Chapter_28_Solution_tips223.#Picture2the2Problem:#Two#coh e re n t#waves#inte rfe re #a t#a n #o b s er v at io n #p oin t #1 6 1 #meters#from #o n e #s ou r ce #a n d #2 9 5 #meters#from#the#other#source.##Strategy:#The#longest#wavelength#that#will#give#constructive#interference#has#a#length#equal#to#the#path#length#difference.###Solution:#Calculate#the#path#length#difference:##295 m 161 m 134 mλ=Δ = − =l#2216.#Picture2the2Problem:#The#figure #s h ows#a#doub le #slit #tha t#produces#the#first#bright#fringes#at#angles#of#±35°#when#laser#light#of#wavelength#670#nm#illuminat es #th e#sl its.####Strategy:#Solve#equation#28E1#for#the#slit#separation#with#m#=#1#for#the#first#bright#fringe.#Solution:#Find#d#for#m#=#1:2sinθ= mλdd =mλsinθ=1 670 nm( )sin 35= 1.2 µm2#Insight:#The#same#slit#spacing#would#have#been#found#using#the#m#=#−1#fringe#at#θ#=#−35°.##222.#Picture2the2Problem:#A#twoEslit#interference#pattern#pro duced #by#546#nm #light#is#observed#on#a#screen#that#is#8.75#m#from#the#slits .##The#firstEorder#maximum#is#5.36#cm#above#the#central#maxima.##Strategy:#Use#equation#28E3#to#find#the#angle#for#the#firstEorder#maximum,#and#then#insert#the#angle#into#equation#28E1#to#solve#for#the#slit#separation#distance.##Solution:#1.2Find#the#angle#to#the#first#bright#fringe:#115.36 cmtan tan 0.351875 cmyLθ−−== =°##2.#Solve#equation#28E1#for#d:#( )( )1 546 nm89.1 msin sin 0.351mdλµθ== =°##Insight:#Because#the#height#of#the#ma xim a#is#m uch #less#than #the#distan ce#to#the #screen ,#the#sm all#angle #appro xim ation #is#valid#(tan sin ).θθθ≈≈#The#slit#separation#distance#th en #b e co mes#( )( )( )1546 nm8.75 m89.1 m.5.36 cmmLdyλµ== =##234.##Picture2the2Problem:#The#ima ge #s h o w s #lig h t#a t #n or mal#incidenc e #re fle c tin g #off#an#oil#film#(n#=#1.38)#that#is#floating#on #a#water#(n#=#1.33)#surface.###Strategy:#Set#the#phase#change#(in#terms#of#w av eleng th)#for #refle ctio n #at #the #airEfilm#interface #as #12,#and#the#phase#change#for#reflection#at#the#filmEwater#interface#as #2.ntλ#Subtract#these#two#phase#differences#and#set#the#result#equal#to#12#to#find#the#minim um#thickness#that#gives#destructive #interferen ce.###Solution:#1.2Set#the#phase#difference#equal#to#12:##set21 122ntλ−=2#2.#Solve#for#the#thickness:##( )521 nm189 nm22 21.38ntnλλ=== =##Insight:#The#next#larger#thickness#that#will#exclude#green#light#(λ#=#521#nm)#reflection#is#378#nm.##47.##Picture2the2Problem:#Light#that#has#a#wavelength#of#676#nm#passes#through#a#7.64Eµm#slit#to#a#screen#that#is#1.85#m#away.#The#cente r #o f#t h e #fi rs t #b r ig h t #fringe#is#a#dista n ce #y#above#the#central#maximum.###Strategy:#The#first#bright#fringe#above#the#central#maximum#is#halfway#between#the#first#and#second#minima.#Solve#equation#28E12#for#the#angle#θ1.5#that#corresponds#to#m#=#1.5.#Then#use#equation#28E3#to#calculate#the#linear#distance#y#on#the#screen#from#the#central#maximum#to#the#first#bright#fringe.##Solution:#1.2Calculate##the#angle#to#the#first##bright#fringe:#( )11.59161.5sin1.5 676 10 msin 7.637.64 10 mWλθ−−−−=×==°×##2.#Calculate#the#distance#y:##( ) ( )1.5tan 185 cm tan 7.63 24.8 cmyLθ== °=##Insight:#The#distance#to#the#first#maximum#can#also#be#calculated#by#taking#the#average#of#the#distances#to#the#first#and#second#minima.##254.##Picture2the2Problem:#The#resolution#of#an#optical#instrument#depends#upon#the#wavelength#of#light.##Strategy:#Use#the#relationship#betwee n #the #res olu tion #of#a n#o pt ical#in stru ment#and#the#w a ve len gt h#λ#of#light,#sin 1.22 Dθλ=#(equation#28E14)#to#answer#the#conceptual#question.###Solution:#The#resolution#of#an#optical#instrument#is#greater#if#the#minimum#angular#separation#θ#is#sma lle r.#The#minimum#angle#θ#decreases#if#the#wavelength#decreases.#Therefore,#greater#resolution#is#obtained#with#the#blue#lig ht .##Insight:#Radio#telescopes,#which#operate#at#very#long#wavelengths,#generally#suffer#from#a#reduced#angular#resolution.##63.##Picture2the2Problem:#655Enm#light#diffracts#as#it#passes#through#a#diffraction#grating#with#787#lines#per#centimeter.##Strategy:#Use#equation#28E16#to#find#the#angles#of#the#first#three#maxim a#(m#=#1,#2,#and#3)#where#the#slit#separation#distance#is#the#inverse#of#the#number#of#lines#per#centimeter#(d#=#1/N).##Solution:#1.2Solve#equation#28E16##for#the#angles#of#the#maxima:#( ) ( )( )( )11 11sinsin sin 655 nm 787 cm sin 0.0515mmNdmN m mλθλθλ−− −−==⎡⎤== =⎣⎦##2.#Find#the#1st#maxima:#( )11sin 0.0515 1 2.95θ−=×=°##3.2Find#the#2nd#maxima:#( )12sin 0.0515 2 5.92θ−=×=°##4.#Find#the#3rd#maxima:#( )13sin 0.0515 3 8 .90θ−=×=°##Insight:#There#are#19#maxima#at#angles#between#2.95°#and#90°.###53.##Picture2the2Problem:#The#resolution#of#the#human#eye#depends#upon#the#diameter#of#the#pupil.##Strategy:#Use#the#relationship#between#the#resolution#of#the#eye#and#the#pupil#diameter#D,#sin 1.22 Dθλ=#(equation#28E14)#to#answer#the#conceptual#question.###Solution:#1.2(a)2On#a#cloudy#day#your#pupil#diameter#will#increase#in#order#to#allow#more#light#to#hit#the#retina.##This#will#decrease#the#angular#size#of#the#diffraction#pattern#created#by#the#pupil#and#increase#the#resolution#as#indicated#by#equation#28E14.##We#conclude#that#your#eyes#have#greater#resolution#on#a#dark#cloudy#day#than#they#do#on#a#bright#sunny#day.###2.2(b)#The#best#explanation#is#I.#Your#eyes#have#greater#resolution#on#a#cloudy#day#because#your#pupils#are#open#wider #to#allow#more#light#to#enter#the#eye.##Statement#II#erroneously#assumes#a#smaller#pupil#would#produce#a#greater#resolution.##Insight:#Part#of#the#reason#that#telescope#designers#strive#for#the#largest#mirror#diameter#(and#hence#the#largest#pu p il#or#aperture#of#the#instrument)#is#to#minimize#the#diffraction#pattern#it#produces#and#maximize#the#resolution.#271.##Picture2the2Problem:#When#white#light#p as s es #th r o u gh #a #g ra t in g#with#760#lin e s#p e r #millimeter,#sp e c tra#are#visible#on#each#side#of#the#central#maximum.##Strategy:#Set#the#angle#equal#to#90°#in#equatio n #2 8 E 16#and#solve#for#the#largest#wavelength#of#light#that#forms#a#secondEorder#(m#=#2)#maximum.#The#slit#separation#d#is#equal#to#the#inverse#of#the#number#of#lines#per#millimeter#N.##Solution:#Solve#equation#28E14#for#the#wavelength:#( )41sinsin sin 906.58 10 mm = 660 nm2 760 mmmmNdmNλθλθλ−−==°== =×##Insight:#Light#that#has#a#wavelength#longer#than#660#nm#will#have#only#one#maximum.#For#example,#light#of#w av