.40) and thus!! = (6.28) # 600 #10 "9 m!! =1.57#10 4 radians Two-slit Interference or Young s Experiment: 5) Consider monochromatic light (550 nm) incident on two parallel, narrow slits of 0.07 mm apart. (a) What is the angular deviation of third-order (m 3) fringe? (b) If the interference pattern is projected upon a screen 5.7 meters away from the slits, what is the corresponding linear separation of the thirdorder fringe from the central maximum? Here we make use equation derived in the text and the notes for the angular position of a maximum, resulting from constructive interference, as a function of angle with respect to the slit plane. Namely: (a) d sin! m = m" where d is the separation of the slits, m is an integer (the number of waves the optical path length differs by), and " is the wavelength of the light. Thus: sin! 3 = 3(550!10"9 m) 7!10 "5 m = and so! 3 =1o.35 (b) Here we compute the height on the screen of the corresponding fringe (order) given the distance of the screen (D): tan! 3 = y 3 D and so y = Dtan! 3 3 = (5.7m)( ) = meters. 6) One slit of a double-slit arrangement is covered with a thin glass plate with index of refraction 1.40 and the other by a thin glass plate with an index of The fringe pattern is found to shift so that the location of the original central maximum (with no glass plates in place) is now occupied by the m =5

3 bright fringe. If the wavelength of illumination is 480 nm and the thickness of the two plates is the same, what are the thicknesses of the two plates? Here we have an example of how the optical path length of one the waves through one slit is increased. Thus: Again d sin! m = m" and since this is also an expression describing the change in the optical path length for light passing through the two slits vs. angle we can simply write:!opl = (5)(480 "10 #9 meters) = t(n 1 # n 2 ) = t(1.70 #1.40) Solving for t: t = 5(480 "10#9 meters) 0.30 = 8"10 #6 meters Chapter 4: Multiple-beam Interference Fringes from Thin Films: 7) The reflection of perpendicularly incident white light by a soap film (n = 1.33) in air has an interference maximum at 600 nm (bright red fringe) and a minimum at 450 nm. Assuming the soap film is uniform, what is the film thickness?

4 Here we once again consider the optical path length difference between waves reflecting off the first surface of the film and those reflecting off the back surface. The key here is remember that a phase change of! will occur at the first reflection (high index to low) but not at the second (low index to high). Thus the condition for constructive interference (bright maxima) is: OPL = (m +1/ 2)" and so if the film thickness is t: 2nt = (m +1/ 2)" Similarly, the condition for destructive interference (dark minima) would be: 2nt = (m +1)" since the blue light has a shorter wavelength and hence larger phase difference. The problem is that we don't know m but we have two equations and two unknowns and so we can solve for both t and m. Solving both equations for 2t and setting them equal yields: (m +1/ 2)(600!10 "9 ) (1.33) = (m +1)(450!10"9 ) (1.33) (m)(600!10 "9 )+ 300!10 "9 = (m)(450!10 "9 )+ (450!10 "9 ) and so: m(600!10 "9 " 450!10 "9 ) = 450!10 "9 " 300!10 "9 Thus: 150!10"9 m = =1 "9 150!10 Solving the second equation for t gives: t = 2" 2n = 450!10"9 meter =1.692!10 "9 meter (2)(1.33) or: 8) When a convex surface (R = 5.0 m) of a plano-convex lens (diameter of 40 mm) is placed in contact with a flat plate Newton s rings are produced if illuminated by light of wavelength 589 nm. How many are seen over the radius of the lens?

5 Here we need to derive an expression for the air gap thickness between the lens surface and the flat plate. This is also equivalent to the thickness of the plano-convex glass lens as a function of radius given its radius of curvature. The geometry is shown in the figure from which we see: s = R! R 2! r 2 then OPL = n " s # s since the gap is in air. At the edge of the lens (r = 20 mm) we have: OPL = (5.0m)! 5 2 m! m = 4.00 "10!3 meters and so since: OPL = (m +1/ 2)! since there is a phase change. Solving for m: m +1/ 2 = 4.00 "10!3 meters 589 "10!9 meters = fringes. Michelson Interferometer: 9) A think film with index of refraction n = 1.40 is placed in one arm of a Michelson interferometer, perpendicular to the optical path. If this causes a shift of 7.0 bright fringes of the pattern produced by light of wavelength 589 nm, what is the thickness of the film? This problem is similar to #6 in that we introduce a change in the optical path length between the two paths and see a shift in the fringes. Thus:!OPL = nt = m! and so solving for t: t = (7.0)(589 "10#9 meters) (1.40) = 2.945"10 #6 meters