Chapter 28 Physical Optics Interference and Diffraction Answers to Even Numbered Conceptual Questions 2 If the slit spacing d were less than the wavelength the condition for a bright fringe equation 28 1 could be satisfied only for the central bright fringe m 0 For nonzero values of m there are no solutions because sin cannot be greater than one In addition equation 28 2 shows that if d is greater than 2 though still less than there will be only one dark fringe on either side of the central bright fringe If d is less than 2 no dark fringes will be observed 4 The locations of bright and dark fringes depend on the wavelength of light Therefore if white light is used in a two slit experiment each bright fringe will show some separation into colors giving a rainbow effect 6 Submerging the two slit experiment in water would reduce the wavelength of the light from to n where n 1 33 is the index of refraction of water Therefore the angles to all the bright fringes would be reduced as we can see from equation 28 1 It follows that the two slit pattern of bright fringes would be more tightly spaced in this case 8 One possible reason is that one of the films may have an index of refraction greater than that of glass whereas the other may have an index of refraction that is less than that of glass If this is the case the phase change in reflection from the film glass interface will be different for the two films This in turn would result in different colors appearing in the reflected light 10 A cat s eye would give greater resolution in the vertical direction because the effective aperture is greater in that direction As shown in equation 28 14 the greater the aperture D the smaller the angle and the greater the resolution 12 In an iridescent object the color one sees is determined by constructive and destructive interference The conditions for interference however depend on path length and path length depends on the angle from which one views the system This is analogous to viewing a two slit system from different angles and seeing alternating regions of constructive and destructive interference A painted object on the other hand simply reflects light of a given color in all directions Solutions to Problems and Conceptual Exercises 1 Picture the Problem Coherent in phase 26 0 m wavelength waves interfere at a location 78 0 m from one source and 143 m from the other source Strategy Subtract the distances from each source to find the difference in path length Divide this distance by the wavelength If the result is an integer value then the waves constructively interfere If the result is a half integer value the waves destructively interfere Solution 1 Calculate the difference in path lengths d d2 d1 143 m 78 0 m 65 m 2 Divide the difference by the wavelength d 65 m 2 5 26 0 m 3 Because the path difference is 2 5 wavelengths the waves interfere destructively at the observation point Insight If the observation point is moved 13 m closer to one of the two sources while staying at the same distance from the other source the waves will constructively interfere Copyright 2010 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 28 1 James S Walker Physics 4th Edition Chapter 28 Physical Optics Interference and Diffraction Copyright 2010 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 28 2 James S Walker Physics 4th Edition Chapter 28 Physical Optics Interference and Diffraction 2 Picture the Problem Coherent in phase 26 0 m wavelength waves interfere at various observation points Strategy Subtract the distances from each source to find the difference in path length Divide this distance by the wavelength If the result is an integer value then the waves constructively interfere If the result is a half integer value the waves destructively interfere Solution 1 a Calculate the difference in path lengths d d2 d1 221 m 91 0 m 130 m 2 Divide the difference by the wavelength d 130 m 5 0 26 0 m 3 Because the path difference is 5 0 wavelengths the waves interfere constructively at the observation point 4 b Calculate the difference in path lengths d d2 d1 135 m 44 0 m 91 m 5 Divide the difference by the wavelength d 91 m 3 5 26 0 m 6 Because the path difference is 3 5 wavelengths the waves interfere destructively at the observation point Insight Whenever the path length difference is an integer the waves will interfere constructively When the path length difference is a half integer value the waves will interfere destructively 3 Picture the Problem Two coherent waves interfere at an observation point 161 meters from one source and 295 meters from the other source Strategy The longest wavelength that will give constructive interference has a length equal to the path length difference Solution Calculate the path length difference l 295 m 161 m 134 m Insight The longest wavelength that will give destructive interference is 268 m 4 Picture the Problem The figure shows a car traveling at 17 m s perpendicular to the line connecting two radio transmitters The car picks up a maximum signal when it is at point A Strategy Set the wavelength equal to the difference in distances between the two towers Use the Pythagorean theorem to calculate the distance between the car and each tower as a function of time Set the difference in distance length equal to half a wavelength for destructive interference and solve for the time Solution 1 a Find the l 450 m 150 m 300 m difference in path lengths 2 b Find the distance to the 2 l 1 450 m v2t2 far tower as a function of time 3 Find the distance to the near tower as a function of time l 2 150 m v2t2 4 Set the difference equal to half a wavelength l 1 l 2 450 m v2t2 150 m v2t2 5 Rearrange the expression square both sides and evaluate 2 2 2 2 2 2 450 m v t 150 m 150 m v t 2 2 2 2 set 150 m 2 2 450 m 2 v2t2 150 m 2 2 300 m 150 m 2 v2t2 2 150 m v2t2 Copyright 2010 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 28 3