Phyic Exam #3 March 4, 20 Name Multiple Choice /6 Problem # /2 Problem #2 /2 Problem #3 /2 Problem #4 /2 Total /00
PartI:Multiple Choice:Circlethebetanwertoeachquetion.Anyothermark willnotbegivencredit.eachmultiple choicequetioniworth4pointforatotalof 6point..Coherent light pae through a rectangular aperture (a ingle lit) of height h and width w. If h of the lit were halved and the w wa doubled, the correponding height and width of the diffraction pattern on a creen far away will change according to a. it height will double and width will halve b. it height and width will double c. it height and width will halve d. it height will halve and width will double 2. Aplaticringithrownverticallyupward(indicatedbythearrow)througha regionofpacecontainingamagneticfield,hownbelowright.dueto electromagneticinductiononlythedirectionofthecurrentintheringi a.clockwie. b.counterclockwie. c.alternatingbothclockwieandcounterclockwie. d.zero. 3. Light in a tranparent medium with index of refraction n i incident upon a plane interface with a econd medium with index of refraction n 2 (< n ). The incident light make a mall angle θ with repect to the normal to the plane interface. Some of the light i reflected at the interface and ome of it i refracted. Compared to the peed of the refracted light, the peed of the reflected light i a. greater. b. the ame. c. le. d. unable to be determined from the given information. x x x x x x x x x x x x x x x 4. Abeamoflightilinearlypolarizedwithit electricfieldpointingvertically. Youwihtorotateitdirectionofpolarizationby90 (othattheelectricfield ipointinghorizontally)uingoneormoreidealpolarizingheet.toget maximumtranmittedintenity,youhoulduehowmanypolaroidheet? a. One. b. Two. c. Three. d. A many a poible. e. It i not poible to rotate the polarization by 90 o.
PartII:FreeReponeProblem:Thefourproblembelowareworth84point totalandeachubpartiworth7pointeach.pleaehowallworkinordertoreceive partialcredit.ifyourolutionareillegibleorillogicalnocreditwillbegiven.a numberwithnoworkhown(evenifcorrect)willbegivennocredit.pleaeuethe backofthepageifneceary,butnumbertheproblemyouareworkingon.. A ingle loop of wire (with dimenion hown) i held at the top edge of a region of pace where there i a 2T magnetic field that point into the page everywhere and i perpendicular to the plane of the wire loop. The wire loop ha a ma of 00g and a reitance of 0.25Ω. W = 0.35m L = 0.25m a. If the bar i releaed from ret with it bottom edge at the very top edge of the region of magnetic field, what i it x x x x peed jut a the loop top edge completely enter the x x x x region of magnetic field? (Hint: Aume over thi interval x x x x that the net force (and hence the acceleration) acro the x x x x conductor width i zero. When the loop i completely x x x x inide of the field, the net force i not zero.) If you cannot calculate a peed ue v =.0m/. F y : F B F w = IwB mg = 0 IwB = ε R v = mgr w 2 B = 0.kg 9.8 m 2 0.25Ω 2 0.35m 2T ( ) 2 = 0.5 m wb = Bwv R wb = mg b. What are the potential difference and the magnitude of the electric field that are induced over the width of the wire? ε = Bwv = 2T 0.35m 0.5 m = 0.35V E = ΔV Δx = 0.35V 0.35m = V m c. What are the magnitude and direction of the induced current in the loop of wire? I = ε R = 0.35V =.4A and the current flow i CCW to oppoe the change in flux. 0.25Ω
2. A potlight i mounted on the left ide and above a wimming pool and thi light ource end out a beam of light into the pool a hown by the red line. The pool i 2.0m deep with the light i mounted on the left ide of the pool.2m above the urface of the water and the light trike the water urface 2.4m from the left edge of the pool. a. What i the angle of refraction that the light make upon entering the water? (The indice of refraction of air and water are n air =.00 and n water =.33 repectively.) φ θ The angle of incidence i given from the geometry tanφ =.2m 2.4m = 0.5 φ = ( tan 0.5) = 26.6 o θ + φ = 90 θ = 63.4 o The angle of refraction i given from the law of refraction n air inθ = n water inθ 2.00in( 63.4) =.33inθ 2 θ 2 = 42.2 o θ 2 b. How far from the left edge of the pool light trike the bottom of the pool? From the projection of the normal to where the light trike the bottom of the pool we have from the geometry of the ytem: tanθ 2 = x x = 2.0mtan( 42.2) =.8m. Therefore from the left edge we have 2.0m the total ditance = 2.4m +.8m = 4.2m.
c. Given the ytem of converging lene below, what i the final image height if the object i 2.4cm tall and the focal length of len # and len #2 are 27mm and 48mm repectively? Suppoe that the object i placed 450mm to the left of len # and that the lene are eparated by 200mm. For len : + = d i = d o d i f f d o M = d i d o = 76.9mm 450mm = 0.39 = 27mm 450mm and D = d i + d o2 d o2 = 200mm 76.9mm = 23.mm For Len 2: + = d i2 = d o2 d i2 f 2 f 2 d o2 M 2 = d i2 = 44.5mm d o2 23.mm =.9 Therefore the final image height i = 48mm 23.mm f f f 2 f 2 =76.9mm = 44.5mm and if you want, the final M T = M M 2 = h i2 h i2 = M M 2 h o = ( 0.39) (.9) 2.4cm =.8cm h o image propertie are virtual, inverted with repect to the object and reduced in ize.
3. Suppoe that you conduct an experiment to tudy the interference and diffraction of a beam of particle. Suppoe that you have a beam of 0.0MeV proton (thi i their kinetic energy) and thee proton are incident on a pair of double lit of width a and eparation d. On a creen 8.0m from the lit, there i a pattern of contructive interference produced. You meaure the ditance between the center of each adjacent contructive interference and you find they are all evenly paced by 7.2mm, and addition you count viible contructive interference pot within the central diffraction envelope. a. What i the wavelength of the proton ued for the experiment? KE =0MeV = ( γ )m p = ( γ ) 937.MeV γ =.0 γ =.0 = v 2 v = γ 2 c = 0.5c p = γmv =.0.67 0 27 kg 0.5 3 0 8 m λ = h p = 6.63 0 34 J 20 kgm = 8.7 0 5 m 7.60 0 = 7.60 0 20 kgm b. What i the eparation d between the two lit? d y D = λ d = λd = 8.7 0 5 m 8m = 2.2 0 m y 7.2 0 3 m c. What i the width a of one of the lit? # total =3 = 2 d + d a a = 6 a = d 6 = 2.2 0 m = 3.6 0 2 m 6
4. Suppoe that you are conducting an experiment to invetigate the photoelectric effect uing a ceium target (φ C = 2.9eV.) a. If a.0w beam of 400nm photon are ued to illuminate the C target, what will be the maximum KE of the electron produced and auming that the proce of photoelectron production i 00% efficient what will be the photocurrent of electron produced? E = hc λ = 6.63 0 34 J 3 0 8 m ev = 3.eV. Thu the KE of the 400 0 9 m.6 0 9 J ejected electron i KE = E φ = 3.eV 2.9eV = 0.2eV. The photocurrent i calculated from the number of photon per econd, ince the proce i 00% efficient. Here we have W = J = NE N = W E = W 4.97 0 9 J = 2.0 photon 08 I = Ne = 2.0 0 8 photon.6 0 9 C e = 0.32A b. What potential difference, V top, would you need to top the electron from reaching the oppoite plate from where they were ejected? (Hint: It take work to top the ejected electron from reaching the oppoite plate.) KE = ev top V top = KE e = 0.2eV e = 0.2V c. Suppoe intead of the 400nm photon ued in part a, you decide to ue 500nm photon intead. If again you have a.0w beam of 500nm photon incident on the C target, what will be the maximum KE of the electron produced and auming that the proce of photoelectron production i 00% efficient what will be the photocurrent produced? φ = hf min = hc λ max = hc λ max φ = 6.63 0 34 J 3 0 8 m 2.9eV.6 = 4.29 0 7 m = 424nm. 0 9 J ev Since thi i the maximum wavelength, then 500nm light will produce no ejected electron and thu no photocurrent will be produced.
ElectricForce,FieldandPotential ElectricCircuit LightaaWave F = k Q Q 2 r ˆ r 2 c = fλ = ε o µ o E = F q E Q = k Q r r ˆ 2 PE = k Q Q 2 r V ( r) = k Q r E x = ΔV B,A Δx W A,B = qδv A,B MagneticForceandField F = qvb inθ F = IlB inθ τ = NIAB inθ = µb inθ PE = µb coθ B = µ 0I 2πr ε induced = N Δφ B Δt Contant g = 9.8 m 2 2 Nm G = 6.67 0 kg 2 e =.6 0 9 C k = = 9 0 9 4πε o C 2 Nm 2 = N Δ ( BA coθ ) Δt ε o = 8.85 0 2 Nm 2 C 2 ev =.6 0 9 J µ o = 4π 0 7 Tm A c = 3 0 8 m h = 6.63 0 34 J m e = 9. 0 3 kg = 0.5MeV m p =.67 0 27 kg = 937.MeV m n =.69 0 27 kg = 948.3MeV amu =.66 0 27 kg = 93.5MeV N A = 6.02 0 23 Ax 2 + Bx + C = 0 x = B ± B2 4AC 2A PhyicEquationSheet Geometry Circle : C = 2πr = πd A = πr 2 Triangle : A = 2 bh Sphere : A = 4πr 2 V = 4 3 πr3 S( t) = energy time area = cε E 2 o t I = S avg = cε E 2 2 o max P = S c = Force Area S = S o co 2 θ v = εµ = c n θ inc = θ refl n inθ = n 2 inθ 2 f = + d o d i M = h i h o = d i d o M total = N i= M i = c B 2 max 2µ 0 ( ( ) = c B2 t) µ 0 dinθ m = d tanθ m = d y m D = mλ or ( m + 2)λ ainφ m' = atanφ m' = a y m' D = m'λ LightaaParticle&Relativity NuclearPhyic Mic.Phyic0 Formulae F = Δ p Δt = Δ( mv) = ma Δt F = ky F C = m v 2 R r ˆ W = ΔKE = m v 2 2 2 f v i PE gravity = mgy PE pring = 2 ky 2 x f = x i + v ix t + 2 a xt 2 v fx = v ix + a x t v 2 vx = v 2 ix + 2a x Δx ( ) = ΔPE