Snell Law - Optics and Modern Physics - Solved Past Paper, Exams of Physics

Download Snell Law - Optics and Modern Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! Physics 228 - Final Exam Solutions May 9, 2006 Prof. Coleman, Dr. Francis, Prof. Bronzan, Prof. Glashausser, and Prof. Madey 1. Which of the following quantities remains unchanged when light passes from a vacuum into a slab of glass with a 45◦ angle of incidence? a) its frequency b) its wavelength c) its speed d) its direction of travel e) none of these Solution: The frequency remains unchanged. The speed changes because the index of refraction of glass is not 1.0, like the vacuum. Since the speed changes and the frequency remains fixed, the wavelength changes. The direction of travel changes according to Snell’s law. 2. A converging and a diverging lens, each with a focal length of 30 cm, are arranged so that they are separated by 60 cm. If a candle is placed 90 cm to the left of the converging lens, where is the image produced by the diverging lens? f f a) 30 cm to the right of the diverging lens b) 10 cm to the left of the diverging lens c) 18 cm to the left of the diverging lens d) 30 cm to the left of the diverging lens e) 90 cm to the right of the diverging lens Solution: For the converging and diverging lenses, 1 do + 1 di = 1 30 cm , 1 d̄o + 1 d̄i = − 1 30 cm . do = 90 cm, so di = 45 cm. This image is 15 cm to the left of the diverging lens, so d̄o = 15 cm. Then d̄i = -10 cm. The final image is 10 cm to the left of the diverging lens. 3. In a Young’s double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separation between adjacent bright fringes on a screen 5 m from the slits is: a) 0.10 cm b) 0.50 cm c) 1.0 cm d) 0.05 cm e) 0.25 cm Solution: 4. A glass (n = 1.6) lens is coated with a thin film (n = 1.3) to minimize reflection of certain incident light. If λair = 500 nm is the wavelength of the light in air, the least film thickness is: a) 78 nm b) 96 nm c) 162 nm d) 200 nm e) 250 nm Solution: We want to find the solution for the first destructive interfer- ence fringe. Since nair < nfilm < nglass, there will be a phase shift at each reflection, so the relative phase shift between the light reflected from the film and the light reflected from the glass will be zero. This means we can use 2t = 1 2 λfilm = λair 2nfilm which gives us t = λair/(4nfilm) = 96 nm. 9. The stopping potential for electrons ejected by 6.8 × 1014-Hz electromagnetic radiation incident on a certain sample is 1.8 V. The kinetic energy, K, of the most energetic electrons ejected and the work function, φ, of the sample, respectively, are: a) K = 1.8 eV, φ = 2.8 eV b) K = 1.8 eV, φ = 1.0 eV c) K = 1.8 eV, φ = 4.6 eV d) K = 2.8 eV, φ = 1.0 eV e) K = 1.0 eV, φ = 4.6 eV Solution: The stopping potential and the maximum kinetic energy are related as Kmax = eV , so that Kmax = 1.8 eV. Using the photoelectric equation eV = hf − φ → φ = hf −Kmax gives us φ = 1.0 eV. 10. In Compton scattering from stationary electrons, the largest amount of energy will be imparted to the electron when the photon is scattered through: a) 0◦ b) 45◦ c) 90◦ d) 180◦ e) 270◦ Solution: 11. The binding energy of an electron in the n=2 state in a hy- drogen atom is about: a) 3.4 eV b) 13.6 eV c) 10.2 eV d) 1.0 eV e) 27.2 eV Solution: The total energy of the hydrogen atom in the n = 2 state is -3.4 eV (ie, -13.6/4). Thus it takes 3.4 eV to separate the electron in this state from the proton in the H atom, ie, its binding energy is 3.4 eV. 12. A ruby laser delivers a 1-ns pulse of 1.0 MW average power. If the light has a wavelength of 694.3 nm, how many photons are contained in the pulse? a) 3.5 × 1024 b) 5.5 × 1014 c) 3.5 × 1015 d) 7.3 × 1015 e) 1.7 × 1024 Solution: The energy in the pulse in (1.0 × 106 J/s)× (1.0 × 10−9s)= 1.0× 10−3 J. The energy carried by each photon is hf = hc/λ = 2.87 × 10−19 J/photon. Thus the number of photons is (1.0× 10−3 J)/(2.87× 10−19 J/photon)= 3.49× 1015 photons. 13. An electron is confined in an infinite, one dimensional, square potential well of width 0.200 nm. V = { ∞ for x < 0 and x > 0.200nm 0 for 0 < x < 0.200nm The energy of the ground state is: a) 0.142 eV b) 1.50 eV c) 9.40 eV d) 13.6 eV e) 54.4 eV Solution: 14. An electron is in a one-dimensional potential well of width L with zero potential energy in the interior and infinite potential energy at the walls. A graph of its wave function ψ(x) versus x is shown. The value of quantum number n is: a) 0 b) 1 c) 2 d) 3 e) 4 0 L ψ Solution: Since the wavefunction for a particle in a box is Ψn(x) ∝ sin ( nπ L x ) for n = 1, 2, 3, . . ., we can determine n out by looking at the picture. One and one-half wavelengths are fit into the box, so the wavelength is λ = 2L/3. Since each half-wavelength cor- responds to one energy level, this means the quantum energy number is 3. 15. If we think of an electron in a hydrogen atom as being a standing wave, then in the ground state of radius, a0, what is the electron’s de Broglie wavelength? a) 2πa0 b) h/(m2v2) c) a0/(h̄c) d) a0 e) h̄/(mv) Solution: 16. The maximum number of electrons that can be accommodated in an orbital with quantum number ℓ = 3 is: a) 2 b) 3 c) 7 d) 9 e) 14 Solution: For ℓ = 3, mℓ can take on the seven integer values from -3 to +3. For each of these values of mℓ, ms can be + or -1/2. 24. In a certain nuclear fission process, 1 0n+ 235 92 U→ 141 56 Ba + 92 36Kr+3 1 0n Where: m(23592 U) = 235.043924 u m(14156 Ba) = 140.9139 u m(9236Kr) = 91.8973 u m(10n) = 1.008665 u The energy released in this process is about: a) 86 MeV b) 200 MeV c) 2.19 GeV d) 79 MeV e) 120 GeV Solution: The energy released is going to be due to the mass difference between the original nucleus (and the catalyst neutron) and the final products: Q = MLHSc 2 −MRHSc 2 = (m(23592 U) +m(n))c 2 − (m(14156 Ba) +m( 92 36Kr) + 2m(n))c 2 = (m(23592 U) −m( 141 56 Ba) +m( 92 36Kr) + 2m(n))c 2 = 200 MeV 25. Each of the following reactions is forbidden. Determine a con- servation law that is violated for each reaction. I. π− + p+ → p+ + π+ II. p+ + p+ → p+ + π+ III. p+ + µ+ → p+ + p+ + ν̄µ a) I. charge, II. baryon number, III. baryon num- ber b) I. charge, II. lepton number, III. baryon number c) I. charge, II. baryon number, III. lepton number d) I. charge, II. lepton number, III. lepton number e) I. lepton number, II. baryon number, III. baryon num- ber Solution: 26. Consider the following reaction: π− + p → Ko + (?) ; which in terms of quarks is: (ūd) + (uud) → (ds̄) + (?) Which of the following is a candidate for the unknown prod- uct? a) p̄ = (ūūd̄) b) K̄o = (d̄s) c) Ξo = (uss) d) Σ− = (dds) e) Λo = (uds) Solution: 27. A telescope has a diffraction grating with 750 slits per cen- timeter. Two different wavelengths of radiation, λ1 = 900 nm and λ2 = 700 nm fall on the grating. How far apart are their first maxima, in degrees? a) 0◦ b) (1.5 × 10−10)◦ c) 0.0086◦ d) 0.015◦ e) 0.86◦ Solution: 28. Which of the following statements about the “Standard Model” of particle physics is false? a) Bosons are the matter constituents. Fermions mediate the forces between particles. b) The weak force is short-ranged because of the large mass of the Z and W particles. c) Strangeness is not conserved by the weak interaction. d) There are three known generations of leptons and quarks. e) Hadrons are bound states of quarks, or antiquarks. Solution: Bosons mediate the force between particles. 29. Protons are accelerated in a cyclotron with internal field of 0.2 T. If beam exits at a radius of 2 m from the center of the cyclotron, what is the energy of the protons? a) 7.67 MeV b) 14 GeV c) 38.3 MeV d) 1.92 MeV e) 15.3 MeV Solution: The cyclotron angular frequency is ω = eB m = (1.60 × 10−19 C)(0.200 T) 1.67 × 10−27 kg = 1.92 × 107 rad/sec. The speed of the exiting protons is v = ωR = (1.92 × 107 rad/sec) (2.00 m)= 3.84 × 107 m/sec. The kinetic energy of the protons is K = 1 2 mv2 = 1 2 (1.67 × 10−27 kg)(3.84 × 107 m/sec)2 = 1.23 × 10−12 J 1.60 × 10−13 J/MeV = 7.67 MeV. 30. An electron in an atom has quantum numbers n = 2, ℓ = 1, mℓ = −1, and ms = +1/2. What is the magnitude of the orbital angular momentum of this electron? a) 0 b) −h̄ c) √ 3h̄/2 d) h̄/2 e) √ 2h̄ Solution: