1 USEFUL FORMULAE AND DATA 1. Wien s Law: pt = m K 2. v=c/n, is the speed of light in a material with an index of refraction n 3. Snell s Law: n1 sin 1 = n2 sin 2, where subscripts 1 stands for incidence 2 stands for refraction 4. Lens equation: 1/f = (1/do) + (1/di) f is the focal distance. do is the distance of an object point to the lens. di is the distance of the image point to the lens nm=10-9 m Light of 650 nm is incident on two slits. A maximum is seen at an angle of 4.10 and a minimum of What is the order m of the maximum and what is the distance d between the slits? Strategy Use Eqs. (25-10) and (25-11). dsin Solution The condition for the maxima is max m dsin and that for the minima is min ( m1 2). Divide the second equation by the first and solve for the order of the maximum, m. 1 m 1 1 min min min 1 sin 4.78 d sin sin sin 1, so m d sinmax sinmax m 2m 2 sin max 2 sin 4.10 The order of the maximum is Find d. m 3. sin 9 5 max, so csc max 3( m) csc d m d m m. 2. In a double-slit interference experiment, the wavelength is 475 nm, the slit separation is mm, and the screen is 36.8 cm away from the slits. What is the linear distance between adjacent maxima on the screen? Strategy Draw a diagram of the situation and use the double-slit maxima equation. Assume is small. Solution In the figure, a bright fringe of order m appears on the screen at a distance x from the central bright fringe. D, x, and the distance to the mth fringe from the slits form a right triangle. So, x Dtan. When is small (x << D), the approximation sin tan may be used, yielding x Dsin. Solving the double-slit maxima equation for m sin gives sin. Substituting this into the previous d equation gives D x m. So, the linear distance d
2 between adjacent maxima on the screen is Since D x D d 9 (0.368 m)( m) m 1.46 mm. 3. Light incident on a pair of slits produces an interference pattern on a screen 2.50 m from the slits. If the slit separation is cm and the distance between adjacent bright fringes in the pattern is cm, what is the wavelength of the light? Strategy Show that the small-angle approximation is justified. Then, use the result for the slit separation obtained in Example Solution Compare x and D. D 2.50 m 329, so x << D and the small angle approximation is justified. x m Find the wavelength of the light. 2 2 D dx ( m)( m) d, so 456 nm. x D 2.50 m 4. Light of wavelength 589 nm incident on a pair of slits produces an interference pattern on a distant screen in which the separation between adjacent bright fringes at the center of the pattern is cm. A second light source, when incident on the same pair of slits, produces an interference pattern on the same screen with a separation of cm between adjacent bright fringes at the center of the pattern. What is the wavelength of the second source? Strategy x << D (distant screen), so using the small angle approximation is justified. Then, from Example 25.4, d D x. So, for fixed D and d, x. Solution Form a proportion to find the wavelength of the second source. 2 x cm, so 2 (589 nm) 711 nm. x cm The central bright fringe in a single-slit diffraction pattern from light of wavelength 476 nm is 2.0 cm wide on a screen that is 1.05 m from the slit. (a) How wide is the slit? (b) How wide are the first two bright fringes on either side of the central bright fringe? (Define the width of a bright fringe as the linear distance from minimum to minimum.) (a)strategy The distance from the center of the diffraction pattern to the first minimum is half the width of the central bright fringe, or 1.0 cm. Since 1.0 cm << 1.05 m, assume that sin tan xd, where x 1.0 cm and D 1.05 m. Use Eq. (25-12). Solution Find the width of the slit. asin m (1) and sin tan xd, so 9 D (47610 m)(1.05 m) a x m mm. (b) Strategy and Solution For small angles, the minima are approximately evenly spaced. This spacing is half the width of the central bright fringe, or 1.0 cm. If we define the width of a bright fringe as the linear distance from minimum to minimum, then the width of the first two bright fringes on either side of the central bright fringe is 1.0 cm.
3 6. Light from a red laser passes through a single slit to form a diffraction pattern on a distant screen. If the width of the slit is increased by a factor of two, what happens to the width of the central maximum on the screen? Strategy Call the width of the central maximum W. Referring to Figure 25.33, we have W trigonometry, x Dtan, so W 2x 2Dtan. Assume is a small angle. Use Eq. (25-12). x, and by Solution Since is assumed to be small, we have W 2Dtan 2Dsin. From Eq. (25-12), we have sin since m W becomes a If a is replaced with 2a, the new width is Thus, the new width is half the old width. 2L W. a 2L 1 2L 1 Wnew W. 2a 2 a 2 7. Light of wavelength 490 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 3.20 m from the slit. The distance on the screen between the central maximum and the third minimum is 2.5 cm. What is the width of the slit? Strategy Since the distance between the third minimum and the center of the screen is much smaller than the slit-screen distance, the small angle approximation can be used. Refer to Figure 25.33, but let x be the third minimum to center distance. Use Eq. (25-12). Solution From the figure, tan xd. The single-slit diffraction minima are given by asin m. Since x << D, is small, so tan sin and ax D m. Calculate a from this and the given information. 9 md 3( m)(3.20 m) a x m 0.19 mm 8. Light rays from the Sun, which is at an angle of 35 above the western horizon, strike the still surface of a pond. (a) What is the angle of incidence of the Sun s rays on the pond? (b) What is the angle of reflection of the rays that leave the pond surface? (c) In what direction and at what angle from the pond surface are the reflected rays traveling? Strategy The normal is perpendicular to the surface of the pond. Use the laws of reflection. Solution (a) The angle of incidence is i (b) The angle of reflection is r i 55. (c) 55 from the normal is equivalent to 35 above the horizontal. The sun is above the western horizon. So, the reflected rays are traveling at an angle 35 above the surface of the pond to the east.
4 9. Light is traveling in glass (n=1.5) and then emerges into air. Which of the following is a possibly correct light-ray diagram for its emergence into air? A glass air B glass air 10. A beam of light in air is incident upon a stack of four flat transparent materials with indices of refraction 1.20, 1.40, 1.32, and If the angle of incidence for the beam on the first of the four materials is 60.0, what angle does the beam make with the normal when it emerges into the air after passing through the entire stack? Strategy Use Snell s law, Eq. (23-4). Let the index of refraction for air be n, and let n1, n2, n3, and n 4 be 1.20, 1.40, 1.32, and 1.28, respectively. Let be the angle of emergence. Solution Find the angle the beam makes with the normal when it emerges into the air after passing through the entire stack of four flat transparent materials. n sin 60.0 n sin n sin n sin n sin n sin, so At a marine animal park, Alison is looking through a glass window and watching dolphins swim underwater. If the dolphin is swimming directly toward her at 15 m/s, how fast does the dolphin appear to be moving? Strategy Due to refraction of the light coming from the dolphin, the speed of the dolphin appears to be less than the actual speed by a factor of 1 n. Solution The dolphin appears to be moving at a speed of 15 m s 11 m s A light ray in the core (n = 1.40) of a cylindrical optical fiber travels at an angle θ1 = 49.0 with respect to the axis of the fiber. A ray is transmitted through the cladding (n = 1.20) and into the air. What angle θ2 does the exiting ray make with the outside surface of the cladding? Strategy Draw a diagram. Use Snell s law, Eq. (23-4). Solution Find 2. Relate 1 and 3. n1 sin(90 1) n3 sin3 Relate 2 and 3. n sin(90 ) n sin
5 Eliminate n3sin 3 and solve for 2. n sin(90 ) n sin(90 ) n sin sin(90 1) n sin sin( ) Is there a critical angle for a light ray coming from a medium with an index of refraction 1.2 and incident on a medium that has an index of refraction 1.4? If so, what is the critical angle that allows total internal reflection in the first medium? Strategy Total internal reflection occurs when the angle of incidence is greater than or equal to the critical angle. Use Eqs. (23-5). Solution The critical angle is given by sin c nt ni. Since the sine of any angle is less than or equal to 1, the index of refraction of the media in which the transmitted rays travel must be less than or equal to that in which the incident rays travel. Since nt ni, the answer is no, there is no critical angle. 14. (a) Calculate the critical angle for a diamond surrounded by air. (b) Calculate the critical angle for a diamond under water. (c) Explain why a diamond sparkles less under water than in air. Strategy From Table 23.1, the index of refraction for diamond is 2.419, for air it is 1.000, and for water it is Total internal reflection occurs when the angle of incidence is greater than or equal to the critical angle. Use Eq. (23-5a). Solution (a) Calculate the critical angle for diamond surrounded by air. 1nt c sin sin n i (b) Calculate the critical angle for diamond under water. 1nt c sin sin n i (c) Under water, the larger critical angle means that fewer light rays are totally reflected at the bottom surfaces of the diamond. Thus, less light is reflected back toward the viewer. 15. What is the index of refraction of the core of an optical fiber if the cladding has n = 1.20 and the critical angle at the core-cladding boundary is 45.0? Strategy Use Eq. (23-5a). Solution Find the index of refraction of the core of the optical fiber. 1 nt nt nt 1.20 c sin, so sin c and ni n n sin sin 45.0 i i c
6 16. Suppose a star has a surface temperature of 32,500 K. What color would this star appear? You need to assume that the star emits radiation as a blackbody and solve for p in Wien s law. Use Wien s law to determine p. p T= mk = mk/32,500 K= m=89 nm This is in the region of UV, but our vision can perceive the bluish region close the blue-violet and UV. Therefore, the color is bluish. 17. Answer the following: a) What is the temperature if the peak of a blackbody spectrum is at 18.0 nm? Use the same approach as above. T= mk/ m= K b) What is the wavelength at the peak of a blackbody spectrum if the body is at a temperature of 2000 K? = mk/2,000 K= m 18. Answer the following: a) White light is shined on two filters, magenta and yellow. What is the color transmitted through both filters? b) White light is shined on two filters, cyan and yellow. What is the color transmitted through both filters? a) Magenta results from the addition of blue and red, whereas cyan results from the addition of green and blue. Yellow results from the addition of green and red. In other words, R+B=M G+B=C G+R=Y Therefore, the correct answer is red for (a) and green for (b). 19. Some glasses used for viewing 3D movies are polarized, one lens having a vertical transmission axis and the other horizontal. While standing in line on a winter afternoon for a 3D movie and looking through his glasses at the road surface, Maurice notices that the left lens cuts down reflected glare significantly, but the right lens does not. The glare is minimized when the angle between the reflected light and the horizontal direction is 37. (a) Which lens has the transmission axis in the vertical direction? (b) What is Brewster s angle for this case? (c) What is the index of refraction of the material reflecting the light? (a)strategy The glare is due to sunlight reflected from the horizontal road, so the reflected light is polarized mostly in a direction parallel to the ground, or horizontal. Since the light is horizontally polarized, the lens with the horizontal transmission axis lets through the most light; that is, the reflected glare is not significantly reduced. The lens with the vertical transmission axis cuts down reflected glare significantly.
7 Solution The left lens cuts down reflected glare significantly, so it must have the vertical transmission axis. (b)strategy When a ray is incident on a boundary, the angle of incidence for which the reflected ray is totally polarized perpendicular to the plane of incidence is Brewster s angle. Solution Since the glare is minimum when the angle between the reflected light and the horizontal direction is 37, Brewster s angle is (c)strategy Use Eq. (23-6). Solution Find the index of refraction n t of the material reflecting the light. n tan, so tan tan t B nt ni B ni 20. Answer the following: (a) Sunlight reflected from the still surface of a lake is totally polarized when the incident light is at what angle with respect to the horizontal? (b) In what direction is the reflected light polarized? (c) Is any light incident at this angle transmitted into the water? If so, at what angle below the horizontal does the transmitted light travel? (a) Strategy The reflected light is totally polarized when the angle of incidence equals Brewster s angle. Use Eq. (23-6). Solution Compute Brewster s angle. 1nt B tan tan ni The angle with respect to the horizontal is the complement of this angle (b)strategy and Solution For Brewster s angle, the reflected light is polarized perpendicular to the plane of incidence. (c) Strategy When the angle of incidence is Brewster s angle, the incident and transmitted rays are complementary. Solution Find the angle of transmission. t 90i The angle below the horizontal is the complement of this angle An object is placed in front of a concave mirror with a 25.0-cm radius of curvature. A real image twice the size of the object is formed. At what distance is the object from the mirror? Draw a ray diagram to illustrate.
8 Strategy Since the image is real, the image distance is positive and the image is inverted (negative). The image is twice the size of the object, so h 2. h Use the magnification and mirror equations. Solution Relate the object and image locations. h 2h q, so q 2 p. h h p Find the distance of the object from the mirror R 3, so p f (25.0 cm) 18.8 cm. p q p 2p 2p f The object is 18.8 cm in front of the mirror. 22. Answer the following: (a) For a converging lens with a focal length of 3.50 cm, find the object distance that will result in an inverted image with an image distance of 5.00 cm. Use a ray diagram to verify your calculations. (b) Is the image real or virtual? (c) What is the magnification? (a)strategy Use the thin lens equation to find the object distance. Then, draw the ray diagram. Solution Find the object distance qf (5.00 cm)(3.50 cm), so p 11.7 cm. p q f q f 5.00 cm 3.50 cm The diagram is shown. (b)strategy and Solution Light rays actually pass through the image location, so the image is real. (c) Strategy Use the magnification equation. Solution Compute the magnification of the image. q 5.00 cm m p cm
9 23. When an object is placed 6.0 cm in front of a converging lens, a virtual image is formed 9.0 cm from the lens. What is the focal length of the lens? Strategy The object distance is p 6.0 m. Since the image is virtual, the image distance is q 9.0 cm. Use the thin lens equation. Solution Find the focal length of the lens pq (6.0 cm)( 9.0 cm), so f 18 cm. p q f p q 6.0 cm 9.0 cm 24. A diverging lens has a focal length of 8.00 cm. (a) What are the image distances for objects placed at these distances from the lens: 5.00 cm, 8.00 cm, 14.0 cm, 16.0 cm, 20.0 cm? In each case, describe the image as real or virtual, upright or inverted, and enlarged or diminished in size. (b) If the object is 4.00 cm high, what is the height of the image for the object distances of 5.00 cm and 20.0 cm? Strategy All images are virtual since a diverging lens is used. The magnification determines the size and orientation of the image. Use the thin lens and magnification equations. Solution (a)solve the thin lens equation for q , so q 1. p q f 1 1 f p Substitute the given values. 1 1 q 3.08 cm for p 5.00 cm. q 4.00 cm for p 8.00 cm cm 5.00 cm 8.00 cm 8.00 cm For p-values of 14.0 cm, 16.0 cm, and 20.0 cm, the corresponding q values are 5.09 cm, 5.33 cm, and 5.71 cm, respectively. The results are summarized in the table. p q (cm) m q p Real or Orientation Relative (cm) virtual size virtual upright diminished virtual upright diminished virtual upright diminished virtual upright diminished virtual upright diminished (b) Solving the magnification equation for the image height yields h mh. For p 5.00 cm, h 4.00 cm, and m , we have h cm 2.46 m. For p 20.0 cm, h 4.00 cm, and m 0.286, we have h cm 1.14 cm.
10 25. What is the energy of a photon of light of wavelength 0.70 μm? Strategy The energy of a photon of EM radiation with frequency f is E hf. The frequency and wavelength are related by f c. Solution Compute the energy. hc 1240 ev nm E hf ev nm 26. A rubidium surface has a work function of 2.16 ev. (a) What is the maximum kinetic energy of ejected electrons if the incident radiation is of wavelength 413 nm? (b) What is the threshold wavelength for this surface? (a)strategy Use Einstein s photoelectric equation. Solution Calculate the maximum kinetic energy. hc 1240 ev nm max 2.16 ev 0.84 ev 413 nm K hf (b)strategy The threshold wavelength is related to the threshold frequency by 0 f 0 c. Use Eq.(27-8). Solution Find the threshold wavelength. hc hc 1240 ev nm hf0, so nm ev Answer the following: (a) Light of wavelength 300 nm is incident upon a metal that has a work function of 1.4 ev. What is the maximum speed of the emitted electrons? (b) Repeat part (a) for light of wavelength 800 nm incident upon a metal that has a work function of 1.6 ev. (c) How would your answers to parts (a) and (b) vary if the light intensity were doubled? Strategy Assume the electrons are nonrelativistic. Use Einstein s photoelectric equation and f c. Solution (a) Find the maximum speed of the emitted electrons. hc 1 2 Kmax hf mv, so hc 2( J ev) 1240 ev nm 1.4 ev m s. 31 v m kg 300 nm (b) Find the maximum speed of the emitted electrons. hc 1240 ev nm Kmax hf 1.6 ev 0.05 ev 800 nm Kmax 0, which is impossible, so no electrons are ejected from the metal. (c) Increasing the intensity of the light increases the number of electrons ejected from the metal, but does not increase their kinetic energy, since the photon energy remains the same. Therefore, doubling the intensity has no effect on the electrons speed.
11 28. How much energy must be supplied to a hydrogen atom to cause a transition from the ground state to the n = 4 state? Strategy The amount of energy required to cause a transition from the ground state to the n 4 state is equal to the difference in the energy between the two states. The energy for a hydrogen atom in the nth 2 stationary state is given by En ( 13.6 ev) n. E Solution The energy of a hydrogen atom in the n 4 state is ev E ev The energy that must be supplied is E4 E ev ( 13.6 ev) 12.8 ev. 29. How much energy is required to ionize a hydrogen atom initially in the n = 2 state? Strategy The minimum energy for an ionized atom is E ionized 0. Use Eq. (27-24). Solution The energy needed to ionize a hydrogen atom initially in the n 2 state is E ev E Eionized E2 Eionized ev Many lasers, including the helium-neon, can produce beams at more than one wavelength. Photons can stimulate emission and cause transitions between the eV metastable state and several different states of lower energy. One such state is ev above the ground state. What is the wavelength for this transition? If only these photons leave the laser to form the beam, what color is the beam? Strategy The wavelength of a photon is related to its energy by E hc. hc 1240 ev nm Solution The wavelength for this transition is E ev ev Light of wavelength 544 nm appears green in color. 544 nm.