Chapter 33 Interference and Diffraction

Transcription

1 Chapter 33 Interference an Diffraction Conceptual Probles A phase ifference ue to path-length ifference is observe for onochroatic visible light. Which phase ifference requires the least (iniu) path length ifference? (a) 90 o (b) 80 o (c) 70 o () the answer epens on the wavelength of the light. Deterine the Concept The phase ifference δ ue to a path ifference Δr are relate accoring to δ π Δr. Therefore, the least path length ifference correspons to the sallest phase ifference. ( a) is correct. Which of the following pairs of light sources are coherent: (a) two canles, (b) one point source an its iage in a plane irror, (c) two pinholes uniforly illuinate by the sae point source, () two healights of a car, (e) two iages of a point source ue to reflection fro the front an back surfaces of a soap fil. Deterine the Concept Coherent sources have a constant phase ifference. The pairs of light sources that satisfy this criterion are (b), (c), an (e). 3 [SSM] The spacing between Newton s rings ecreases rapily as the iaeter of the rings increases. Explain qualitatively why this occurs. Deterine the Concept The thickness of the air space between the flat glass an the lens is approxiately proportional to the square of, the iaeter of the ring. Consequently, the separation between ajacent rings is proportional to /. 4 If the angle of a wege-shape air fil, such as that in Exaple 3- is too large, fringes are not observe. Why? Deterine the Concept There are two possible reasons that fringes ight not be observe. () The istance between ajacent fringes is so sall that the fringes are not resolve by the eye. () Twice the thickness of the air space is greater than the coherence length of the light. If this is the case, fringes woul be observe in the region close to the point where the thickness of the air space approaches zero. 5 Why ust a fil that is use to observe interference colors be thin? Deterine the Concept Colors are observe when the light reflecte off the front an back surfaces of the fil interfere estructively for soe wavelengths an 3047

2 3048 Chapter 33 constructively for other wavelengths. For this interference to occur, the phase ifference between the light reflecte off the front an back surfaces of the fil ust be constant. This eans that twice the thickness of the fil ust be less than the coherence length of the light. The fil is calle a thin fil if twice its thickness is less than the coherence length of the light. 6 A loop of wire is ippe in soapy water an hel up so that the soap fil is vertical. (a) Viewe by reflection with white light, the top of the fil appears black. Explain why. (b) Below the black region are colore bans. Is the first ban re or violet? (a) The phase change ue to reflection fro the front surface of the fil is 80 ; the phase change ue to reflection fro the back surface of the fil is 0. As the fil thins towar the top, the phase change ue to the path length ifference between the two reflecte waves (the phase ifference associate with the fil s thickness) becoes negligible an the two reflecte waves interfere estructively. (b) The first constructive interference will arise when twice the thickness of the fil is equal to half the wavelength of the color with the shortest wavelength. Therefore, the first ban will be violet (shortest visible wavelength). 7 [SSM] A two-slit interference pattern is fore using onochroatic laser light with a wavelength of 640 n. At the secon axiu fro the central axiu, what is the path-length ifference between the light coing fro each of the slits? (a) 640 n (b) 30 n (c) 960 n () 80 n. Deterine the Concept For constructive interference, the path ifference is an integer ultiple of ; that is, r 640 n is correct. Δ. For, Δr ( ). ( ) 8 A two-slit interference pattern is fore using onochroatic laser light with a wavelength of 640 n. At the first iniu fro the central axiu, what is the path-length ifference between the light coing fro each of the slits? (a) 640 n (b) 30 n (c) 960 n () 80 n. Deterine the Concept For estructive interference, the path ifference is an o-integer ultiple of Δr,,3,5,. For the first ; that is ( )... iniu, an Δr ( 640 n) 30 n. ( b) is correct. 9 A two-slit interference pattern is fore using onochroatic laser light with a wavelength of 450 n. What happens to the istance between the first axiu an the central axiu as the two slits are ove closer together?

3 Interference an Diffraction 3049 (a) The istance increases. (b) The istance ecreases. (c) The istance reains the sae. Deterine the Concept The relationship between the slit separation an the angular position θ of each axiu is given by sin θ, 0,,,... (Equation 33-). Because an sinθ are inversely proportional for a given wavelength an interference axiu (value of ), ecreasing increases sinθ an θ. ( a) is correct. 0 A two-slit interference pattern is fore using two ifferent onochroatic lasers, one green an one re. Which color light has its first axiu closer to the central axiu? (a) Green, (b) re, (c) both axia are in the sae location. Deterine the Concept The relationship between the slit separation, the angular position θ of each axiu, an the wavelength of the light illuinating the slits is given by sin θ, 0,,,... (Equation 33-). Because an sinθ are irectly proportional for a given interference axiu (value of ) an the wavelength of green light is shorter than the wavelength of re light, ( a) is correct. A single slit iffraction pattern is fore using onochroatic laser light with a wavelength of 450 n. What happens to the istance between the first axiu an the central axiu as the slit is ae narrower? (a) The istance increases. (b) The istance ecreases. (c) The istance reains the sae. Deterine the Concept The relationship between the slit with a, the angular position θ of each axiu, an the wavelength of the light illuinating the slit is given by asin θ,,,3,... (Equation 33-). Because a an sinθ are inversely proportional for a given iffraction axiu (value of ), narrowing the slit increases sin an θ. ( a) θ is correct. Equation 33- which is sin θ, an Equation 33-, which is a sin θ, are soeties confuse. For each equation, efine the sybols an explain the equation s application. Deterine the Concept Equation 33- expresses the conition for an intensity axiu in two-slit interference. Here is the slit separation, the wavelength of the light, an integer, an θ the angle at which the interference axiu appears. Equation 33- expresses the conition for an intensity iniu in single-slit iffraction. Here a is the with of the slit, the wavelength of the light, an θ the angle at which the iniu appears, an is

4 3050 Chapter 33 a nonzero integer. 3 When a iffraction grating is illuinate by white light, the first-orer axiu of green light (a) is closer to the central axiu than the first-orer axiu of re light. (b) is closer to the central axiu than the first-orer axiu of blue light. (c) overlaps the secon-orer axiu of re light. () overlaps the secon-orer axiu of blue light. Picture the Proble We can solve sinθ for θ with to express the location of the first-orer axiu as a function of the wavelength of the light. The interference axia in a iffraction pattern are at angles θ given by: Solve for the angular location θ of the first-orer axiu : sin θ where is the separation of the slits an 0,,, θ sin Because green light < re light : θ < θ an green light re light (a) is correct. 4 A ouble-slit interference experient is set up in a chaber that can be evacuate. Using light fro a heliu-neon laser, an interference pattern is observe when the chaber is open to air. As the chaber is evacuate, one will note that (a) the interference fringes reain fixe. (b) the interference fringes ove closer together. (c) the interference fringes ove farther apart. () the interference fringes isappear copletely. Deterine the Concept The istance on the screen to th bright fringe is given by: nl y where L is the istance fro the slits to the screen, n is the wavelength of the light in a eiu whose inex of refraction is n, an is the separation of the slits. The separation of the interference y y ( ) fringes is given by: + + nl nl nl Because the inex of refraction of a vacuu is slightly less than the inex of refraction of air, the reoval of air increases n an, hence, y y. (c) is correct.

5 Interference an Diffraction [SSM] True or false: (a) (b) (c) () (e) When waves interfere estructively, the energy is converte into heat energy. Interference patterns are observe only if the relative phases of the waves that superipose reain constant. In the Fraunhofer iffraction pattern for a single slit, the narrower the slit, the wier the central axiu of the iffraction pattern. A circular aperture can prouce both a Fraunhofer iffraction pattern an a Fresnel iffraction pattern. The ability to resolve two point sources epens on the wavelength of the light. (a) False. When estructive interference of light waves occurs, the energy is no longer istribute evenly. For exaple, light fro a two-slit evice fors a pattern with very bright an very ark parts. There is practically no energy at the ark fringes an a great eal of energy at the bright fringe. The total energy over the entire pattern equals the energy fro one slit plus the energy fro the secon slit. Interference re-istributes the energy. (b) True. (c) True. The with of the central axiu in the iffraction pattern is given by θ sin where a is the with of the slit. Hence, the narrower the slit, the a wier the central axiu of the iffraction pattern. () True. (e) True. The critical angle for the resolution of two sources is irectly proportional to the wavelength of the light eitte by the sources ( α c. ). D 6 You observe two very closely-space sources of white light through a circular opening using various filters. Which color filter is ost likely to prevent your resolving the iages on your retinas as coing fro two istinct sources? (a) re (b) yellow (c) green () blue (e) The filter choice is irrelevant.

6 305 Chapter 33 Deterine the Concept The conition for the resolution of the two sources is given by Rayleigh s criterion: α. D (Equation 33-5), where α c is the c critical angular separation an D is the iaeter of the aperture. The larger the critical angle require for resolution, the less likely it is that you can resolve the sources as being two istinct sources. Because α c an are irectly proportional, the filter that passes the shorter wavelength light woul be ost likely to resolve the sources. ( ) is correct. 7 Explain why the ability to istinguish the two healights of an oncoing car, at a given istance, is easier for the huan eye at night than uring aylight hours. Assue the healights of the oncoing car are on uring both aytie an nighttie hours. Deterine the Concept The conition for the resolution of the two sources is given by Rayleigh s criterion: α c. D (Equation 33-5), where α c is the critical angular separation, D is the iaeter of the aperture, an is the wavelength of the light coing fro the objects, in this case healights, to be resolve. Because the iaeter of the pupils of your eyes are larger at night, the critical angle is saller at night, which eans that at night you can resolve the light as coing fro two istinct sources when they are at a greater istance. Estiation an Approxiation 8 It is claie that the Great Wall of China is the only an ae object that can be seen fro space with the nake eye. Check to see if this clai is true, base on the resolving power of the huan eye. Assue the observers are in low- Earth orbit that has an altitue of about 50 k. Picture the Proble We ll assue that the iaeter of the pupil of the eye is 5.0 an use the best-case scenario (the iniu resolvable with varies irectly with the wavelength of the light reflecting fro the object) that the wavelength of light is 400 n (the lower liit for the huan eye). Then we can use the expression for the iniu angular separation of two objects than can be resolve by the eye an the relationship between this angle an the with of an object an the istance fro which it is viewe to support the clai. Relate the with w of an object that can be seen at an altitue h to the critical angular separation α c : w tanα c w htanα c h

7 Interference an Diffraction 3053 The iniu angular separation α c of two point objects that can just be resolve by an eye epens on the iaeter D of the eye an the wavelength of light: Substitute for α c in the expression for w to obtain: α c. D w h tan. D Substitute nuerical values an evaluate w in for an altitue of 50 k: w in 400n 5.0 ( 50k) tan. 4 This clai is probably false. Because the iniu with that is resolvable fro low-earth orbit (50 k) is 4 an the with of the Great Wall is 5 to 8 high an 5 wie, so this clai is likely false. However, it is easily seen using binoculars, an pictures can be taken of it using a caera. This is because both binoculars an caeras have apertures that are larger than the pupil of the huan eye. (The Chinese astronaut Yang Liwei reporte that he was not able to see the wall with the nake eye uring the first Chinese anne space flight in 003.) 9 [SSM] (a) Estiate how close an approaching car at night on a flat, straight stretch of highway ust be before its healights can be istinguishe fro the single healight of a otorcycle. (b) Estiate how far ahea of you a car is if its two re taillights erge to look as if they were one. Picture the Proble Assue a separation of.5 between typical autoobile healights an tail lights, a nighttie pupil iaeter of 5.0, 550 n for the wavelength of the light (as an average) eitte by the healights, 640 n for re taillights, an apply the Rayleigh criterion. (a) The Rayleigh criterion is given by Equation 33-5: The critical angular separation is also given by: α c. D where D is the separation of the healights (or tail lights). L α where is the separation of hea lights (or tail lights) an L is the istance to approaching or receing autoobile.

8 3054 Chapter 33 Equate these expressions for α c to obtain: Substitute nuerical values an evaluate L: L D. L D. ( 5.0 )(.5 ) L.( 550 n) (b) For re light: ( 5.0 )(.5 ) L.( 640 n) k 9.6 k 0 A sall louspeaker is locate at a large istance to the east fro you. The louspeaker is riven by a sinusoial current whose frequency can be varie. Estiate the lowest frequency for which your ears woul receive the soun waves exactly out of phase when you are facing north. Picture the Proble If your ears receive the soun exactly out of phase, the waves arriving at your ear that is farthest fro the speaker ust be traveling onehalf wavelength farther than the waves arriving at your ear that is nearest the speaker. The lowest frequency correspons to the longest wavelength. Assue that the speaker an your ears are on the sae line an let the istance between your ears be about 0 c. Take the spee of soun in air to be 343 /s. The frequency receive by your ears is given by: Let Δr be the istance between your ears (the path ifference) to obtain: f v Δr Δr Substituting for yiels: f v Δ r Substitute nuerical values an 343 /s f 0.86 khz evaluate f: ( 0 c ) or between 0.80 an 0.90 khz. [SSM] Estiate the axiu istance a binary star syste coul be resolvable by the huan eye. Assue the two stars are about fifty ties further apart than the Earth an Sun are. Neglect atospheric effects. (A test siilar to this eye test was use in ancient Roe to test for eyesight acuity before entering the ary. A noral eye coul just barely resolve two well-known close-together stars in the sky. Anyone who coul not tell there were two stars was rejecte. In this case, the stars were not a binary syste, but the principle is the sae.)

9 Interference an Diffraction 3055 Picture the Proble Assue that the iaeter of a pupil at night is 5.0 an that the wavelength of light is in the ile of the visible spectru at about 550 n. We can use the Rayleigh criterion for the separation of two sources an the geoetry of the Earth-to-binary star syste to erive an expression for the istance to the binary stars. If the istance between the binary stars is represente by an the Earth-star istance by L, then their angular separation is given by: The critical angular separation of the two sources is given by the Rayleigh criterion: For α α c : Substitute nuerical values an evaluate L: α L α c. D L D. L D. ( 5.0 )( 50)(.5 0 ) L.( 550 n) 3 c y k c y 5 Phase Difference an Coherence Light of wavelength 500 n is incient norally on a fil of water.00 μ thick. (a) What is the wavelength of the light in the water? (b) How any wavelengths are containe in the istance t, where t is the thickness of the fil? (c) What is the phase ifference between the wave reflecte fro the top of the air water interface an the wave reflecte fro the botto of the water air interface in the region where the two reflecte waves superpose? Picture the Proble The wavelength of light in a eiu whose inex of refraction is n is the ratio of the wavelength of the light in air ivie by n. The nuber of wavelengths of light containe in a given istance is the ratio of the istance to the wavelength of light in the given eiu. The ifference in phase between the two waves is the su of a π phase shift in the reflecte wave an a phase shift ue to the aitional istance travele by the wave reflecte fro the botto of the water air interface. (a) Express the wavelength of light in water in ters of the wavelength of light in air: n 500n.33 air water water 376n

10 3056 Chapter 33 (b) Relate the nuber of wavelengths N to the thickness t of the fil an the wavelength of light in water: N t water ( μ) n 5.3 (c) Express the phase ifference as the su of the phase shift ue to reflection an the phase shift ue to the aitional istance travele by the wave reflecte fro the botto of the water air interface: Substitute for N an evaluate δ: δ δ reflection t π + water + δ δ π ra + π aitionalistance travele π π + π N ( 5.3 ra).64π ra.6π ra or, subtracting.64π ra fro π ra, δ 0.4π ra. ra 3 [SSM] Two coherent icrowave sources both prouce waves of wavelength.50 c. The sources are locate in the z 0 plane, one at x 0, y 5.0 c an the other at x 3.00 c, y 4.0 c. If the sources are in phase, fin the ifference in phase between these two waves for a receiver locate at the origin. Picture the Proble The ifference in phase epens on the path ifference Δr accoring to δ π. The path ifference is the ifference in the istances of (0, 5.0 c) an (3.00 c, 4.0 c) fro the origin. Relate a path ifference Δr to a phase shift δ: Δr δ π The path ifference Δr is: Δr 5.0c ( 3.00 c) + ( 4.0c) 0.68c Substitute nuerical values an evaluate δ: Interference in Thin Fils 0.68c δ.50c π.9 ra 4 A wege-shape fil of air is ae by placing a sall slip of paper between the eges of two flat plates of glass. Light of wavelength 700 n is

11 Interference an Diffraction 3057 incient norally on the glass plates, an interference fringes are observe by reflection. (a) Is the first fringe near the point of contact of the plates ark or bright? Why? (b) If there are five ark fringes per centieter, what is the angle of the wege? Picture the Proble Because the th fringe occurs when the path ifference t equals wavelengths, we can express the aitional istance travele by the light in air as an. The thickness of the wege, in turn, is relate to the angle of the wege an the istance fro its vertex to the th fringe. (a) The first fringe is ark because the phase ifference ue to reflection by the botto surface of the top plate an the top surface of the botto plate is 80 (b) The th fringe occurs when the path ifference t equals wavelengths: Relate the thickness of the air wege to the angle of the wege: Substitute for t to obtain: Substitute nuerical values an evaluate θ : t t θ t xθ x where we ve use a sall-angle approxiation to replace an arc length by the length of a chor. x θ θ x x θ 5 c 4 ( 700 n).75 0 ra 5 [SSM] The iaeters of fine fibers can be accurately easure using interference patterns. Two optically flat pieces of glass of length L are arrange with the wire between the, as shown in Figure The setup is illuinate by onochroatic light, an the resulting interference fringes are observe. Suppose that L is 0.0 c an that yellow soiu light (wavelength of 590 n) is use for illuination. If 9 bright fringes are seen along this 0.0-c istance, what are the liits on the iaeter of the wire? Hint: The nineteenth fringe ight not be right at the en, but you o not see a twentieth fringe at all. Picture the Proble The conition that one sees fringes requires that the path ifference between light reflecte fro the botto surface of the top slie an the top surface of the botto slie is an integer ultiple of a wavelength of the light.

12 3058 Chapter 33 The th fringe occurs when the path ifference equals wavelengths: Because the nineteenth (but not the twentieth) bright fringe can be seen, the liits on ust be: Substitute nuerical values to obtain: ( ) ( < < + ) where n 9 or 5.5μ < < 5.8μ ( ) < < ( 9 + ) 590 n 6 Light that has a wavelength equal to 600 n is use to illuinate two glass plates at noral incience. The plates are c in length, touch at one en, an are separate at the other en by a wire that has a raius of How any bright fringes appear along the total length of the plates? Picture the Proble The light reflecte fro the top surface of the botto plate (wave in the iagra) is phase shifte relative to the light reflecte fro the botto surface of the top plate (wave in the iagra). This phase ifference is the su of a phase shift of π (equivalent to a / path ifference) resulting fro reflection plus a phase shift ue to the aitional istance travele. glass plate glass plate wire t Relate the extra istance travele by wave to the istance equivalent to the phase change ue to reflection an to the conition for constructive interference: t +,,3,... or 5 t, 3,,... an t ( + ) where 0,,,, 0 t r an is the wavelength of light in air.

13 Interference an Diffraction 3059 Solving for gives: Solve for the highest value of : Substitute nuerical values an evaluate : Because we start counting fro 0, the nuber of bright fringes is ax +. t where 0,,, an 0 t r ( ) r ax int int 4r where r is the raius of the wire. ( ) 600n ax int int( 66.) 66 N ax A thin fil having an inex of refraction of.50 is surroune by air. It is illuinate norally by white light. Analysis of the reflecte light shows that the wavelengths 360, 450, an 60 n are the only issing wavelengths in or near the visible portion of the spectru. That is, for these wavelengths, there is estructive interference. (a) What is the thickness of the fil? (b) What visible wavelengths are brightest in the reflecte interference pattern? (c) If this fil were resting on glass with an inex of refraction of.60, what wavelengths in the visible spectru woul be issing fro the reflecte light? Picture the Proble (a) We can use the conition for estructive interference in a thin fil to fin the thickness of the fil. (b) an (c) Once we ve foun the thickness of the fil, we can use the conition for constructive interference to fin the wavelengths in the visible portion of the spectru that will be brightest in the reflecte interference pattern an the conition for estructive interference to fin the wavelengths of light issing fro the reflecte light when the fil is place on glass with an inex of refraction greater than that of the fil.

14 3060 Chapter 33 (a) Express the conition for estructive interference in the thin fil: Solving for yiels: 5 t +, 3 ' ' ', ',... or t ',',3',... or t ' () n where,, 3, an is the wavelength of the light in the fil. nt Substitute for the issing wavelengths to obtain: nt 450 n an nt 360 n + Divie the first of these equations by the secon an siplify to obtain: Solving for gives: nt 450 n + 360n nt + 4 for 450 n Solve equation () for t to obtain: t n Substitute nuerical values an evaluate t: 4 t ( 450n) (.50) 600n (b) The conition for constructive interference in the thin fil is: t + ' ',',3',... or 3 5 t ', ', ',... + ( ) ' where is the wavelength of light in the oil an 0,,, Substitute for to obtain: t ( + ) n nt + where n is the inex of refraction of the fil. Substitute nuerical values an siplify to obtain: ( )( 600n) n +

15 Interference an Diffraction 306 Substitute nuerical values for an evaluate to obtain the following table: (n) Fro the table, we see that the only wavelengths in the visible spectru are 70 n, 54 n, an 400 n. (c) Because the inex of refraction of the glass is greater than that of the fil, the light reflecte fro the fil-glass interface will be shifte by (as is the wave reflecte fro the top surface) an the conition for estructive interference becoes: Solving for yiels: 5 t, 3 ' ', or t + ' ( ) n,... where n is the inex of refraction of the fil an 0,,, nt + Substitute nuerical values an siplify to obtain: ( )( 600n) n + Substitute nuerical values for an evaluate to obtain the following table: (n) Fro the table we see that the issing wavelengths in the visible spectru are 70 n, 54 n, an 400 n. 8 A rop of oil (refractive inex of.) floats on water (refractive inex of.33). When reflecte light is observe fro above, as shown in Figure 33-4 what is the thickness of the rop at the point where the secon re fringe, counting fro the ege of the rop, is observe? Assue re light has a wavelength of 650 n. Picture the Proble Because there is a phase change ue to reflection at both the air-oil an oil-water interfaces, the conition for constructive interference is that twice the thickness of the oil fil equal an integer ultiple of the wavelength of light in the fil.

16 306 Chapter 33 The conition for constructive interference is: Substitute for to obtain: Substitute nuerical values an evaluate t: t ',',3',... or t ' () where is the wavelength of light in the oil an,, 3, t t n n ( )( 650n) t 533n. ( ) 9 [SSM] A fil of oil that has an inex of refraction of.45 rests on an optically flat piece of glass with an inex of refraction of.60. When illuinate by white light at noral incience, light of wavelengths 690 n an 460 n is preoinant in the reflecte light. Deterine the thickness of the oil fil. Picture the Proble Because there is a phase change ue to reflection at both the air-oil an oil-glass interfaces, the conition for constructive interference is that twice the thickness of the oil fil equal an integer ultiple of the wavelength of light in the fil. Express the conition for constructive interference: Substitute for to obtain: t ',',3',... ' () where is the wavelength of light in the oil an 0,,, nt t n where n is the inex of refraction of the oil. Substitute for the preoinant wavelengths to obtain: nt 690 n an nt 460 n + Divie the first of these equations by the secon an siplify to obtain: Solve equation () for t: nt 690 n + 460n nt + t n

17 Substitute nuerical values an evaluate t: t Interference an Diffraction 3063 ( )( 690n) (.45) 476n 30 A fil of oil that has an inex of refraction.45 floats on water. When illuinate with white light at noral incience, light of wavelengths 700 n an 500 n is preoinant in the reflecte light. Deterine the thickness of the oil fil. Picture the Proble Because the inex of refraction of air is less than that of the oil, there is a phase shift of π ra ( ) in the light reflecte at the air-oil interface. Because the inex of refraction of the oil is greater than that of the glass, there is no phase shift in the light reflecte fro the oil-glass interface. We can use the conition for constructive interference to eterine for 700 n an then use this value in our equation escribing constructive interference to fin the thickness t of the oil fil. Express the conition for constructive interference between the waves reflecte fro the airoil interface an the oil-glass interface: t + ' ',',3',... or 3 5 t ', ', ',... + () ( ) ' where is the wavelength of light in the oil an 0,,, Substitute for ' an solve for nt to obtain: + Substitute the preoinant nt 700n wavelengths to obtain: + an nt 500n + 3 Divie the first of these equations by the secon to obtain: 700n 500n nt + nt Solve equation () for t: t ( + ) n Substitute nuerical values an evaluate t: t 700n.45 ( + ) ( ) 603n

18 3064 Chapter 33 Newton s Rings 3 [SSM] A Newton s ring apparatus consists of a plano-convex glass lens with raius of curvature R that rests on a flat glass plate, as shown in Figure The thin fil is air of variable thickness. The apparatus is illuinate fro above by light fro a soiu lap that has a wavelength of 590 n. The pattern is viewe by reflecte light. (a) Show that for a thickness t the conition for a bright (constructive) interference ring is t ( + ) where 0,,,... (b) Show that for t << R, the raius r of a fringe is relate to t by r tr. (c) For a raius of curvature of 0.0 an a lens iaeter of 4.00 c. How any bright fringes woul you see in the reflecte light? () What woul be the iaeter of the sixth bright fringe? (e) If the glass use in the apparatus has an inex of refraction n.50 an water replaces the air between the two pieces of glass, explain qualitatively the changes that will take place in the bright-fringe pattern. Picture the Proble This arrangeent is essentially ientical to a thin fil configuration, except that the fil is air. A phase change of 80 ( ) occurs at the top of the flat glass plate. We can use the conition for constructive interference to erive the result given in (a) an use the geoetry of the lens on the plate to obtain the result given in (b). We can then use these results in the reaining parts of the proble. (a) The conition for constructive interference is: t +,,3,... or 3 5 t,,,... ( + ) where is the wavelength of light in air an 0,,, Solving for t yiels: t ( + ), 0,,,... () (b) Fro Figure 33-4 we have: ( R t) R r + or R r + R Rt + t For t << R we can neglect the last ter to obtain: (c) Square equation () an substitute for t fro equation () to obtain: R r + R Rt r Rt () r ( + ) R R r

19 Interference an Diffraction 3065 Substitute nuerical values an evaluate : (.00c) ( 0.0)( 590n) an so there will be bright fringes. () The iaeter of the th fringe is: D r ( + ) R Noting that 5 for the sixth fringe, substitute nuerical values an evaluate D: D ( 5 + )( 0.0)( 590n).4c (e) The wavelength of the light in the fil becoes air /n 444 n. The separation between fringes is reuce (the fringes woul becoe ore closely space.) an the nuber of fringes that will be seen is increase by a factor of A plano-convex glass lens of raius of curvature.00 rests on an optically flat glass plate. The arrangeent is illuinate fro above with onochroatic light of 50-n wavelength. The inexes of refraction of the lens an plate are.60. Deterine the raii of the first an secon bright fringe fro the center in the reflecte light. Picture the Proble This arrangeent is essentially ientical to a thin fil configuration, except that the fil is air. A phase change of 80 ( ) occurs at the top of the flat glass plate. We can use the conition for constructive interference an the results fro Proble 3(b) to eterine the raii of the first an secon bright fringes in the reflecte light. The conition for constructive interference is: t +,,3,... or 3 5 t,,,... ( + ) where is the wavelength of light in air an 0,,, Solving for t gives: t ( + ), 0,,,... In Proble 3(b) it was shown that: r tr Substitute for t to obtain: r ( + ) R

20 3066 Chapter 33 The first fringe correspons to 0: r ( 50n)(.00) 0.7 The secon fringe correspons to : r ( 50n)(.00) Suppose that before the lens of Proble 3 is place on the plate, a fil of oil of refractive inex.8 is eposite on the plate. What will then be the raii of the first an secon bright fringes? Picture the Proble This arrangeent is essentially ientical to a thin fil configuration, except that the fil is oil. A phase change of 80 ( ) occurs at lens-oil interface. We can use the conition for constructive interference an the results fro Proble 3(b) to eterine the raii of the first an secon bright fringes in the reflecte light. The conition for constructive interference is: t + ' ',',3',... or 3 5 t ', ', ',... + ( ) ' where is the wavelength of light in the oil an 0,,, n where is the wavelength of light in air. Substitute for an solve for t: t ( + ), 0,,,... In Proble 3(b) it was shown that: r tr Substitute for t to obtain: r R ( + ) n The first fringe correspons to 0: r ( 50n)(.00) The secon fringe correspons to : r 3 ( 50n)(.00) Two-Slit Interference Patterns 34 Two narrow slits separate by.00 are illuinate by light of wavelength 600 n, an the interference pattern is viewe on a screen.00

21 Interference an Diffraction 3067 away. Calculate the nuber of bright fringes per centieter on the screen in the region near the center fringe. Picture the Proble The nuber of bright fringes per unit istance is the reciprocal of the separation of the fringes. We can use the expression for the istance on the screen to the th fringe to fin the separation of the fringes. Express the nuber N of bright fringes per centieter in ters of the separation of the fringes: N () Δy Express the istance on the screen to the th an ( + )st bright fringe: y L L an ( ) y + + Subtract the first of these equations fro the secon to obtain: L Δ y Substitute in equation () to obtain: N L Substitute nuerical values an N n evaluate N: ( )(.00) 8.33c 35 [SSM] Using a conventional two-slit apparatus with light of wavelength 589 n, 8 bright fringes per centieter are observe near the center of a screen 3.00 away. What is the slit separation? Picture the Proble We can use the expression for the istance on the screen to the th an ( + )st bright fringes to obtain an expression for the separation Δy of the fringes as a function of the separation of the slits. Because the nuber of bright fringes per unit length N is the reciprocal of Δy, we can fin fro N,, an L. Express the istance on the screen to the th an ( + )st bright fringe: y L L an ( ) y + + Subtract the first of these equations fro the secon to obtain: Because the nuber of fringes per unit length N is the reciprocal of Δy: L Δy NL L Δy

22 3068 Chapter 33 Substitute nuerical values an evaluate : ( 8c )( 589n)( 3.00) Light of wavelength 633 n fro a heliu neon laser is shone norally on a plane containing two slits. The first interference axiu is 8 c fro the central axiu on a screen away. (a) Fin the separation of the slits. (b) How any interference axia is it, in principle, possible to observe? Picture the Proble We can use the geoetry of the setup, represente to the right, to fin the separation of the slits. To fin the nuber of interference axia that, in principle, can be observe, we can apply the equation escribing two-slit interference axia an require that sinθ. θ θ L y 0 8 c Because << L, we can approxiate sinθ as: sinθ () sinθ Fro the right triangle whose sies sinθ 8c are L an y we have: ( ) + ( 8c ) Substitute nuerical values in equation () an evaluate : 633n μ 9.3μ (b) The equation escribing two-slit interference axia is: Because sinθ eterines the axiu nuber of interference fringes that can be seen: Substitute nuerical values an evaluate ax : Because there are 4 fringes on either sie of the central axiu: sin θ, 0,,,... ax ax 9.9 μ ax 4 because ust 633n be an integer. ( ) + 9 N ax + 4

23 Interference an Diffraction Two narrow slits are separate by a istance. Their interference pattern is to be observe on a screen a large istance L away. (a) Calculate the spacing between successive axia near the center fringe for light of wavelength 500 n, when L is.00 an is.00 c. (b) Woul you expect to be able to observe the interference of light on the screen for this situation? (c) How close together shoul the slits be place for the axia to be separate by.00 for this wavelength an screen istance? Picture the Proble We can use the equation for the istance on a screen to the th bright fringe to erive an expression for the spacing of the axia on the screen. In (c) we can use this sae relationship to express the slit separation. (a) Express the istance on the screen to the th an ( + )st bright fringe: y L L an ( ) y + + Subtract the first of these equations fro the secon to obtain: L Δ y () Substitute nuerical values an evaluate Δy: Δy ( 500n)(.00).00c 50.0 μ (b) Accoring to the Raleigh criterion you coul resolve the, but not by uch. (c) Solve equation () for to obtain: L Δy Substitute nuerical values an evaluate : ( 500n)(.00) Light is incient at an angle φ with the noral to a vertical plane containing two slits of separation (Figure 33-43). Show that the interference axia are locate at angles θ given by sin θ + sin φ /. Picture the Proble Let the separation of the slits be. We can fin the total path ifference when the light is incient at an angle φ an set this result equal to an integer ultiple of the wavelength of the light to obtain the given equation. Express the total path ifference: The conition for constructive interference is: Δl sinφ + sinθ Δl where is an integer.

24 3070 Chapter 33 Substitute to obtain: Divie both sies of the equation by to obtain: sin φ sinθ + sin φ + sinθ 39 [SSM] White light falls at an angle of 30º to the noral of a plane containing a pair of slits separate by.50 μ. What visible wavelengths give a bright interference axiu in the transitte light in the irection noral to the plane? (See Proble 38.) Picture the Proble Let the separation of the slits be. We can fin the total path ifference when the light is incient at an angle φ an set this result equal to an integer ultiple of the wavelength of the light to relate the angle of incience on the slits to the irection of the transitte light an its wavelength. Express the total path ifference: The conition for constructive interference is: Substitute to obtain: Divie both sies of the equation by to obtain: Δl sinφ + sinθ Δl where is an integer. sin φ + sinθ sin φ + sinθ Set θ 0 an solve for : sinφ Substitute nuerical values an siplify to obtain: (.50 μ) sin 30.5μ Evaluate for positive integral values of : (n) Fro the table we can see that 65 n an 47 n are in the visible portion of the electroagnetic spectru.

25 Interference an Diffraction Two sall louspeakers are separate by 5.0 c, as shown in Figure The speakers are riven in phase with a sine wave signal of frequency 0 khz. A sall icrophone is place a istance.00 away fro the speakers on the axis running through the ile of the two speakers, an the icrophone is then ove perpenicular to the axis. Where oes the icrophone recor the first iniu an the first axiu of the interference pattern fro the speakers? The spee of soun in air is 343 /s. Picture the Proble The iagra shows the two speakers, S an S, the central-bright iage an the first-orer iage to the left of the central-bright iage. The istance y is easure fro the center of the central-bright iage. We can apply the conitions for constructive an estructive interference fro two sources an use the geoetry of the speakers an icrophone to fin the istance to the first interference iniu an the istance to the first interference axiu. Relate the istance Δy to the first iniu fro the center of the central axiu to θ an the istance L fro the speakers to the plane of the icrophone: y θ S S y tanθ y L tanθ () L L Interference inia occur where: sin θ ( + ) where 0,,, 3, Solve for θ to obtain: ( ) + θ sin Relate the wavelength of the soun waves to the spee of soun v an the frequency f of the soun: Substitute for in the expression for θ to obtain: Substituting for θ in equation () yiels: v f θ sin ( ) v + f ( ) + v L tan sin () f y

26 307 Chapter 33 Noting that the first iniu correspons to 0, substitute nuerical values an evaluate Δy: ( )( 343/s) ( 5.0c)( 0kHz) y st in (.00) tan sin 37c The axia occur where: For iffraction axia, equation () becoes: sin θ where,, 3, y L tan sin v af Noting that the first axiu correspons to, substitute nuerical values an evaluate y: ( )( 343/s) ( 5.0c)( 0kHz) y st ax (.00) tan sin 94c Diffraction Pattern of a Single Slit 4 Light that has a 600-n wavelength is incient on a long narrow slit. Fin the angle of the first iffraction iniu if the with of the slit is (a).0, (b) 0.0, an (c) Picture the Proble We can use the expression locating the first zeroes in the intensity to fin the angles at which these zeroes occur as a function of the slit with a. The first zeroes in the intensity occur at angles given by: sinθ θ sin a a (a) For a.0 : 600n θ sin ra (b) For a 0.0 : θ 600n sin ra (c) For a 0.00 : θ 600n sin ra

27 Interference an Diffraction Plane icrowaves are incient on the thin etal sheet that has a long, narrow slit of with 5.0 c in it. The icrowave raiation strikes the sheet at noral incience. The first iffraction iniu is observe at θ 37º. What is the wavelength of the icrowaves? Picture the Proble We can use the expression locating the first zeroes in the intensity to fin the wavelength of the raiation as a function of the angle at which the first iffraction iniu is observe an the with of the plate. The first zeroes in the intensity occur at angles given by: Substitute nuerical values an evaluate : sinθ asinθ a ( 5.0c) sin37 3.0c 43 [SSM] Measuring the istance to the oon (lunar ranging) is routinely one by firing short-pulse lasers an easuring the tie it takes for the pulses to reflect back fro the oon. A pulse is fire fro Earth. To sen the pulse out, the pulse is expane so that it fills the aperture of a 6.00-in-iaeter telescope. Assuing the only thing spreaing the bea out is iffraction an that the light wavelength is 500 n, how large will the bea be when it reaches the Moon, k away? Picture the Proble The iagra shows the bea expaning as it travels to the oon an that portion of it that is reflecte fro the irror on the oon expaning as it returns to Earth. We can express the iaeter of the bea at the oon as the prouct of the bea ivergence angle an the istance to the oon an use the equation escribing iffraction at a circular aperture to fin the bea ivergence angle. telescope L irror D Relate the iaeter D of the bea when it reaches the oon to the istance to the oon L an the bea ivergence angle θ : D θ L ()

28 3074 Chapter 33 The angle θ subtene by the first iffraction iniu is relate to the wavelength of the light an the iaeter of the telescope opening telescope by: Because θ <<, sinθ θ an: Substitute for θ in equation () to obtain: sinθ. θ..l D telescope telescope telescope Substitute nuerical values an evaluate D: D.54c 6.00in in 0 c 8. ( ) ( 500n) k Interference-Diffraction Pattern of Two Slits 44 How any interference axia will be containe in the central iffraction axiu in the interference iffraction pattern of two slits if the separation of the slits is exactly 5 ties their with? How any will there be if the slit separation is an integral ultiple of the slit with (that is na) for any value of n? Picture the Proble We nee to fin the value of for which the th interference axiu coincies with the first iffraction iniu. Then there will be N fringes in the central axiu. The nuber of fringes N in the central axiu is: Relate the angle θ of the first iffraction iniu to the with a of the slits of the iffraction grating: N () sin θ a Express the angle θ corresponing to the th interference axia in ters of the separation of the slits: sin θ

29 Interference an Diffraction 3075 Because we require that θ θ, we can equate these expressions to obtain: Substituting for an siplifying yiels: a a 5 a 5 a Substitute for in equation () to obtain: N ( 5) 9 If na: na n an N n a a 45 [SSM] A two-slit Fraunhofer interference iffraction pattern is observe using light that has a wavelength equal to 500 n. The slits have a separation of 0.00 an an unknown with. (a) Fin the with if the fifth interference axiu is at the sae angle as the first iffraction iniu. (b) For that case, how any bright interference fringes will be seen in the central iffraction axiu? Picture the Proble We can equate the sine of the angle at which the first iffraction iniu occurs to the sine of the angle at which the fifth interference axiu occurs to fin a. We can then fin the nuber of bright interference fringes seen in the central iffraction axiu using N. (a) Relate the angle θ of the first iffraction iniu to the with a of the slits of the iffraction grating: Express the angle θ 5 corresponing to the th fifth interference axia axiu in ters of the separation of the slits: Because we require that θ θ 5, we can equate these expressions to obtain: Substituting the nuerical value of yiels: sin θ a 5 sinθ 5 5 a a a μ (b) Because 5: N ( 5) 9

30 3076 Chapter A two-slit Fraunhofer interference iffraction pattern is observe using light that has a wavelength equal to 700 n. The slits have withs of 0.00 an are separate by 0.0. How any bright fringes will be seen in the central iffraction axiu? Picture the Proble We can equate the sine of the angle at which the first iffraction iniu occurs to the sine of the angle at which the th interference axiu occurs to fin. We can then fin the nuber of bright interference fringes seen in the central iffraction axiu using N. The nuber of fringes N in the central axiu is: Relate the angle θ of the first iffraction iniu to the with a of the slits of the iffraction grating: N () sin θ a Express the angle θ corresponing to the th interference axia in ters of the separation of the slits: sin θ Because we require that θ θ, we can equate these expressions to obtain: Substitute for in equation () to obtain: a a N a Substitute nuerical values an evaluate N: N ( 0.0 ) Suppose that the central iffraction axiu for two slits has 7 interference fringes for soe wavelength of light. How any interference fringes woul you expect in the iffraction axiu ajacent to one sie of the central iffraction axiu? Picture the Proble There are 8 interference fringes on each sie of the central axiu. The seconary iffraction axiu is half as wie as the central one. It follows that it will contain 8 interference axia.

31 Interference an Diffraction Light that has a wavelength equal to 550 n illuinates two slits that both have withs equal to an separations equal to 0.5. (a) How any interference axia fall within the full with of the central iffraction axiu? (b) What is the ratio of the intensity of the thir interference axiu to one sie of the center interference axiu to the intensity of the center interference axiu? Picture the Proble We can equate the sine of the angle at which the first iffraction iniu occurs to the sine of the angle at which the th interference axiu occurs to fin. We can then fin the nuber of bright interference fringes seen in the central iffraction axiu using N. In (b) we can use the expression relating the intensity in a single-slit iffraction pattern to phase π constant φ a sinθ to fin the ratio of the intensity of the thir interference axiu to one sie of the center interference axiu. (a) The nuber of fringes N in the central axiu is: Relate the angle θ of the first iffraction iniu to the with a of the slits of the iffraction grating: N () sin θ a Express the angle θ corresponing to the th interference axia in ters of the separation of the slits: sin θ Because we require that θ θ, we can equate these expressions to obtain: Substitute in equation () to obtain: a a N a Substitute nuerical values an evaluate N: ( 0.5) N (b) Express the intensity for a singleslit iffraction pattern as a function of the phase ifference φ: sin φ I I0 () φ π where φ a sinθ

32 3078 Chapter 33 For 3: 3 sinθ 3 an π π 3 φ a sinθ3 a 6π a Substitute nuerical values an π φ 6π evaluate φ: Solve equation () for the ratio of I 3 to I 0 : I 3 sin φ I 0 φ Substitute nuerical values an evaluate I 3 /I 0 : I I 3 0 6π sin 5 6π Using Phasors to A Haronic Waves 49 [SSM] Fin the resultant of the two waves whose electric fiels at a r given location vary with tie as follows: E sin iˆ A 0 ωt an r 3 E A sin( ω t + )iˆ. 3 0 π Picture the Proble Chose the coorinate syste shown in the phasor iagra. We can use the stanar ethos of vector aition to fin the resultant of the two waves. y δ E r E r E r x The resultant of the two waves is of the for: r r E E sin( ω t + δ )i () The agnitue of E r is: ( A 0 ) + ( 3A0 ) 3. 6A0 E r The phase angle δ is: δ tan 3A A ra

33 Substitute for E r an δ in equation () to obtain: Interference an Diffraction 3079 r E 3.6A0 sin( ωt 0.98 ra)i 50 Fin the resultant of the two waves whose electric fiels at a given r r location vary with tie as follows: E A sinω iˆ an E A sin( ω t + )iˆ. 4 0 t π Picture the Proble Chose the coorinate syste shown in the phasor iagra. We can use the stanar ethos of vector aition to fin the resultant of the two waves. E r y E r δ 60 x E r The resultant of the two waves is of the for: r r E E sin( ω t + δ )i () Apply the law of cosines to the agnitues of the scalars to obtain: r E or r E r r r r E + E E cos0 E r r r r E + E E cos0 E Substitute for E r an E r an evaluate E r to obtain: E r ( 4A 0 ) + ( 3A0 ) ( 4A0 )( 3A0 ) cos A0 Applying the law of sines yiels: Solve for δ to obtain: Substitute nuerical values an evaluate δ: sinδ sin0 r r E E δ sin δ sin r E sin0 r E 3A0 sin ra 6.08A0

34 3080 Chapter 33 Substitute for E r an δ in equation () to obtain: r E 6.A0 sin( ωt ra)i Rearks: We coul have use the law of cosines to fin R an the law of sines to fin δ. 5 Monochroatic light is incient on a sheet with a long narrow slit (Figure 33-45). Let I 0 be the intensity at the central axiu of the iffraction pattern on a istant screen, an let I be the intensity at the secon intensity axiu fro the central intensity axiu. The istance fro this secon intensity axiu to the far ege of the slit is longer than the istance fro the secon intensity axiu to the near ege of the slit by approxiately.5 wavelengths. What is the ratio if I to I 0? Picture the Proble We can evaluate the expression for the intensity for a single-slit iffraction pattern at the secon seconary axiu to express I in ters of I 0. The intensity at the secon seconary axiu is given by: sin φ I I 0 φ where π φ a sinθ I I 0 sin φ φ At this secon seconary axiu: 5 π 5 asin θ an φ 5π Substitute for φ an evaluate I : I 0 I I 0 5π sin 5π Monochroatic light is incient on a sheet that has three long narrow parallel equally space slits a istance apart. (a) Show that the positions of the interference inia on a screen a large istance L away fro the sheet that has the three equally space slits (spacing, with >> ) are given approxiately by y L 3 where,, 4, 5, 7, 8, 0,... that is, is not a ultiple of 3. (b) For a screen istance of.00, a light wavelength of 500 n, an a source spacing of 0.00, calculate the with of the principal interference axia (the istance between successive inia) for three sources.