WAVE OPTICS. Conceptual Questions Because ym

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1 WAVE OPTICS Conceptual Questions.. The initial light pattern is a ouble-slit interference pattern. It is centere behin the ipoint of the slits. The slight ecrease in intensity going outwar fro the ile inicates that the light fro each of the iniviual slits is not unifor but slowly ecreases towar the eges of the screen. If the right slit is covere, light coes through only the left slit. Without a secon slit, there is no interference. Instea, we get siply the sprea-out pattern of light iffracting through a single slit, such as in the center of the photograph of Figure.. The intensity is a axiu irectly behin the left slit, an as we iscerne fro the intensities in the ouble-slit pattern the single-slit intensity faes graually towar the eges of the screen. L.. Because y increasing an L increases the fringe spacing. Increasing ecreases the fringe spacing. Suberging the experient in water ecreases an ecreases the fringe spacing. So the answers are (a) an (c)..3. The fringe separation for the light intensity pattern of a ouble-slit is eterine by y L/. (a) If increases, y will ecrease. (b) If ecreases, y will increase. (c) If L ecreases, y will ecrease. () Since the ot is in the bright fringe, the path length ifference fro the two slits is 000 n.4. (a) The equation for gratings oes not contain the nuber of slits, so increasing the nuber of slits can t affect the angles at which the bright fringes appear as long as is the sae. So the nuber of fringes on the screen stays the sae. (b) The nuber of slits oes not appear in the equation for the fringe spacing, so the spacing stays the sae. (c) ecreases; the fringes becoe narrower. () The equation for intensity oes contain the nuber of slits, so each fringe becoes brighter: I ax N I.5. a. Several seconary axia appear. For asin p, the first inia fro the central axiu require asin, which ust be less than. p Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher. -

2 - Chapter.6. (a) The with increases because. (b) The with ecreases because. (c) Alost uniforly gray with no inia..7. When the experient is ierse in water the frequency of the light stays the sae. But the spee is slower, L so the wavelength is saller. When the wavelength is saller the fringes get closer together because y.8. (a) Because the hole is so large we o not use the wave oel of light. The ray oel says we treat the light as if it travels in straight lines, so increasing the hole iaeter will increase the size of the spot on the screen. (b) When the hole is saller than we nee to consier iffraction effects; aking the hole 0% larger will ecrease the iaeter of the spot on the screen..9. Moving a irror by 00 n increases or ecreases the path length by 400 n. Since one irror is ove in an the other out each increases the path length ifference by 400 n. The total is a path length ifference of 800 n, so the spot will still be bright..0. If the wavelength is change to / there will still be crests everywhere there were crests before (plus new crests halfway between, where there use to be troughs). If the interferoeter was set up to isplay constructive interference originally, then crests were arriving fro each ar of the interferoeter together at the etector. That will still be true when the wavelength is halve. So the central spot reains bright. Exercises an Probles Section. The Interference of Light.. Moel: Two closely space slits prouce a ouble-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure.3(b). It is syetrical with the 3 fringes on both sies of an equally istant fro the central axiu. Solve: The bright fringes occur at angles such that 9 sin 0,,, 3, 3(600 0 ) 80 sin ra 003 ra 3 (80 0 ) ra.. Moel: Two closely space slits prouce a ouble-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure.3(b). Solve: The bright fringes are locate at positions given by Equation.4, sin. For the 3 bright orange 9 fringe, the interference conition is sin 3(600 0 ). For the 4 bright fringe the conition is sin4 4 Because the position of the fringes is the sae, sin sin 4 3(600 0 ) (600 0 ) 450 n.3. Moel: Two closely space slits prouce a ouble-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure.3(b). It is syetrical, with the fringes on both sies of an equally istant fro the central axiu. Solve: The two paths fro the two slits to the bright fringe iffer by r r r, where r (500 n) 000 n Thus, the position of the bright fringe is 000 n farther away fro the ore istant slit than fro the nearer slit. Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

3 Wave Optics Moel: Two closely space slits prouce a ouble-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure.3(b). Solve: The forula for fringe spacing is L 3 9 L L y 80 (6000 ) 3000 The wavelength is now change to 400 n, an L /, being a part of the experiental setup, stays the sae. Applying the above equation once again, L 9 y (4000 )(3000) L.5. Visualize: The fringe spacing for a ouble slit pattern is y We are given L.0 an 600 n We also see fro the figure that y 3 c Solve: Solve the equation for. Assess: 0.36 is a typical slit spacing. 3 9 L (6000 )(0 ) 036 y 0.6. Moel: Two closely space slits prouce a ouble-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure.3(b). Solve: The fringe spacing is 9 L L (589 0 )(50 0 ) y 0 y Moel: Two closely space slits prouce a ouble-slit interference pattern. Visualize: The interference fringes are equally space on both sies of the central axiu. The interference pattern looks like Figure.3(b). Solve: In the sall-angle approxiation ( ) Since 00, we have ra Moel: Two closely space slits prouce a ouble-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure.3(b). Solve: The ark fringes are locate at positions given by Equation.9: L y 0,,, 3, L L 3 5 y 3 4 (60 0 ) y n 00 0 Section.3 The iffraction Grating.9. Moel: A iffraction grating prouces an interference pattern. Visualize: The interference pattern looks like the iagra in Figure.8. Solve: The bright constructive-interference fringes are given by Equation.5: sin 0,,, Likewise, sin 0055 an 3. 9 ()(550 0 ) sin (400 )/000 Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

4 -4 Chapter.0. Moel: A iffraction grating prouces a series of constructive-interference fringes at values of eterine by Equation.5. Solve: We have iviing these two equations, sin 0,,, 3, sin 00 an sin sin sin Moel: A iffraction grating prouces an interference pattern. Visualize: The interference pattern looks like the iagra in Figure.8. Solve: The bright constructive-interference fringes are given by Equation.5: ()(600 0 ) 6 sin 890 sin sin(39 5 ) 9 The nuber of lines in per illieter is 3 6 ( 0 )/( 89 0 ) Moel: A iffraction grating prouces an interference pattern. Visualize: The interference pattern looks like the iagra of Figure.8. Solve: The bright interference fringes are given by sin 0,,, 3, 6 The slit spacing is / an. For the re an blue light, sin 9 5 sin re 6 blue 6 The istance between the fringes, then, is y y re y blue where So, y c y y re blue (5 )tan95 05 (5 )tan Moel: A iffraction grating prouces an interference pattern. Visualize: The interference pattern looks like the iagra in Figure.8. Solve: The bright fringes are given by Equation.5: sin 0,,, 3, sin () / sin Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

5 Wave Optics -5 The angle can be obtaine fro geoetry as follows: (03 )/ tan 0080 tan (0.080) Using sin sin , Moel: A iffraction grating prouces a series of constructive-interference fringes at values of that are eterine by Equation.5. Solve: For the 3 axiu of the re light an the 5 axiu of the unknown wavelength, Equation.5 gives unknown sin 3(660 0 ) sin 5 The 5 fringe an the 3 fringe have the sae angular positions. This eans 5 3. equations, unknown Section.4 Single-Slit iffraction 9 5 3(660 0 ) 396 n unknown.5. Moel: A narrow single slit prouces a single-slit iffraction pattern. Visualize: The intensity pattern for single-slit iffraction will look like Figure.4. Solve: The inia occur at positions L y p a p L L L L (6330 )(5 ) 4 So y y y a a a a y iviing the two.6. Moel: A narrow slit prouces a single-slit iffraction pattern. Visualize: The intensity pattern for single-slit iffraction will look like Figure.4. Solve: The with of the central axiu for a slit of with a 00 is L (0 ) w 0 a Visualize: We are given L 0 an w L Solve: Solve w for a. a L (0 ) 0 a 0 50 w Assess: This is a typical slit with. L.8. Visualize: Use Equation.: w We are given 600 n an L.0. We see fro the figure a that w 0 c Solve: Solve the equation for a. L (6000 )(0 ) a 04 w 000 Assess: 0.4 is in the range of typical slit withs. 9 Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

6 -6 Chapter.9. Moel: A narrow slit prouces a single-slit iffraction pattern. Visualize: The intensity pattern for single-slit iffraction will look like Figure.4. Solve: The with of the central axiu for a slit of with a 00 is 9 L (500 0 )(0 ) w a Moel: The spacing between the two builings is like a single slit an will cause the raio waves to be iffracte. Solve: Raio waves are electroagnetic waves that travel with the spee of light. The wavelength of the 800 MHz waves is 8 30 /s Hz To investigate the iffraction of these waves through the spacing between the two builings, we can use the general conition for coplete estructive interference: asin p p,, 3, where a is the spacing between the p builings. Because the with of the central axiu is efine as the istance between the two p inia on either sie of the central axiu, we will use p an obtain the angular with fro 0375 asin sin sin 43 a 5 Thus, the angular with of the wave after it eerges fro between the builings is ( 43 ) Moel: The crack in the cave is like a single slit that causes the ultrasonic soun bea to iffract. Visualize: Solve: The wavelength of the ultrasoun wave is 340 /s khz Using the conition for coplete estructive interference with p, 003 asin sin Fro the geoetry of the iagra, the with of the soun bea is w y (00 tan ) 00 tan Assess: The sall-angle approxiation is alost always vali for the iffraction of light, but ay not be vali for the iffraction of soun waves, which have a uch larger wavelength. Section.5 Circular-Aperture iffraction.. Moel: Light passing through a circular aperture leas to a iffraction pattern that has a circular central axiu surroune by a series of seconary bright fringes. Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

7 Wave Optics -7 Solve: The with of the central axiu for a circular aperture of iaeter is 9 44 L (44)(500 0 )(0 ) w Moel: Light passing through a circular aperture leas to a iffraction pattern that has a circular central axiu surroune by a series of seconary bright fringes. Visualize: The intensity pattern will look like Figure.5. Solve: Accoring to Equation.3, the angle that locates the first iniu in intensity is ()(.5 0 ) ra These shoul be roune to 0.05 ra Moel: Light passing through a circular aperture leas to a iffraction pattern that has a circular central axiu surroune by a series of seconary bright fringes. Visualize: The intensity pattern will look like Figure.5. Solve: Fro Equation.4, 3 w (0 0 )( 00 ) L 78 c (633 0 ).5. Moel: Light passing through a circular aperture leas to a iffraction pattern that has a circular central axiu surroune by a series of seconary bright fringes. Visualize: The intensity pattern will look like Figure.5. Solve: Fro Equation.4, the iaeter of the circular aperture is Section.6 Interferoeters 9 44L 44(6330 )(40 ) 05 w Moel: An interferoeter prouces a new axiu each tie L increases by causing the path-length ifference r to increase by. Visualize: Please refer to the interferoeter in Figure.0. Solve: Fro Equation.33, the nuber of fringe shifts is L ( 000 ) 30, Moel: An interferoeter prouces a new axiu each tie L increases by causing the path-length ifference r to increase by. Visualize: Please refer to the interferoeter in Figure.0. Solve: Fro Equation.33, the wavelength is 6 L (00 0 ) n Moel: An interferoeter prouces a new axiu each tie L increases by causing the path-length ifference r to increase by. Visualize: Please refer to the interferoeter in Figure.0. Solve: Fro Equation.33, the istance the irror oves is 9 (33,98)( ) L c Assess: Because the wavelength is known to 6 significant figures an the fringes are counte exactly, we can eterine L to 6 significant figures. Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

8 -8 Chapter.9. Moel: An interferoeter prouces a new axiu each tie L increases by causing the path-length ifference r to increase by. Visualize: Please refer to the interferoeter in Figure.0. Solve: For soiu light of the longer wavelength ( ) an of the shorter wavelength ( ), L L ( ) We want the sae path ifference ( L L) to correspon to one extra wavelength for the soiu light of shorter wavelength ( ). Thus, we cobine the two equations to obtain: 5890 n ( ) ( ) n 5890 n Thus, the istance by which M is to be ove is n L Moel: Two closely space slits prouce a ouble-slit interference pattern with the intensity graph looking like Figure.3(b). The intensity pattern ue to a single slit iffraction looks like Figure.4. Both the spectra consist of a central axiu flanke by a series of seconary axia an ark fringes. Solve: (a) The light intensity shown in Figure P.30 correspons to a ouble-slit aperture. This is because the fringes are equally space an the ecrease in intensity with increasing fringe orer occurs slowly. (b) Fro Figure P.30, the fringe spacing is y 0 c 00 Therefore, 9 L L (600 0 )(5 ) y 05 y Moel: Two closely space slits prouce a ouble-slit interference pattern with the intensity graph looking like Figure.3(b). The intensity pattern ue to a single slit iffraction looks like Figure.4. Both the spectra consist of a central axiu flanke by a series of seconary axia an ark fringes. Solve: (a) The light intensity shown in Figure P.3 correspons to a single slit aperture. This is because the central axiu is twice the with an uch brighter than the seconary axiu. (b) Fro Figure P.3, the separation between the central axiu an the first iniu is y.0 c.0 0. Therefore, using the sall-angle approxiation, Equation. gives the conition for the ark iniu: y p 9 pl L (5 )(600 0 ) a 05 y Moel: Two closely space slits prouce a ouble-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure.3(b). Solve: In a span of fringes, there are gaps between the. The forula for the fringe spacing is 3 9 L 50 (6330 )(30 ) y 0.40 Assess: This is a reasonable istance between the slits, ensuring L / Solve: Accoring to Equation.7, the fringe spacing between the fringe an the fringe is y L/. y can be obtaine fro Figure P.33. The separation between the fringes is.0 c, iplying that the separation between the two consecutive fringes is (.0 c) 0.50 c. 4 Thus, 4 3 L y (000 )(0500 ) y L 67 c Assess: A istance of 67 c fro the slits to the screen is reasonable. Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

9 Wave Optics Solve: Accoring to Equation.7, the fringe spacing between the fringe an the fringe is y L/. y can be obtaine fro Figure P.33. Because the separation between the fringes is.0 c, two consecutive fringes are y ( 0 c) 0 50 c 4 apart. Thus, 3 L y (00 0 )(0500 ) y n L Solve: The intensity of light of a ouble-slit interference pattern at a position y on the screen is Iouble 4I cos y L where I is the intensity of the light fro each slit alone. At the center of the screen, that is, at y 0, I I Fro Figure P.33, is I 3 W/ I ouble at the central axiu is 4 ouble W/. So, the intensity ue to a single slit.36. Prepare: The path length ifference is affecte by this rotation. See accopanying figure.. We are given The new part of the path length ifference is r sin while the ol piece is still r sin, so the total path length ifference ust obey (sin sin ) for constructive interference. However, looking at the central axiu, 0, so sin sin This is inepenent of an. Solve: y Ltan Ltan[sin ( sin )] (.00 )tan[sin ( sin.0 )] 75 c So the central axiu oves by.75 c on the screen Assess: This sees like a large shift for a rotation, but is probably reasonable for L..37. Moel: Two closely space slits prouce a ouble-slit interference pattern. The interference pattern is syetrical on both sies of the central axiu. Visualize: Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

10 -0 Chapter In figure (a), the interference of laser light fro the two slits fors a constructive-interference central ( 0) fringe at P. The paths P an P are equal. When a glass piece of thickness t is inserte in the path P, the interference th between the two waves yiels the 0 ark fringe at P. Note that the glass piece is not present in figure (a). Solve: The nuber of wavelengths in the air-segent of thickness t is t The nuber of wavelengths in the glass piece of thickness t is t t nt n glass / The path length has thus increase by wavelengths, where th Fro the 0 ark fringe to the one-half of a fringe. Hence, ( n ) t st ark fringe is 9 fringes an fro the st ark fringe to the 9 9 t 9 9 (6330 ) 9 ( n ) t 0 ( n ) Moel: A iffraction grating prouces an interference pattern. Visualize: The interference pattern looks like the iagra in Figure.8. Solve: 500 lines per on the iffraction grating gives a spacing between the two lines of 3 6 ( 0 )/ The wavelength iffracte at angle 30 in orer is 9 th 0 bright fringe is /500 sin 6 (00 )sin n We re tol it is visible light that is iffracte at 30, an the wavelength range for visible light is n. Only gives a visible light wavelength, so 500 n..39. Moel: Each wavelength of light is iffracte at a ifferent angle by a iffraction grating. Solve: Light with a wavelength of 50.5 n creates a first-orer fringe at y.90 c. This light is iffracte at angle 90 c tan c We can then use the iffraction equation sin, with, to fin the slit spacing: The unknown wavelength creates a first orer fringe at 505 n 50 n sin sin(365 ) y 3.60 c, or at angle 360 c tan c With the split spacing now known, we fin that the wavelength is sin (50 n)sin(3 9 ) n Assess: The istances to the fringes an the first wavelength were given to 4 significant figures. Consequently, we can eterine the unknown wavelength to 4 significant figures..40. Moel: Assue the incient light is coherent an onochroatic. Solve: (a) Where the path length ifference is between ajacent slits there will be constructive interference fro all three slits, an I ax N I applies. ax 3 9 I I I Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

11 Wave Optics - (b) For the case where the path length ifference is / ajacent slits will prouce estructive interference, so the thir slit will just give a total intensity of I..4. Moel: We will assue that the listeners i not hear any other lou spots between the center an.4 on each sie,. We ll use the iffraction grating equations. First solve Equation.6 for an insert it into Equation.5. We nee the wavelength of the soun waves, an we ll use the funaental relationship for perioic waves to get it. v 340 /s 3 4 c f 0000 Hz We are also given L 0 an y 4 Solve: y 4 tan tan 04 ra L 0 (This ay be sall enough to use the sall-angle approxiation, but we are alost finishe with the proble without it, so aybe we can save the approxiation an try it in the assess step.) () c sin sin(04 ra) Assess: These nubers see reasonable given the size (wavelength) of soun waves. With the sall-angle approxiation, sin tan, (0034 ) 4 c / y L 4 /0 This is alost the sae, but roune own to two significant figures rather than up. The we copute seee alost sall enough to use the approxiation, an it woul probably be OK for any applications when the angle is this sall..4. Moel: A iffraction grating prouces an interference pattern. Visualize: The interference pattern looks like the iagra of Figure.8. Solve: (a) A grating iffracts light at angles sin /. The istance between ajacent slits is / n The angle of the fringe is 500 n sin n The istance fro the central axiu to the bright fringe on a screen at istance L is y Ltan ( ) tan (Note that the sall angle approxiation is not vali for the axia of iffraction gratings, which alost always have angles 0. ) There are two bright fringes, one on either sie of the central axiu. The istance between the is y y 58 3 (b) The axiu nuber of fringes is eterine by the axiu value of for which sin oes not excee because there are no physical angles for which sin. In this case, (500 n) sin 667 n We can see by inspection that,, an 3 are acceptable, but 4 woul require a physically ipossible sin 4 Thus, there are three bright fringes on either sie of the central axiu plus the central axiu itself for a total of seven bright fringes..43. Moel: Assue the screen is centere behin the slit. We actually want to solve for, but given the other ata, it is unlikely that we will get an integer fro the equations for the ege of the screen, so we will have to truncate our answer to get the largest orer fringe on the screen. o Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

12 - Chapter Visualize: Refer to Figure.7. Use Equation.5: sin, an Equation.6: y Ltan We are given 50 n, L 0, an 500 As entione above, we are not guarantee that a bright fringe will occur exactly at the ege of the screen, but we will kin of assue that one oes an set y 0 ; if we o not get an integer for then the fringe was not quite at the ege of the screen an we will truncate our answer to get an integer. Solve: Solve Equation.6 for an insert it in Equation.5. Solve Equation.5 for tan y L sin sin tan y sin tan 8 L 50 n 0 Inee, we i not get an integer, so truncate.8 to get This eans we will see three fringes, one for 0, an one on each sie for Assess: Our answer fits with the stateent in the text: Practical gratings, with very sall values for, isplay only a few orers..44. Moel: A iffraction grating prouces an interference pattern that is eterine by both the slit spacing an the wavelength use. The visible spectru spans the wavelengths 400 n to 700 n. Solve: Accoring to Equation.6, the istance y fro the center to the th axiu is y Ltan. The angle of iffraction is eterine by the constructive-interference conition sin, where 0,,, 3, The with of the rainbow for a given fringe orer is thus w y re y violet. The slit spacing is For the re wavelength an for the orer, sin () sin sin Fro the equation for the istance of the fringe, y Ltan (0 )tan(4 83 ) 9 56 c re Likewise for the violet wavelength, sin 388 y 6 violet (0 )tan(3 88 ) 494 c The with of the rainbow is thus 9.56 c 49.4 c 43.4 c 43 c..45. Moel: A iffraction grating prouces an interference pattern that is eterine by both the slit spacing an the wavelength use. Solve: (a) If blue light (the shortest wavelengths) is iffracte at angle, then re light (the longest wavelengths) is iffracte at angle 30. In the first orer, the equations for the blue an re wavelengths are b sin sin( 30 ) r Cobining the two equations we get for the re wavelength, b b r (sin cos30 cos sin30 ) (08660sin 050cos ) (050) b r b b r b (050) (050) ( ) ( ) Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

13 Wave Optics -3 9 r 08660b b Using b an r 700 0, we get 85 0 This value of correspons to lines/ (b) Using the value of fro part (a) an 589 0, we can calculate the angle of iffraction as follows: 7 9 sin () (85 0 )sin Moel: A iffraction grating prouces an interference pattern that is eterine by both the slit spacing an the wavelength use. Solve: An 800 line/ iffraction grating has a slit spacing 3 6 Figure P.46, the angle of iffraction is given by y 0436 tan sin 0400 L 0 Using the constructive-interference conition sin, ( 00 )/ Referring to sin 6 (5 0 )(0400) 500 n We can obtain the sae value of by using the secon-orer interference fringe. We first obtain : y tan sin 0800 L 0 Using the constructive-interference conition, sin 6 (5 0 )(0800) 500 n Assess: Calculations with the first-orer an secon-orer fringes of the interference pattern give the sae value for the wavelength..47. Moel: A iffraction grating prouces an interference pattern that is eterine by both the slit spacing an the wavelength use. Solve: Fro Figure P.46, 0436 tan sin Using the constructive-interference conition sin, Thus, the nuber of lines per illieter is sin 3557 ()(600 0 ) 50 0 sin(3557 ) lines/ 50 0 Assess: The sae answer is obtaine if we perfor calculations using inforation about the secon-orer bright constructive-interference fringe..48. Moel: A narrow slit prouces a single-slit iffraction pattern. Visualize: The iffraction-intensity pattern fro a single slit will look like Figure.4. Solve: The ark fringes are locate at pl yp p = 0,,, 3, a Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

14 -4 Chapter The locations of the first an thir ark fringes are Subtracting the two equations, L 3L y y3 a a L L (589 0 )(075 ) ( y y ) a a y 3 3 y.49. Visualize: The relationship between the iffraction grating spacing, the angle at which a particular orer of constructive interference occurs, the wavelength of the light, an the orer of the constructive interference is sin. Also note N / Solve: The first orer iffraction angle for green light is 7 6 sin ( / ) sin (55 0 /0 0 ) sin (075) 078 ra 6 Assess: This is a reasonable angle for a first orer axiu..50. Moel: The peacock feather is acting as a reflection grating, so we ay use Equation.5: sin, with (we are tol it is first-orer iffraction), 470 n, an 5 06 ra Solve: Solve Equation.5 for 9 ()(470 0 ) 8 sin sin(0.6 ra) Assess: The answer is sall, but plausible for the bans on the barbules..5. Moel: Assue the incient light is coherent an onochroatic. Visualize: The istances given in the table are y since the easureent is between the two first orer fringes. Also note that in this proble. Solve: The equation for iffraction gratings is sin where is the istance (in ) between slits; we seek, Fro the nuber of lines per. sin istance y Ltan an y, where istance is the istance between first orer fringes as given in the table, we arrive at istance sintan L istance This leas us to believe that a graph of sintan L vs. will prouce a straight line whose slope is an whose intercept is zero. Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

15 Wave Optics -5 We see fro the spreasheet that the linear fit is excellent an that the slope is an the intercept is very sall. Because the nuber of lines per is always reporte as an integer we roun this answer to 800 lines/. Assess: This nuber of lines per is a typical nuber for a ecent grating..5. Moel: Sall angle approxiations woul not apply here. Visualize: We see fro the figure that if y L then 45. Solve: Fro sin we get but 45, an sin45. sin Assess: This large iffraction angle correspons to a wavelength that is about the sae size as the slit spacing. Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

16 -6 Chapter.53. Moel: A narrow slit prouces a single-slit iffraction pattern. Visualize: The iffraction-intensity pattern fro a single slit will look like Figure.4. Solve: These are not sall angles, so we can t use equations base on the sall-angle approxiation. As given by Equation.9, the ark fringes in the pattern are locate at asin p, where p,, 3, For the first iniu of the pattern, p. Thus, a p sin sin p For the three given angles the slit with to wavelength ratios are a a a (a):, (b): 5, (c): 30 sin30 60 sin60 90 sin90 Assess: It is clear that the saller the a/ ratio, the wier the iffraction pattern. This is a conclusion that is contrary to what one ight expect..54. Moel: A narrow slit prouces a single-slit iffraction pattern. Visualize: The iffraction-intensity pattern fro a single slit will look like Figure.4. Solve: As given by Equation.9, the ark fringes in the pattern are locate at asin p, where p,, 3, For the iffraction pattern to have no inia, the first iniu ust be locate at least at 90. Fro the constructive-interference conition asin p, we have p p a a 633 n sin sin90 p.55. Moel: A narrow slit prouces a single-slit iffraction pattern. Visualize: The ark fringes in this iffraction pattern are given by Equation.: pl yp p,, 3, a We note that the first iniu in the figure is 0.50 c away fro the central axiu. We are given a 0.0 n an L 5 Solve: Solve the above equation for ya 3 p (050 0 )(00 0 ) 670 n pl ()( 5 ) Assess: 670 n is in the visible range..56. Moel: A narrow slit prouces a single-slit iffraction pattern. Solve: The ark fringes in this iffraction pattern are given by Equation.: pl yp p,, 3, a We note fro Figure P.55 that the first iniu is 0.50 c away fro the central axiu. Thus, ay 3 p (050 )(050 0 ) L.3 n p 9 ()(600 0 ) Assess: This is a typical slit to screen separation..57. Moel: Light passing through a circular aperture leas to a iffraction pattern that has a circular central axiu surroune by a series of seconary bright fringes. Solve: Within the sall angle approxiation, which is alost always vali for the iffraction of light, the with of the central axiu is w y 44 L p p Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

17 Wave Optics -7 Fro Figure P.55, w.0 c, so 9 44L.44(500 0 )(.0 ) 0. w (.0 0 ) Assess: This is a typical size for an aperture to show iffraction..58. Moel: Light passing through a circular aperture leas to a iffraction pattern that has a circular central axiu surroune by a series of seconary bright fringes. Solve: Within the sall-angle approxiation, the with of the central axiu is w 44 L Because w, we have L L (44)(6330 )(050 ) Moel: Light passing through a circular aperture leas to a iffraction pattern that has a circular central axiu surroune by a series of seconary bright fringes. Solve: (a) Because the visible spectru spans wavelengths fro 400 n to 700 n, we take the average wavelength of sunlight to be 550 n. (b) Within the sall-angle approxiation, the with of the central axiu is 9 (550 0 )(3 ) 4 L w 44 (0 ) (44) Moel: The antenna is a circular aperture through which the icrowaves iffract. Solve: (a) Within the sall-angle approxiation, the with of the central axiu of the iffraction pattern is w.44 L/. The wavelength of the raiation is 8 3 c 3 0 /s 44( )(30 w 0 ) 90 f 9 0 Hz 0 That is, the iaeter of the bea has increase fro.0 to 95, a factor of 458. (b) The average icrowave intensity is P 000 W I 0 5 W / area (95 ).6. Moel: The laser light is iffracte by the circular opening of the laser fro which the bea eerges. Solve: The iaeter of the laser bea is the with of the central axiu. We have L 44L 44(53 0 )(384 0 ) w 0 50 w 000 In other wors, the laser bea ust eerge fro a laser of iaeter 50 c..6. Moel: Two closely space slits prouce a ouble-slit interference pattern. Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

18 -8 Chapter Visualize: Solve: (a) The bright fringes are separate fro the 0 central axiu by (b) The light s frequency is s s is T. 9 L (6000 )(0 ) y f c / Hz. Thus, the perio is 5 T / f.00 0 s. A elay of (c) The wave passing through the glass is elaye by of a cycle. Consequently, the two waves are not in phase as 4 they eerge fro the slits. The slits are the sources of the waves, so there is now a phase ifference 0 between the two sources. A elay of a full cycle ( t T) woul have no effect at all on the interference because it correspons to a phase ifference 0 ra Thus a elay of 4 of a cycle introuces a phase ifference 0 4 ( ) ra () The text s analysis of the ouble-slit interference experient assue that the waves eerging fro the two slits were in phase, with 0 0 ra. Thus, there is a central axiu at a point on the screen exactly halfway between the two slits, where 0 ra an r 0. Now that there is a phase ifference between the sources, the central axiu still efine as the point of constructive interference where 0 ra will shift to one sie. The wave leaving the slit with the glass was elaye by 4 of a perio. If it travels a shorter istance to the screen, taking 4 of a perio less than the wave coing fro the other slit, it will ake up for the previous elay an the two waves will arrive in phase for constructive interference. Thus, the central axiu will shift towar the slit with the glass. How far? A phase ifference 0 woul shift the fringe pattern by y 3.0, aking the central axiu fall exactly where the bright fringe ha been previously. This is the point where r (), exactly copensating for a phase shift of π at the slits. Thus, a phase shift of 0 4 ( ) will shift the fringe pattern by 4 (3 ) The net effect of placing the glass in the slit is that the central axiu (an the entire fringe pattern) will shift 0.75 towar the slit with the glass..63. Moel: A iffraction grating prouces an interference pattern, which looks like the iagra of Figure.8. Solve: (a) Nothing has change while the aquariu is epty. The orer of a bright (constructive interference) fringe is relate to the iffraction angle by sin, where 0,,, 3, The space between the slits is For, sin sin Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

19 Wave Optics -9 (b) The path-ifference between the waves that leas to constructive interference is an integral ultiple of the wavelength in the eiu in which the waves are traveling, that is, water. Thus, 633 n 633 n sin n water.64. Moel: An interferoeter prouces a new axiu each tie L increases by causing the pathlength ifference r to increase by. Visualize: Please refer to the interferoeter in Figure.0. Solve: The path-length ifference between the two waves is r L L. The conition for constructive interference is r, hence constructive interference occurs when 7 ( L L ) L L where 63.8 n is the wavelength of the heliu-neon laser. When the irror M is ove back an a hyrogen ischarge lap is use, 00 fringes shift again. Thus, where 6565 n Subtracting the two equations, That is, M is now 4. L L ( L L ) L L 600( ) 600( ) 6 L L 4 0 closer to the bea splitter..65. Moel: The gas increases the nuber of wavelengths in one ar of the interferoeter. Each aitional wavelength causes one bright-ark-bright fringe shift. Solve: Fro the equation in Challenge Exaple.9, the nuber of fringe shifts is ()(00 0 ) ( n ) (0008 ) vac.66. Moel: The piece of glass increases the nuber of wavelengths in one ar of the interferoeter. Each aitional wavelength causes one bright-ark-bright fringe shift. Solve: We can rearrange the equation in Challenge Exaple.9 to fin that the inex of refraction of glass is 9 vac (500 0 n)(00) n.50 3 (00 0 ).67. Moel: The ars of the interferoeter are of equal length, so without the crystal the output woul be bright. Visualize: We nee to consier how any ore wavelengths fit in the electro-optic crystal than woul have occupie that space (6.70 ) without the crystal; if it is an integer then the interferoeter will prouce a bright output; if it is a half-integer then the interferoeter will prouce a ark output. But the wavelength we nee to consier is the wavelength insie the crystal, not the wavelength in air. n n We are tol the initial n with no applie voltage is.5, an the wavelength in air is 000 Solve: The nuber of wavelengths that woul have been in that space without the crystal is (a) With the crystal in place (an n.5) the nuber of wavelengths in the crystal is / Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

20 -0 Chapter which shows there are a half-integer nuber ore wavelengths with the crystal in place than if it weren t there. Consequently the output is ark with the crystal in place but no applie voltage. (b) Since the output was ark in the previous part, we want it to be bright in the new case with the voltage on. That eans we want to have just one-half ore extra wavelengths in the crystal (than if it weren t there) than we i in the previous part. That is, we want 4.00 extra wavelengths in the crystal instea of 3.5, so we want wavelengths in the crystal ( 000) 070 n / n 670 Assess: It sees reasonable to be able to change the inex of refraction of a crystal fro.5 to.597 by applying a voltage..68. Moel: A iffraction grating prouces an interference pattern like the one shown in Figure.8. We also assue that the sall-angle approxiation is vali for this grating. Solve: (a) The general conition for constructive-interference fringes is sin 0,,, 3, When this happens, we say that the light is iffracte at an angle. Since it is usually easier to easure istances rather than angles, we will consier the istance y fro the center to the th axiu. This istance is y Ltan. In the sall-angle approxiation, sin tan, so we can write the conition for constructive interference as y L y L The fringe separation is L y y y (b) Now the laser light falls on a fil that has a series of slits (i.e., bright an ark stripes), with spacing L Applying once again the conition for constructive interference: y L L sin y L L/ The fringe separation is y y y. That is, using the fil as a iffraction grating prouces a iffraction pattern whose fringe spacing is, the spacing of the original slits..69. Moel: The laser bea is iffracte through a circular aperture. Visualize: Solve: (a) No. The laser light eerges through a circular aperture at the en of the laser. This aperture causes iffraction, hence the laser bea ust graually sprea out. The iffraction angle is sall enough that the laser bea appears to be parallel over short istances. But if you observe the laser bea at a large istance it is easy to see that the iaeter of the bea is slowly increasing. Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

21 Wave Optics - (b) The position of the first iniu in the iffraction pattern is ore or less the ege of the laser bea. For iffraction through a circular aperture, the first iniu is at an angle (6330 ) ra (c) The iaeter of the laser bea is the with of the iffraction pattern: 44L 44(6330 )(30 ) w () At L k 000, the iaeter is L 44(6330 )(000 ) w Visualize: To fin the location y where the intensity is I use Equation.4: Iouble 4Icos y. L L Then ivie by the istance to the first iniu y0 to get the fraction esire. Solve: First set I ouble I. ouble 4 cos I I y L I 4Icos y L cos y 4 L cos y L L y cos Now set up the ratio that will give the esire fraction. L cos y cos y L Assess: The fraction ust be less than, an 3 sees reasonable..7. Moel: A iffraction grating prouces a series of constructive-interference fringes at values of that are eterine by Equation.5. Solve: (a) The conition for bright fringes is sin. If changes by a very sall aount, such that changes by, then we can approxiate / as the erivative / sin cos sin (b) We can now obtain the first-orer an secon-orer angular separations for the wavelengths n an n. The slit spacing is Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

22 - Chapter The first-orer ( ) angular separation is ra 00 The secon orer ( ) angular separation is ra Moel: The intensity in a ouble-slit interference pattern is eterine by iffraction effects fro the slits. Solve: (a) For the two wavelengths an passing siultaneously through the grating, their first-orer peaks are at L ( ) L y y Subtracting the two equations gives an expression for the separation of the peaks: (b) For a ouble-slit, the intensity pattern is L y y y Iouble 4I cos y L The intensity oscillates between zero an 4 I, so the axiu intensity is 4 I. The with is easure at the point where the intensity is half of its axiu value. For the intensity to be Iax I for the peak: L I 4I cos yhalf cos yhalf yhalf yhalf L L L 4 4 The with of the fringe is twice y half. This eans L wyhalf But the location of the peak is y L/, so we get w y. (c) We can exten the result obtaine in (b) for two slits to w y / N The conition for barely resolving two iffraction fringes or peaks is w y in. Fro part (a) we have an expression for the separation of the first-orer peaks an fro part (b) we have an expression for the with. Thus, cobining these two pieces of inforation, y inl y L yin in N LN NL N () Using the result of part (c), in N lines Moel: A iffraction grating prouces an interference pattern like the one shown in Figure.8, when the incient light is noral to the grating. The equation for constructive interference will change when the light is incient at a nonzero angle. Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

23 Wave Optics -3 Solve: (a) The path ifference between waves an on the incient sie is r sin The path ifference between the waves an on the iffracte sie, however, is r sin The total path ifference between waves an is thus r r (sin sin ). Because the path ifference for constructive interference ust be equal to, (b) For 30 the angles of iffraction are.74. Solve: (a) (sin sin ) 0,,, 9 ()(500 0 ) sin sin (0 0 )/600 9 ( )(500 0 ) sin sin (0 0 )/600 We have two incoing an two iffracte light rays at angles an an two wave fronts perpenicular to the rays. We can see fro the figure that the wave travels an extra istance r sin to reach the reflection spot. Wave travels an extra istance r sin fro the reflection spot to the outgoing wave front. The path ifference between the two waves is r r r (sin sin ) (b) The conition for iffraction, with all the waves in phase, is still r. Using the results fro part (a), the iffraction conition is sin sin,, 0,,, Negative values of will give a ifferent iffraction angle than the corresponing positive values. (c) The zeroth orer iffraction fro the reflection grating is Fro the iffraction conition of part (b), this iplies sin0 sin an hence 0 That is, the zeroth orer iffraction obeys the law of reflection the angle of reflection equals the angle of incience. () A 700 lines per illieter grating has spacing are given by n The iffraction angles (500 n) sin sin sin sin n Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.

24 -4 Chapter M not efine 0 not efine (e).75. Moel: iffraction patterns fro two objects can just barely be resolve if the central axiu of one iage falls on the first ark fringe of the other iage. Solve: (a) Using Equation.4 with the with equal to the aperture iaeter, 44L 7 w 44 L (44)(550 0 )(0.0 ) 0 5 (b) We can now use Equation.3 to fin the angle between two istant sources that can be resolve. The angle (550 0 ) ra (c) The istance that can be resolve is 9 3 (000 ) (000 )( 9 0 ra) 3 Copyright 03 Pearson Eucation, Inc. All rights reserve. This aterial is protecte uner all copyright laws as they currently exist. No portion of this aterial ay be reprouce, in any for or by any eans, without perission in writing fro the publisher.