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UH PHYS 1302 - HW_solutions_26

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HW#Solutions#Chapter#26#(Das,#1302)##1.##Picture#the#Problem:#The#ima ge #s h o w s #a #la se r #b ea m#with#an#a n g le #o f#28°#between#the#incident#and#reflected#rays.#The#image#also#shows#the#path#of#the#laser#beam#when#the#mirror#is#rotated#by#5°.###Strategy:#Use#the#law#of#reflection#to#set#the#angle#between#the#incident#and#reflected#rays#equal#to#twice#the#incident#angle,#and#then#solve#for#the#incident#angle.#After#the#m irror#is#rotated ,#se t #th e #new#ang le #between #the#incident#and#reflected#w ave s#equ al#to#tw ice#the#ne w #inciden t#angle,#which#is#5°#greater#than#the#initial#incident#angle.##Solution:#1.#Calculate#the#initial#incident#angle:## θi+θr= 2θi= 28°θi= 14°##2.#Calculate#the#final#angle#b e tween#inciden t#a n d #refl ec te d#b e ams:# !θi+!θr= 2!θi= 2θi+ 5°( )= 2 14° + 5°( )= 38°##Insight:#Because#the#incident#and#reflecte d#ang les#are#eq ual,#increas ing#the #incident#an gle#by #an#an gle# Δθ#will#increase#th e#a n g le#b e tween#inciden t#a n d #refl ec te d #ray s #by #2Δθ.###18.##Picture#the#Problem:#A#spheric al #mirror#is#mad e #fr om#a#section #of #a#s p h e re #o f#ra d iu s #0 .8 6 #m.##Strategy: Calculate the focal point for the convex side using equation 26-2 and the focal point for the concave side using equation 26-3. #Solution:#1.#(a)#Calculate#the#convex#focal#length:# f = −12R = −120.86 m( )= − 43 cm##2.#(b)#Calculate#the#concave#focal#length:## f =12R =120.86 m( )= 43 cm##Insight:#The#magnitudes#of#the#focal#lengths#are#equal,#but#the#convex#mirror#has#a#negative#focal#length.##20.##Picture#the#Problem:#The#ima ge #sho ws#pa rallel#rays#from#the#Sun#converging#at#the#focal#point#of#a#concave#piece#of#glass.###Strategy: Solve equation 26-3 for the radius of curvature, given that the rays converge at the focal point. #Solution:#Calculate#the#radius:# R = 2 f = 2 15 cm( )= 30 cm##Insight: If the glass were turned around so that the light was reflecting off the convex portion, a virtual image of the Sun would appear to be 15 cm behind the glass. ##21.##Picture#the#Problem:#You#hold#a#shiny#tablespoon#at#arm’s#length#and#look#at#the#back#side#of#the#spoon.##Strategy: Consider the images formed by convex mirrors when answering the conceptual questions. #Solution:#1.#(a)#The#back#of#the#spoon#behaves#like#a#convex#mirror,#and#convex#mirro r s#always#fo r m#upright#images.###2.#(b)#Convex#mirrors#always#form#images#that#are#reduced#in #s ize .###3.#(c)#The#fact#that#no#light#passes#through#the#image#(which#is#behin d#the#sp oon )# means#that#the#image#is# virtual.##Insight:#Figure#26O17#illustrates#the#upright,#reduced,#virtual#image#formed#by#a#convex#mirror.###22.##Picture#the#Problem:#You#hold#a#shiny#tablespoon#at#arm’s#length#and#look#at#the#fron t#sid e#of#th e #spo o n.###Strategy: Consider the images formed by concave mirrors when answeringthe conceptual questions. #Solution:#1.#(a)#Looking#at#the#front#side#of#a#spoon#means#you#are#looking#at#a#con ca ve #m irro r.#In #ad ditio n,#ho ld ing #the #sp oo n#a t#arm’s#length#me an s#th at#you #are#outside#the#focal#point#of#the#mirror—clearly,#the#focal#length#of#the#front#side#of#a#spoon #is#only#a#few#centimeters.#The#situation,#then,#is###like that illustrated in the figure at the right. It follows that your image is inverted. #2. (b) Referring to the figure above, we see that your image is reduced in size. #3. (c) Light rays pass through the image, and it therefore is a real image. #Insight: Figure 26-18 and Example 26-3 illustrate how a single concave mirror can form either real or virtual images. #26.##Picture#the#Problem:#An#object #is#p l ac e d #in #fro n t #of #a#c o n ca v e #mirror.#The#mirror#produces#a#real,#magnified#image.####Strategy:#Use#equation#26O3#to#calculate#the#focal#length#from#the#radius#of#curvature.#Then#use#equation#26O6#to#calculate#the#image#distance.##Solution:#1.#Calculate#f#:# f =12R =1240.0 cm( )= 20.0 cm##2.#Solve#Eq.#26O6#for# di:# di=1f−1do"#$%&'−1=120.0 cm−130.0 cm"#$%&'−1= 60.0 cm##Insight:#As#shown#in#the#ray#diagram,#the#image#is#60.0#cm#in#front#of#the#mirror.##29.##Picture#the#Problem:#An#object # is #p la ce d#2 .0 #m #in #fro n t#o f#a#concave#mirror#that#has#a#focal#length#of#0.50#m.#The#mirror#produces#a#real,#reduced#image.###Strategy:#Solve#equation#26O6#for#the#image#distance.#Then#solve#equation#26O8#for#the#magnification.##Solution:#1.#Calculate#the#image#distance:# di=1f−1do"#$%&'−1=10.50 m−12.0 m"#$%&'−1= 0.67 m##2.#Calculate#the#magnification:# m = −dido= −12.0 m10.50 m−12.0 m"#$%&'−1= − 0.33##Insight:#Because#the#image#distan ce#is#positive,#this#is#a#real#imag e#in#front#of#the#m irror.#Since#the #m agn ification#is#negative,#we#know#that#the#image#is#inverted.##31.##Picture#the#Problem:#An#object #is#p l ac e d #2 .0 #m#in#front#of#a#convex#mirror#that#has#a#focal#length#of#−#0.50#m.#The#mirror#produces#a#virtual,#reduced#image.###Strategy:#Solve#equation#26O6#for#the#image#distance.#Then#solve#equation#26O8#for#the#magnification.##Solution:#1.#Calculate#the#image#distance:# di=1f−1do"#$%&'−1=1− 0.50 m−12.0 m"#$%&'−1= − 0.40 m##2.#Calculate#the#magnification:# m = −dido= −12.0 m1− 0.50 m−12.0 m"#$%&'−1= 0.20##Insight:#Because#the#image#distan ce#is#nega tive,#this#is#a#virtual#image #that#is#beh ind#the #m irror.#We #know #th at#the#image#is#up r igh t #be c au s e#th e #m a g n ifica tio n #is#p o sitiv e .# ##39.##Picture#the#Problem:#A#convex #mirror#produ c e s#a #v irt u a l,#upright,#and#reduced#image#of#an#object.###Strategy:#Use#the#magnification#equation#(equation#26O8)#to#calculate#the#image#distance#from#the#mirror,#and#then#solve#equation#26O6#for#the#focal#length.##Solution:#1.#(a)#Calculate#the#image#distance:# di= −mdo= −1436 cm( )= − 9.0 cm##2.#(b)#Calculate#the#focal#length#from#equation#26O6:## f =1do+1di!"#$%&−1=136 cm+1− 9.0 cm!"#$%&−1= −12 cm##Insight:#As#expected,#the#image#distance#is#negative,#indicating#a#virtual#image,#and#the#focal#length#is#negative,#indicating #a#c on v e x #m irr o r.##46.##Picture#the#Problem:#When#rays#of#color#A#and#color#B#are#sent#through#a#prism,#color#A#is#bent#more#than#color#B.##Strategy: Recall that the amount that a ray refracts depends upon the index of refraction of the material that it enters. #Solution: Color B travels more rapidly. The reason is that color B, which is “bent” less than color A, must have the smaller index of refraction. Because the


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