1. Newton's Laws provide a good description of the flight of a baseball because:

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1 1. Newton's Laws provide a good description of the flight of a baseball because: Solution: Newton's laws can be used provided that the velocities are small compared to c (so that relativistic effects can be neglected) and the objects are large compared to atomic scales (so that quantum effects can be neglected). What's wrong with the other answers? The nonzero value of Planck's constant is what's responsible for the quantum behavior of objects. But because its value in standard units is small, quantum effects are not generally noticeable on macroscopic scales. An inertial frame is a frame in which you can apply the relevant laws of physics. Depending on the situation, you may need to apply the laws of quantum mechanics or relativity. QM and relativity can be applied to baseballs, but their predictions are indistinguishable from those of Newton's laws. The Compton wavelength of a baseball would be h/mc, where M is the mass of the baseball. Since this is huge compared to, say, an electron, the Compton wavelength would be tiny. A) Its speed is small compared to c and its size is large compared to atomic scales. B) Planck's constant is nonzero. C) The earth is an inertial reference frame. D) Quantum mechanics and relativity cannot be applied to baseballs. E) The Compton wavelength of a baseball is large compared to atomic scales. 2. In the two figures below, the solid arrow represents the object, and the dashed arrow represents the image. The dashed rectangle represents a lens. Situation 1 Situation 2 In situation 1 and 2, the lenses are respectively: Solution: In situation 1, the image is virtual (on the same side as the object). Only a diverging lens can produce and image that is virtual, upright, and reduced. (See the lecture slides from 3/17 for a ray diagram of this situation.) In situation 2, the image is real (on the opposite side from the object). Only a converging lens can produce a real image. The situation shown here, where the image is inverted and reduced, would happen if the object is beyond 2f. See again the 3/17 lecture slides. A) diverging, diverging. B) converging, converging. C) diverging, converging. D) converging, diverging. E) impossible to determine.

2 3. A red laser pointer has a wavelength of 650 nm. The laser light emerges from a pupil with a diameter of 3.2 mm. What is the width of the spot when the laser is pointed at a screen 18 m away? Solution: A diagram of this situation is shown below: D θ y L As the laser light passes through the opening with diameter D, it diffracts into an angle that's given (in the small-angle approximation) by θ = 1.22λ/D = y/l. The diameter of the spot is twice y, so we have (Spot width) = L D m m 18 m 8.9 mm. A) 4.5 mm B) 7.3 mm C) 8.9 mm D) 3.2 mm E) 6.0 mm 4. A soap bubble is illuminated by a combination of red light (λ=736 nm) and green light (λ=552 nm). The red light is strongly reflected and the green light is strongly absorbed. red green.0.0 What is the minimum thickness of the bubble?.33 Solution: First, note that there is a half-wavelength phase shift for the wave reflected from the front of the bubble, because it reflects off a medium with a higher index. There is no phase shift for the wave reflected from the back of the bubble. However, this wave travels an additional distance of 2t, where t is the thickness of the bubble. To get constructive interference for the red light, we need this extra distance to be an odd number of half-wavelengths, i.e. 2 t ' 2, 3 ' 2, 5 ' 2,... etc., where λ =λ/n is the wavelength of the light in the film. So the possible thicknesses are 138 nm, 415 nm, 692 nm, etc. To get destructive interference for the green light, we need 2t=mλ, where m is an integer. So the possible thicknesses are 0, 207 nm, 415 nm, 622 nm,... The first possibility that satisfies both conditions is 415 nm. A) 138 nm B) 415 nm C) 552 nm D) 622 nm E) 276 nm

3 5. The muon is an elementary particle with the same charge as the electron, but its mass is 206 times larger. Suppose that a muon and a proton come together to form a muonic hydrogen atom. If the ionization energy of an ordinary hydrogen atom is 13.6 ev, what is the ionization energy of muonic hydrogen? (you derived this h 2 n 2 formula in MP #22). So if we replace the electron with a muon whose mass is 206 times larger, the ionization energy will correspondingly increase to 206(-13.6 ev) = 2800 ev. Solution: The Bohr energy levels are E n 2 2 m k 2 e 4 A) 2800 ev B) 0.95 ev C) ev D) 195 ev E) 13.6 ev Problems 6 and 7 refer to the following situation. A sheet of clean, photosensitive metal with work function φ 0 is illuminated by light with intensity I 0 and wavelength λ 0. When this happens, electrons are ejected from the surface at a rate R 0 electrons per second. You observe that the most energetic of these electrons have a kinetic energy of KE If the intensity of the incident light is increased, electrons now leave the surface at rate R and with maximum kinetic energy KE. Which statement is true? Solution: When the intensity is raised (keeping the wavelength fixed), the number of photons per second striking the metal surface increases. Therefore the rate at which electrons are ejected increases. The maximum KE is determined by the frequency of the incident light and the work function of the metal, so it does not change here. A) R >R 0 and KE >KE 0. B) R >R 0 and KE <KE 0. C) R =R 0 and KE >KE 0. D) R <R 0 and KE =KE 0. E) R >R 0 and KE =KE Which of the following changes to the experiment would cause the maximum kinetic energy of the electrons to increase? (For each choice, assume all the other parameters of the experiment remain constant.) Solution: The max KE is KE max hf, where f=c/λ is the frequency of the light and φ is the work function of the metal. So we can increase the max KE by decreasing the wavelength. Increasing the intensity would have no effect on the max KE, and using a metal with a larger work function would reduce it. A) Increase the light intensity to 2I 0. B) Increase the wavelength to 2λ 0. C) Decrease the wavelength to λ 0 /2. D) Choose a metal with a larger work function. E) None of these.

4 8. In Bohr's model of the hydrogen atom, the allowed energy levels are E n = -(13.6 ev)/n 2, where n is the principal quantum number. If a hydrogen atom undergoes a transition from the n=4 to the n=3 state, it will: Solution: A photon is emitted when the atom makes a transition to a lower energy level. The energy is E E i E f 13.6 ev 0.66 ev. A) emit a 0.85 ev photon. B) emit a 0.66 ev photon. C) absorb 1.51 ev of energy. D) emit a 1.51 ev photon. E) absorb 0.66 ev of energy On February 1, 2003, the space shuttle Columbia disintegrated upon re-entry to the Earth's atmosphere, killing its crew of seven astronauts. Flight engineers immediately suspected that the tragedy was caused by damaged heat-resistant tiles on the left wing, which had been struck by a piece of debris during liftoff. NASA was subsequently criticized for failing to use spy satellites to photograph the shuttle while it was still in orbit, to try to determine the extent of the damage. Suppose that the damaged area of the wing was roughly 30 cm in diameter. What is the greatest distance at which a spy satellite with a 1.2 m diameter lens could detect the damage? Assume that the photograph would be taken in yellow light (the peak of the solar spectrum), with a wavelength of 580 nm. Solution: This problem involves Rayleigh's criterion. In the small angle approximation, we can resolve objects with an angular separation of θ=1.22λ/d, where D is the diameter of the objective lens. But we also have θ= x/l, where x is the width of the damaged area and l is the distance from the space shuttle to the satellite. Equating these two expressions, we have l = D x/1.22λ=(1.2m)(0.30m)/( m) = 510 km. A) 600 km B) 430 km C) 310 km D) 510 km E) 850 km

5 10. Which of the following figures most accurately depicts the passage of white light through a prism? Solution: The light refracts as it enters and leaves the prism. Because of dispersion, violet light is bent more than red. A) B) red red white.5 violet white.5 violet C) D) red violet white.5 violet white.5 red E) red white.5 violet 11. Captain Kirk and his crew are cruising around the galaxy in the USS Enterprise at a constant speed of Warp Factor 5. They report that they have been maintaining this speed for 48 days. During this time, 76 days have elapsed on earth. How fast is Warp Factor 5? Solution: The crew is moving with their clock, so they report the proper time, t 0. This is related to the time t elapsed on earth via t 0 t and solving for v, we obtain v c 1 t 0 t 2 c c. 1 v 2 c 2. Squaring both sides A) 0.70c B) 0.91c C) 0.61c D) 0.78c E) 0.82c

6 12. A famous problem that helped lead to the development of quantum mechanics was the socalled ultraviolet catastrophe. The catastrophe refers to the failure of classical physics to explain what phenomenon? Solution: Classical physics could not account for the blackbody spectrum of hot, glowing objects. It predicted an unbounded increase of radiation at short wavelengths, hence the name ultraviolet catastrophe. This problem was resolved by Planck's quantum hypothesis. For more details, see the lecture slides from 3/31. A) The photoelectric effect. B) Electron diffraction. C) Young's double-slit experiment. D) Compton scattering. E) The spectrum of hot, glowing objects. 13. Without her contact lenses, a student reports that she cannot see objects clearly if they are less than 65 cm away. What should be the refractive power of her contact lenses (in diopters) in order for her to see objects as close as 25 cm away? Solution: Because the student cannot see nearby objects, she is farsighted, with a near point of 65 cm. To allow her to see an object 25 cm away, we must create a virtual image of that object at a distance of 65 cm. In other words, we must apply the thin lens equation with d 0=25 cm and d i=-65 cm (with a minus sign because the image is virtual). Thus 1 1 f 0.25 m m 2.5 m diopters. A) 2.5 B) 5.5 C) -2.5 D) -5.5 E) A sharpshooter is taking aim at the center of a target. He fires the bullet from his rifle, which of course has a circular aperture through which the bullet exits the rifle. Unfortunately, he misses the bullseye. He claims this happened because diffraction changed the path of the bullet. Is this a plausible explanation? Solution: The de Broglie wavelength of any macroscopic object is far too small for the wavelike nature of the object to be apparent. For example, the de Broglie wavelength of a 10g bullet moving at 200 m/s is h p h mv m, which is trillions of times smaller than the diameter of a proton. But for diffraction to be important, the wavelength must be comparable to the size of the aperture. So diffraction could not possibly occur here. The sharpshooter simply missed. A) Yes. B) No, because the de Broglie wavelength of the bullet is too small. C) No, because the velocity of the bullet is much less than c. D) No, because the diameter of the aperture is too small. E) No, because only a rectangular slit could cause diffraction.

7 15. The Compton scattering formula is ' h m e c 1 cos. In this expression, is Solution: The diagram for the Compton scattering of a photon off an electron is shown below. The angle θ is the angle between the incident and scattered photon. scattered electron incident photon θ scattered photon A) The angle between the incident photon and the scattered electron. B) The angle between the scattered photon and the scattered electron. C) Equal to n D, where n is an integer. D) The angle to the first minimum. E) The angle between the incident photon and the scattered photon. 16. A light ray in the core (.40) of an optical fiber travels at an angle of θ 1 =49 with respect to the axis of the fiber. The ray is transmitted through the cladding (.20) and into the air. What angle θ 2 does the exiting ray make with respect to the outside surface of the cladding? Solution: A diagram of this situation is shown below. n cl 90 θcl 90 θ 2 1 Axis θcl 90 θcl θ2 θ 2 air cladding core n 1 90 θ 1 θ 1 We want a formula that will relate θ 2 to θ 1. Snell's law involves the angles that the rays make with respect to the normal (not to the axis), so we have for the refraction at the core-cladding boundary: n 1 sin 90 1 n cl cos cl. At the cladding-air boundary, we have n cl sin 90 cl n 1 cos 1 n cl sin 90 cl 1 sin 90 2 n cl cos cl cos. 2 Both expressions involve n cl cos. cl Equating them, we have n cos 1 1 cos 2, and thus 1 2 cos 1 n 1 cos 1 cos 1.4 cos Note that n 2 does not appear in the answer (although it does determine which rays will undergo total internal reflection). A) 35 B) 8 C) 23 D) 13 E) 40

8 17. In the graphs below, the dashed line shows the nonrelativistic momentum for a certain particle as a function of its velocity. (Since the nonrelativistic momentum is mv, the graph is a straight line.) In which graph does the solid line best represent the relativistic momentum for this particle? Solution: The relativistic momentum is p mv 1 v 2 c 2, which is very close to the Newtonian value for v<<c. But it increases more rapidly as v/c becomes significant, and tends to infinity as v approaches c. p A) B) p non relativistic v/c non relativistic v/c C) D) p p non relativistic v/c non relativistic v/c E) p non relativistic v/c

9 18. A standard 35mm slide measures 24.0 mm by 36.0 mm. Suppose a slide projector produces a 60.0 cm by 90.0 cm image of the slide on a screen. The focal length of the lens is 12.0 cm. How far from the lens is the screen? Solution: We know the focal length and the magnification, so use the magnification equation to eliminate d o from the thin lens equation: m 1 1 m. So d f d i d o d i d i d i f 1 m. You can obtain the i magnification either from the ratio of the heights or the ratio of the widths. But notice that the magnification here is negative, because the image is real (m=-d i/d o, where the distances are both positive). So here m=-(60 cm)/(2.4 cm) = -25. Thus d i m m 3.12 m. A) 2.04 m B) 3.78 m C) 3.00 m D) 3.12 m E) 4.64 m 19. A beam of monochromatic laser light passes through a diffraction grating with 500 lines/cm. When the light strikes a screen 12 m away, the first-order maximum is 33 cm away from the central bright spot. What is the wavelength of the light? Solution: For a diffraction grating, the angle to the first bright spot is given by sin d, where d is the spacing between the lines of the grating. (Here there are 50,000 lines/m so d= m.) In the small angle x approximation, which is valid here, sin, where x is the L spacing between the central bright spot and the first-order max, and L is the distance to the screen. So we have x d x m 0.33 m 550 nm. d L L 12 m A) 550 nm B) 533 nm C) 455 nm D) 600 nm E) 500 nm

10 20. The α particles used by Geiger and Marsden to study the nucleus had kinetic energy E and mass m. The speed of the particles was much less than c. What was the de Broglie wavelength of these particles? Solution: The de Broglie wavelength is h/p, where p is the momentum. We know the kinetic energy, E mv p 2 2 mv2 2 m 2 m. So p h 2 me, and 2 me. A) hc E B) h mc C) h 4 0 me D) he 2 m E) h 2 me Do not worry about your troubles with mathematics; I can assure you that mine are still greater. --Albert Einstein

1. Newton's Laws provide a good description of the flight of a baseball because:

1. Newton's Laws provide a good description of the flight of a baseball because: 1. Newton's Laws rovide a good descrition of the flight of a baseball because: A) Its seed is small coma to c and its size is large coma to atomic scales. B) Planck's constant is nonzero. C) The earth