University of California, Berkeley Physics H7C Fall 1999 (Strovink) SOLUTION TO FINAL EXAMINATION

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1 University of California Berkeley Physics H7C Fall 999 (Strovink SOUTION TO FINA EXAMINATION Directions. Do all six problems (weights are indicated. This is a closed-book closed-note exam except for three 8 inch sheets containing any information yo wish on both sides. Yo are free to approach the proctor to ask qestions bt he or she will not give hints and will be obliged to write yor qestion and its answer on the board. Calclators are allowed bt not essential roots circlar fnctions etc. may be left nevalated if yo do not know them. Use a blebook. Do not se scratch paper otherwise yo risk losing part credit. Cross ot rather than erase any work that yo wish the grader to ignore. Jstify what yo do. Box or circle yor answer.. (35 points One circlarly polarized photon is trapped between two parallel perfectly condcting plates separated by a distance which are parallel to the photon s electric and magnetic fields. a. (0 points If the photon has the smallest definite energy possible nder these circmstances at a point halfway between the plates describe a possible motion of its electric field vector. Soltion. The electric field of the light E is parallel to the condcting plates so E mst vanish at the plates. Therefore we have E(0t= E( t =0. Of corse E(zt is a soltion to the wave eqation: ( z E(zt =0. c t In order to satisfy both the wave eqation and the bondary conditions we choose where E(zt =E 0 sin (kz exp ( iωt (ˆx ± iŷ k = nπ where n =... The smallest definite energy is achieved when k = π/ (conseqently ω = πc/ in which case we have for the electric field: E(zt =E 0 sin ( πz ( πct exp i (ˆx ± iŷ. Halfway between the plates (z = we find that E( t=e 0 exp ( i πct (ˆx ± iŷ. So E moves in a circle with anglar freqency ω = πc/. b. (0 points Same for its magnetic field vector. Soltion. FromMaxwell s eqations Therefore we have E = B t. B y t B x t = E x z = E y z. Fromthe above relations it follows that the spatial dependence of B(zt is given by cos (πz/. Halfway between the plates B =0. c. (5 points For this plate configration making no restriction on the photon energy evalate the density of photon states d N d dλ where N is the nmber of states and λ is the photon wavelength. Take into accont the possible states of circlar polarization.

2 Soltion. We have frompart (a. that k = nπ/ so in terms of wavelength λ (recalling k =π/λ: n = λ. Since we have possible polarization states the total nmber of states N as a fnction of wavelength is N = 4 λ. The derivative with respect to wavelength is dn d λ = 4 λ and so we obtain for the density of photon states d N d λ d = 4 λ.. (35 points A linearly (ˆx polarized plane EM wave travelling along ẑ is incident on an opaqe baffle located in the plane z = 0. The baffle has two slits ct in it which are of infinite extent in the ŷ direction. In the ˆx direction the slit widths are each a and their center-to-center distance is d. (Obviosly d>a bt yo may not assme that d a. The top and bottomslits are each an eqal distance from x =0. The diffracted image is viewed on a screen located in the plane z = where d; also λ d where λ is the EM wavelength. Qarter-wave plates are placed in each slit. They are identical except that the top plate s slow (high-index axis is along (ˆx+ŷ/ (+45 with respect to the ˆx axis while the bottomplate s slow axis is along (ˆx ŷ/ ( 45 with respect to the ˆx axis. a. (5 points What is the state of polarization of the diffracted light that hits the center of the screen at x = y = 0? Explain. Soltion. ight that exits the top slit (slow axis at +45 is in a state ( ( i i 0 = ( i or left-hand circlar polarization; light that exits the bottomslit (slow axis at 45 isina state of polarization ( ( i = ( i 0 i or right-hand circlar polarization. At the center of the screen light fromeach slit contribtes eqally; the state of polarization is proportional to ( + ( = ( ; i i 0 it is ˆx polarized like the incident beam. (See Fowles page 34 and Table. for the Jones vectors and matrices. b. (0 points At what diffracted angle θ x does the first minimm of the irradiance occr? Soltion. Right- and left-hand polarized states are orthogonal; they do not interfere. To see this formally (thogh this is not reqired as part of the soltion conslt Fowles Eq. 3.; the interference termthere is proportional to E E ( i ( =+i =0. i Since there is no interference between the light fromthe top and bottomslit the reslting irradiance is jst twice that expected froma single slit of width a. According to Fowles Eq. 5.8 this pattern is proportional to ( sin β β where in this problem s notation and k =π/λ. β = π or β = ka sin θ x The first minimm occrs at sin θ x = λ a.

3 3. (35 points A lens has an f-nmber (ratio of focal length to diameter eqal to F. The lens is sed to concentrate snlight on a ball whose diameter is eqal to the diameter of the sn s image. The ball is convectively and condctively inslated bt it freely radiates energy otward so that its temperatre can approach an eqilibrimvale T b. a. (0 points The sn sbtends a half-angle of radians. Is the size of its image diffraction-limited i.e. determined largely by the effects of diffraction? Make an orderof-magnitde argment assming that the lens is a typical camera lens with a radis of order 0 m. Soltion. The image of a distant point sorce formed at the focal plane of a lens is actally a Franhofer diffraction pattern where the apertre is the lens opening. The image becomes diffraction limited when the size of the image is near the size of an Airy disk. Fromthis condition we have the Rayleigh criterion: θ >.λ D where θ is the half-angle sbtended by the sn ( rad λ is the wavelength of the light and D is the diameter of the lens. We can assme λ 600 nmfor the sn D = 0 m so the image is not diffraction limited if θ> rad which is clearly satisfied in this problem. Ths the image is not diffraction limited. (Note that this qestion reqires only an orderof-magnitde analysis. So yo don t need to know anything abot the details of Rayleigh s criterion or Airy disks to get fll credit; all that yo need to say is that the diffraction angle is of order λ/d which here is mch smaller than the sn s anglar width. b. (5 points Assming the sn to be a blackbody of temperatre T calclate the ball s temperatre T b. Neglect reflection by the lens. (Hint: yor answer shold depend only on T and F. Soltion. The sn radiates total power P S = σt 4 S 4πR S where T S is the sn s srface temperatre and R S is the sn s radis. The lens collects a fraction of this light power given by Ω Ω = π(d/ 4πR ES where R ES is the distance fromthe earth to the sn. The entirety of this light is focsed on the ball. The ball re-radiates power P b = σt 4 b 4πR b. In eqilibrimwe have the light power absorbed by the ball eqal to the light power radiated by the ball. Setting the two eqal we obtain TS 4 D 4 R S R ES = T b r b. To relate this reslt to F we note that r b /f = /F where f is the focal length. Also we have r b /f R S /R ES θ where θ is the half-angle sbtended by the sn. Employing the above relations in the eqation relating the sn s temperatre to the ball s temperatre: So we find that Tb 4 = TS 4 d θ 4rb = 6F T 4 S. T b = T S F. 4. (30 points Yo are given a Hamiltonian H = (R + R where R and are two operators sch that [ R] =E 0

4 with E 0 a constant. Yo are also given an eigenfnction E (x ofh sch that H E = E E where E another constant is the energy eigenvale. Prove that H(R E =(E + E 0 (R E i.e. R is a raising operator. Soltion. H(R RR + RR RR RR +RR [ R]R +RR E0 R +RR E0 R + RR RR + RR + RR E0 R + R[ R]+RR + RR E0 R + R(R + R E0 R + R(H =(E + E 0 R. 5. (35 points. Consider a harmonic oscillator potential V (x = mω 0x in one dimension. An even nmber N of particles of mass m are placed in this potential. There are no special interactions between the particles no significant mtal electrostatic replsion gravitational attraction etc. compared to the strength of their interaction with the harmonic potential itself. Yo may se what yo already know abot the levels of a harmonic oscillator. The systemis in its grond state i.e. T =0 Kelvin. Calclate the total energy E of the N-particle system relative to the bottom of the well for the cases a. (0 points The N particles are distingishable. Soltion. The energy levels of a harmonic oscillator are ( E n = n + hω 0 where n =0... Nothing prevents mtally noninteracting distingishable particles fromoccpying the same spatial wavefnction. At T =0 this will be the grond state n = 0. Then E = N hω 0. b. (0 points The N particles are identical bosons. Soltion. Same as (a.. All the identical bosons are in the grond state at T =0. c. (5 points The N particles are identical spin fermions. Soltion. Each spatial wavefnction can accommodate two identical spin fermions one with spin p one with spin down. At T = 0 the lowest occpied state is the grond state with energy E 0 = hω 0 while the highest-energy occpied state has (Fermi energy eqal to ( N E F = E 0 + hω 0. The total energy is N times the average energy which is the mean of E 0 and E F : E = N E 0 + E F = N ( ( N E 0 + hω 0 = N 4 hω 0.

5 6. (30 points In the rest frame S of a star ignoring the gravitational redshift some of the photons emitted by the star arise froma particlar atomic transition with an nshifted wavelength λ. When these photons are observed on earth they are shifted to longer wavelength λ = λ + λ becase the star is receding fromthe earth with velocity β 0 c de to the Hbble expansion of the niverse. Astronomers measre this redshift by means of the parameter z definedby z λ λ. For light the relativistic Doppler shift is ω = ω γ 0 ( β 0 cos θ. a. (0 points In the neighboring star limit β 0 show that β 0 is approximately eqal to the measred z. Soltion. ω ω = γ 0 ( β 0 cos θ = πc λ /λ /λ = γ 0 ( β 0 cos θ λ λ = γ 0( β 0 cos θ λ λ = γ 0 ( β 0 cos θ z = γ 0 ( β 0 cos θ. For a receding star cos θ = and so z = γ 0 ( + β 0. When β 0 γ 0 to second order in β 0. Then z +β 0 =β 0. b. (0 points In the distant star limit γ 0 derive an expression for γ 0 in terms of the measred z. Soltion. When γ 0 β 0. Then from the soltion to (a. z γ 0 γ 0 +z. Fll credit is given with or withot the term. c. (0 points The observation of Spernova 987A marked the dawn of a new astronomy in which hmans are able to detect fermions (netrinos as well as bosons (photons from(spatially or temporally resolved sorces otside the solar system. Abot a dozen sch netrinos were detected in each of two hge ndergrond water tanks. The photons fromspernova 987A were redshifted by z 0 5. Taking the Hbble constant to be H yr for how many years did the netrinos from SN987A travel before hmans observed them? Soltion. FromRohlf Eq. (9.7 (necessary for solving assigned problem9.8 the velocity v with which SN987A is receding fromearth is v = H 0 d where d is its present distance fromearth. Netrinos are nearly massless and travel essentially at the speed of light c. Using the reslt of part (a. the travel time T of the netrinos from SN987A was T = d c = v H 0 c = β 0 H 0 z H yr yr.

L = 2 λ 2 = λ (1) In other words, the wavelength of the wave in question equals to the string length,

L = 2 λ 2 = λ (1) In other words, the wavelength of the wave in question equals to the string length, PHY 309 L. Soltions for Problem set # 6. Textbook problem Q.20 at the end of chapter 5: For any standing wave on a string, the distance between neighboring nodes is λ/2, one half of the wavelength. The