Copyright 2012 Nelson Education Ltd. Unit 4: The Wave Nature of Light U4-10

Transcription

1 Unit 4 Review, pages Knowlege. (b). (a) 3. () 4. (c) 5. () 6. (b) 7. (c) 8. (a) 9. (a) 0. (a). (a). (c) 3. (b) 4. () 5. (a) 6. (b) 7. (c) 8. (a) 9. False. Raio waves go through fewer cycles per secon than X-rays. 0. False. If a group of waves travels at the same spee, light waves with higher frequencies have shorter wavelengths, an light waves with lower frequencies have longer wavelengths.. False. In a ripple tank, a ecrease in the frequency of the two vibrating sources causes a ecrease in the number of noal lines.. True 3. False. The wave moel of light can explain all wave properties of light. 4. True 5. True 6. True 7. True 8. False. For a iffraction grating, the angular separation of the maxima is generally large because the slit spacing is so small. 9. True 30. The law of reflection states that a wave is reflecte from a surface such that the angle of incience equals the angle of reflection. 3. Spee an wavelength etermine the frequency of a wave. 3. In materials in which the inex of refraction is high, the spee of a wave ecreases. 33. A noe is a point in a staning wave where the isplacement is always zero. An antinoe is a point in a staning wave halfway between two noes at which the largest isplacement, or amplitue, occurs. A noal line is a line or curve along which estructive interference results in zero isplacement, that is, where a series of crests an troughs meet resulting in zero amplitue. 34. The cm opening will prouce the largest iffraction because it is the smallest opening compare to the wavelength of the water wave. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-0

2 35. When a wave encounters a barrier that has an opening that is similar in size to the wavelength, the part of the wave that passes through the opening will iffract into the region beyon the barrier. 36. The wavelength an the slit with are the quantities that etermine the amount of iffraction of a wave ue to a single slit. 37. When the source separation in a ripple tank increases, the noal lines become more tightly space. 38. Accoring to Newton s particle moel, the spee of the light particles increase as they entere a meium with a greater inex of refraction. As the particles move faster, their path move closer to the normal in the secon meium. 39. I think that the wave moel provies a better explanation of polarization. Polarize light consists of transverse isturbances that can be aligne with the polarization axis of a polarizing substance. Particles cannot account for such alignments, but waves can. 40. Huygens s principle states that each point on a wave front is the source of a spherical wavelet that spreas out at the wave s spee. 4. Early attempts to emonstrate interference of light were ifficult because the sources of light that were use were out of phase. In aition, it was ifficult to prouce two sources of monochromatic light with exactly the same frequency. 4. To solve the problem of phase ifference in the light sources, Young use a single source (sunlight) for his experiment. As a result, changes in phase were uniform, allowing him to observe interference. 43. The wave theory coul not explain how light travelle through the vacuum of space without a meium. 44. If the two slits in Young s experiment ha a smaller separation, the fringes in the interference pattern woul be more sprea out. 45. The frequency of a light wave is the same in a vacuum as it is in a piece of glass. 46. To form Newton s rings, make an air wege with two pieces of glass, an shine light on the wege. The change in the phase of the light as it reflects between the glass piece of the air wege, along with the graually increasing path length through the air, causes estructive interference between light waves, thus forming the ark rings. 47. If I increase the wavelength of light passing through the aperture of an optical instrument, then the instrument s resolution ecreases. 48. In single-slit iffraction, as the slit with ecreases, the istance between maxima an minima increases. 49. The more slits there are in a iffraction grating, the greater the sprea, or ispersion, of the wavelengths of light. 50. The spee of light rays of ifferent wavelengths in a vacuum is same. Therefore, the ratio of their spees is :. 5. Three methos of polarizing light are polarization by reflection, polarization by absorption, an polarization by scattering. 5. The optical activity of molecules in a solution changes the orientation of the electric wave in light. By passing polarize light through the solution, the polarization of the light rotates through a certain angle. By turning the secon polarizer until the light passing through the solution has maximum intensity, the angle of optical activity can be etermine an measure. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-

3 53. Brewster s angle is the angle of incience at which the irection of a reflecte portion of a light wave is perpenicular to the irection of the refracte portion of the wave. If unpolarize light is incient at Brewster s angle, the reflecte light is completely polarize. Unerstaning 54. Answer may vary. Sample answer will inclue some of the following information. Both light waves an water waves are transverse an reflect from a barrier, refract when they change spees, iffract aroun barriers an through openings, an interfere constructively an estructively. They iffer in that there are two transverse wave components to light (oscillating electric an magnetic fiels), an water waves require a meium (water), whereas light oes not. Light can be polarize, but water waves cannot be polarize. 55. Accoring to Snell s law, the ratio of the inex of refraction of the incient meium to the inex of refraction of the refracting meium equals the ratio of the sines of the angle of refraction to the angle of incience: n i n R sin! R sin! i. For total internal reflection, sinθ R, so n i n R sin! i. In aition, sinθ i, so n i n R >. This can only be true if n i is greater than n R. 56. After passing through a flat plane of glass, all wavelengths of light emerge in the same irection of propagation. A prism s triangular shape means that ifferent wavelengths emerge in ifferent irections, proucing ifferent colours. 57. (a) At a point where constructive interference takes place, the waves will combine to form waves with amplitues greater than the amplitue of each iniviual wave. The boat will therefore rise higher. (b) At a point where estructive interference takes place, the waves will combine to form waves with amplitues that are smaller than the amplitue of each iniviual wave. The boat will therefore rop after estructive interference. If the estructively interfering waves are of equal amplitue, the boat will remain vertically stationary on the water. 58. We can hear soun aroun a corner because the soun waves iffract aroun the corner. We cannot see light aroun a corner because light has a much smaller wavelength than soun, so it iffracts much less. 59. Newton assume that corpuscles of light move faster in a meium with a higher inex of refraction. This was consistent with the behaviour of particles that move closer to the normal at higher spees. Later, it was emonstrate that light actually moves slower in meia with a higher inex of refraction. 60. (a) Three properties of light accounte for by both moels are rectilinear propagation, reflection, an refraction. (b) Three properties of light not accounte for by the particle moel are interference, iffraction, an polarization. (c) Light oes not nee a meium in which to travel. 6. Raio waves have much longer wavelengths than visible light, so it woul be easier to create two slits for interference an to measure the separations of the fringes. Only the invisibility of the waves woul make the emonstration challenging. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-

4 6. Sample answer: The observation of ouble-slit interference was more convincing as evience for the wave theory of light because particle theory ha an explanation for iffraction in terms of particle collisions, but was unable to explain ouble-slit interference. 63. Incanescent light bulbs o not prouce light waves that are in phase or at the same frequency over a perio of time. As a result, no interference pattern will exist for sufficient time to allow the human eye to observe. 64. For a thin film covering a piece of glass, the light rays reflect at both the air film bounary an the film glass bounary, so both reflections change phase. They remain in phase after reflecting, an constructive interference happens when twice the thickness of the film is equal to whole-number multiples of the wavelength of light use. For a thin film surroune by air, the light rays change phase when they reflect from the air film bounary, but not when they reflect at the film air bounary insie the film. As a result, if twice the thickness of the film equals whole-number multiples of the wavelength of light use, then the light rays will be out of phase an will interfere estructively. For constructive interference to take place in this situation, twice the thickness of the film must be a half wavelength longer. Therefore, the equation for constructive interference in one situation is ientical to the equation for estructive interference in the other. 65. At the central maximum, the amount of estructive interference that takes place between pairs of out-of-phase wave fronts is at a minimum. At the first-orer maximum, more of the iniviual wave fronts interfere estructively in pairs, so the overall intensity of the recombine light is much less than at the central maximum. 66. The ratio is! w. 67. Sample answer: Resolution is the ability of an optical evice to separate two images. Diffraction ecreases resolution. By using a larger aperture, iffraction is reuce an resolution is improve. 68. Sample answer: A iffraction grating allows a more precise measurement of the iffraction of ifferent wavelengths than a prism. Increasing the number of slits gives a sharper peak, an each colour has multiple peaks that allow for multiple measurements. 69. If I immerse the apparatus for the investigation in water, the wavelength of the light ecreases. As a result, the fringes will be space closer together. 70. Polarization is evience that light can be oriente in one irection. In the wave theory, this is the irection of oscillation of the electromagnetic waves. Polarization experiments can also emonstrate the superposition of ifferent waves. 7. (a) Intensity is greatest when the polarizer axis is parallel to the electric fiel of the light, allowing the light to pass through most easily. So at 0, the intensity is at its maximum. (b) Intensity is smallest when the polarizer axis is perpenicular to the electric fiel of the light, allowing little to no light to pass through. So at 90, the intensity is zero, which is the minimum. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-3

5 (c) Malus s law states that the change in intensity epens on the square of the cosine of the angle between the axis of the polarizer an the electric fiel of the light wave: I out cos!i in. This prouces a maximum value at 0 an zero intensity at Answers may vary. Sample answer: Analysis an Application 73. Given:! 5 ; n air.0003; n re.459; n blue.467 Require: Δθ, the ifference between θ re an θ blue Analysis: Rearrange the equation nsinθ nsinθ to calculate θ for both the re an the blue light; sin! n sin! n Solution: Re light:. Then subtract the ifference. Blue light: sin! n sin! n n! sin sin! & % ( $ ' n (.0003)sin 5 & sin $ %.459 ' ( sin! n sin! n n! sin sin! & % ( $ '! re (two extra igits carrie)! blue (two extra igits carrie) The ifference in the angles of refraction is Δ θ Δ θ 0. Statement: The ifference in the angles of refraction is Given: θ 35.0 ; n.0003; n.50 Require: θ n (.0003)sin 5 & sin $ %.467 ' ( Analysis: Rearrange the equation nsinθ nsinθ to solve for θ ; sin! n sin!. n Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-4

6 Solution: sin! n sin! n n! sin sin! & % ( $ ' n (.0003)sin35.0 & sin $ %.50 ' (!.5 Statement: The angle of refraction of the light through the glass is Given: θ 5.0 ; n.0003; n.33 Require: θ Analysis: Rearrange the equation nsinθ nsinθ to solve for θ ; sin! n sin!. n Solution: sin! n sin! n n! sin sin! & % ( $ ' n (.0003)sin5.0 & sin $ %.33 ' (! 8.5 Statement: The angle of refraction of the water is Given: θ.5 ; n.0003; n.33 Require: θ Analysis: Rearrange the equation nsinθ nsinθ to solve for θ ; sin! n sin!. n Solution: sin! n sin! n n! sin sin! & % ( $ ' n (.0003)sin.5 & sin $ %.33 ' (! 6.7 Statement: The angle of refraction is 6.7. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-5

7 77. Given: niamon.4; nair.0003 Require: θ c, critical angle for iamon air bounary n Analysis: θc sin n Solution: n air θc sin niamon.0003 sin.4 θ 4.4 c Statement: The critical angle for a iamon air bounary is Given: θ c 4.4 ; nair.0003 Require: n acrylic n Analysis: Rearrange the equation sinθ c to solve for inex of refraction; n n n. sin! c Solution: n n sin! c.0003 sin 4.4 n.48 Statement: The inex of refraction of acrylic is Given: n.0003; n.60;! 485 nm Require: λ n λ Analysis: Rearrange the equation to solve for wavelength;! n λ n!. n Solution:! n! n (.0003)(485 nm).60! 303 nm Statement: The wavelength of cyan light in the flint glass is 303 nm. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-6

8 80. Given: two-source interference; λ.4 m; 8.6 m; n 3 Require: θ 3 for estructive interference Analysis: sinθn n λ;! n sin n & $ % ' ( ) & $ % ' ( Solution:! n sin n & $ % ' ( ) & $ % ' ( sin 3 & $ % ' (.4 m & $ % 8.6 m ' (! 3 4 Statement: The angle of the thir noal line is (a) Given: two-source interference; P S 3.0 cm m; P S 5.5 cm 0 m; n Require: λ Analysis: PS PS n n n λ ;! P S P S n n n Solution:! P n S P n S n m m m (two extra igits carrie)! 0.07 m Statement: The wavelength of the waves is 0.07 m. (b) Given: v 7.5 m/s; λ m Require: f Analysis: Rearrange the universal wave equation, v Solution: f v! 7.5 m /s m f 450 Hz Statement: The frequency of the waves is 450 Hz. fλ, to solve for frequency; f v! Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-7

9 8. Given: ouble-slit interference; x 7 x 5.8 cm m; L 3.50 m;.!0 4 m Require: λ Analysis: The istance between the bright fringes correspons to 6 fringe separations: x7 x 6Δ x. Determine Δ x, an then rearrange the equation wavelength;! x L Solution: x 7! x 6x x x 7! x ! m 6 x !3 m (two extra igits carrie)! x L Lλ Δ x to solve for (. 0$4 m)( $3 m) 3.50 m 6.0 $7 m! 60 nm Statement: The wavelength of the source is 60 nm. This wavelength correspons to re light. 83. Given: ouble-slit interference; n ark fringe;! 5. Require: λ Analysis: Rearrange the equation Solution: n! sin n sinθn n λ to etermine λ ; n!. sin n sin5.! 7 Statement: The ratio of the slit separation to the wavelength is 7:. 84. Given: ouble-slit interference; 0.08 mm m; n 5 ark fringe;! 5 8. Require: λ Analysis: sinθn n λ;! sin n n Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-8

10 Solution:! sin n n (.8 $05 m)sin $0 7 m! 570 nm Statement: The wavelength of the light is 570 nm. 85. Given: ouble-slit interference; λ 638 nm m; m 3 bright fringe;! Require: Analysis: sinθ m mλ; m! sin m Solution: m! sin m (3)(6.38 0$7 m) sin $5 m Statement: The slit separation is m. 86. Given: ouble-slit interference; λ 633 nm m; 0.00 mm m; L.0 m Require: x, n ark fringe position Analysis: x n n! % $ & ' L( ; n Solution: x n n! % $ & ' L(! % $ & ' (.0 m)(6.33)0!7 m) x.00 )0!4 m x 6.65)0!3 m Statement: The first-orer ark fringe occurs at 6.65! 0 3 m from the central maximum. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-9

11 87. (a) Given: ouble-slit interference; 0.04 mm m; L 4.00 m;!x 5.5 cm 5.50 m Require: λ Lλ Analysis: Rearrange the equation Δ x to etermine the wavelength;! x L Solution:! x L (4. 0$5 m)(5.50 $ m ) (4.00 m ) $7 m (two extra igits carrie)! 580 nm Statement: The wavelength of the light is 580 nm. 8 7 (b) Given: c m/s; λ m Require: f Analysis: c fλ ; f c! Solution: f c! m s m f Hz Statement: The frequency of the light is 5.! 0 4 Hz. 88. Given: ouble-slit interference; λ 639 nm m; L.80 m; n ; mm 4.8! 0 5 m Require: x, n ark fringe position Lλ Analysis: xn n Solution: x n n! % L( $ & '! % $ & ' (.80 m)(6.39 ) 0!7 m ) 4.8 ) 0!5 m x.9 ) 0! m Statement: The first ark fringe occurs at.9! 0 m from the central maximum. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-0

12 89. Given: ouble-slit interference; 0.5 mm m; L.0 m; x! x.0 cm.0 0! m Require: λ Analysis: The istance between the fringes correspons to 0 fringe separations: x x 0Δ x. Determine Δ x an rearrange the equation wavelength;! x L. Solution: x! x 0x x x! x 0.0 0! m 0 x 0!3 m! x L Lλ Δ x to solve for (5 0$4 m)( 0 $3 m ) (.0 m ) 5 0 $7 m! 500 nm Statement: The wavelength of the light use is 500 nm. 90. Given: ouble-slit interference; λ 480 nm m; L.0 m; 3.0 mm 3.0! 0 3 m Require: separation between m 8 bright fringe an n 3 ark fringe mlλ Lλ Analysis: Use the equations xm with m 8 an xn n with n 3 to locate the two fringe positions. Then calculate the ifference of these positions to etermine the require separation. Solution: x m ml! x n n! % $ & ' L( 8(.0 m)( m ) 3! % m $ & ' (.0 m)(4.8 )0!7 m ) x m (one extra igit carrie) 3.0 )0!3 m x )0!4 m The separation is 4 4 x8 x m m m Statement: The separation between the thir ark fringe an the eighth bright fringe is.8!0 3 m. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-

13 9. Given: ouble-slit interference; λ m;.0 mm m; L.0 m Require: separation between m 5 bright fringe an n 3 ark fringe mlλ Lλ Analysis: Use the equations xm with m 5 an xn n with n 3 to locate the two fringe positions. Then calculate the ifference of these positions to etermine the require separation. Solution: x m ml! (5)(.0 m)(6.507 m ) m x m (one extra igit carrie) x n n! % L( $ & ' 3! % $ & ' (.0 m)(6.5 ) 0!7 m ).0 ) 0!3 m x 3.65 ) 0!3 m (two extra igits carrie) Calculate the separation: 3 3 x x m.65 0 m m Statement: The separation between the thir ark fringe an the fifth bright fringe is m. 9. Given: ouble-slit interference; λ 656 nm m; L.50 m; m 4 bright fringe; x mm 4.80! 0 m Require: mlλ Analysis: Rearrange the equation xm to etermine the slit separation; ml!. x m Solution: ml! x m (4)(.50 m)( m ) m m 5 Statement: The slit separation is m. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-

14 93. Given: ouble-slit interference uner water; λ 465 nm m; 5.00! 0 4 m; L 50.0 cm m; n.0003; n.33 Require: Δ x n λ Analysis: ;! n λ n! Lλ ; Δ x n Solution:! n! n (.0003)( m) m (two extra igits carrie)!x L 4 Statement: The fringe separation istance is m. 94. Given: n oil.40;! 40 nm nm (0.500 m)( $7 m ) $4 m!x $4 m Require: t Analysis: Use the formula for constructive interference of two waves when phase! m + $ % & '! m + $ % & ' changes occur in only one reflection. Use m 0; t ; t. n oil n oil! m + $ % & ' Solution: t n oil! 0 + $ % & ' n oil (0.5)(4.0 (0)7 m).40 ( ) t 7.3(0 )8 m 8 Statement: The minimum thickness of the oil slick is m. 95. Given: n coating.38;! 58 nm nm Require: t Analysis: Use the formula for estructive interference of two waves when phase changes! m + λ m + $ light occur at both reflections an m 0; t % & ' ; t n coating n coating Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-3

15 ! m + $ % & ' Solution: t n coating! 0 + $ % & (5.8 (0)7 m).38 ( ) t.05(0 )7 m Statement: The minimum thickness of the coating is.05! 0 7 m. 96. Given: n coating.6;! 587 nm m Require: t Analysis: The light unergoes a phase change at the air coating bounary interface but not at the coating glass bounary. We o not want to see the light, so use the formula for estructive interference of the two waves an use n an n ; t n! n! ; t n coating n coating Solution: For n : For n : n! t t ()! n coating n coating ()( m) (.6) t.80 7 m ()( m) (.6) t m Statement: The two smallest possible non-zero values for the coating thickness are.8!0 7 m an 3.6!0 7 m Given: λ 566 nm m; t nm. 0 m Require: n coating Analysis: Assume that the light unergoes a phase change at both the air coating bounary an the coating glass bounary. Since we o not want to see the light, use the formula for estructive interference of the two waves an m 0;! m + $ % & '! m + $ % & ' t ; n n coating coating t Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-4

16 ! m + $ % & ' Solution: n coating t! 0 + $ % & 5.66 (0)7 m.(0 )7 m n coating.7 ( ) Statement: The inex of refraction of the coating is.7. Note that our assumption about phase change is correct. 98. Given: n coating.36;! 54 nm m Require: t Analysis: The light unergoes a phase change at both the air coating bounary an the coating glass bounary. We see the green light, so use the formula for constructive interference of the two waves an n ; t n! light n coating ; t n! light n coating. Solution: t n! n coating ()( m).36 ( ) t m 7 Statement: The thickness of the alcohol film is.93 0 m. 99. Given: n soap.33; t 09 nm.09!0 7 m Require:! Analysis: The light unergoes a phase change at the air soap bounary but not at the soap air bounary. We want to see the light, so use the formula for constructive! m + $ % & ' interference; t ;! tn soap n soap m +. To etermine the longest wavelength, % $ & ' choose the smallest value for m: m 0. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-5

17 tn soap Solution:! m + % $ & ' ( ) (0)7 m 0 + % $ & ' ( )! 5.80 (0 nm Statement: The wavelength of the light that is most constructively reflecte is nm. 00. Given:! 68 nm m; there are 36 cycles of alternating light patterns in L Require: t Analysis: Δ x L ;!x L 36 t ; t L! x Solution: t L! x L ( 6.8 0$7 m) % L ( ' & 36 * ) t.0 $5 m Statement: The thickness of the plastic sheet is.! 0 5 m. 0. Given:! 59 nm m; t m Require: n, the number of bright fringes Analysis: n L L ;!x!x t n L!x L L t n t Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-6

18 Solution: n t! ( ) m m n 88 Statement: The number of bright fringes seen along the wege is Given:! 53 nm m; L 5.4 cm.54 0 m; Δx.4 mm m Require: t Analysis:!x L t t L!x t L!x Solution: t L! x (.54 0$ m)(5.3 0 $7 m ) ()(.4 0 $3 m ) t.9 0 $5 m Statement: The thickness of the metal foil is.9! 0 5 m. 03. (a) Sample answer: The large ark space in the soap film at the top is uniform in thickness. It is also the right thickness to cause estructive interference for reflecte yellow light. (b) As time goes on, the film will get thinner at the top an thicker towar the bottom as gravity pulls the soap own. I expect to see bans of yellow an ark rifting ownwar. (c) The two ajacent bright bans correspon to using m 0 an m in the! m + $ % & '! m + $ % & ' constructive interference formula, t ; t. n soap n soap Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-7

19 Subtract t at m 0 from t at m :! + $ % & '! 0 + $ % & '! 3 ( $ % & ' ( n soap n soap n soap! + $ % & '! 0 + $ % & ' ( ' n soap n soap n soap The ifference in thickness of the soap film in the first two bans is () Given: n soap.33; n air.00;! 587 nm m λ n Require: t Analysis: Assume the light source is between the loop an your eye. The light unergoes a phase change at the air soap bounary but not at the soap air bounary. To have a ark ban, use the formula for estructive interference of the two waves. The lowest ark ban is the thir ark ban, so n 3. nλ t n soap nλ t n soap Solution: t n! n soap (3)( m) (.33) t m Statement: The thickness of the soap film in the lowest ark ban is 6.6! 0 7 m. 04. Given:! 45 nm 4.5! 0 7 m; w 0.35 mm 3.5! 0 4 m; L.0 m Require: Δy Analysis: The istance between ajacent minima is Δy ; multiply Δy by. wδy λ L Lλ Δ y w soap. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-8

20 Solution:!y L w ( ) 4.50 $7 m.0 m!y $3 m ( ) 3.50 $4 m 3 Statement: The with of the central maximum is m. 05. Given: w 5.60!0-4 m; L 3.00 m; y 3.5 mm 3.5!0 3 m Require: λ Δ Analysis: λ w y L Solution:! wy L ( ) 3.50 $3 m $4 m $7 m 3.00 m ( )! 650 nm Statement: The wavelength of the light is 650 nm (a) Given: w.08 0 m; λ 589 nm m Require: θ Analysis: sin! n n w $ n '! n sin % & w ( ) $ n ' Solution:! n sin % & w ( ) ( ) $ () 5.89 *0 7 m! sin & % &.08 *0 6 m! 33.0 ' ) ( ) Statement: The angle of the first minimum is (b) By examining the calculation in part (a), we see that replacing n by n or higher leas to a value for sinθ that is greater than.0. Therefore, there is no secon minimum Given: λ.5 0 m; θ4 8.4 ; n 4 Require: w Analysis: sin! n n w ; w n! sin n Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-9

21 Solution: w n! sin n (4)(.50$6 m) sin8.4 w 3.0 $5 m Statement: The with of the slit is 3.! 0 5 m Given: λ 639 nm m; w 4. 0 m; L 3.50 m Require: Δy λ Analysis: Δ y L w Solution:!y L w ( ) ( ) ( ) 3.50 m $7 m 4. 0 $4 m!y 5.30 $3 m 3 Statement: The separation between the maxima is m. 09. Sample answer: X-rays have very high energy an will pass through any mirror in a traitional telescope. This problem is avoie by having the rays come in at an angle so that they bounce off an initial mirror an then bounce of more mirrors, after which they come to a focus. Telescopes are esigne so that each mirror is most efficient at reflecting a particular range of X-ray wavelengths. A large aperture is not necessary for this process. 0. Sample answer: The spacing between fringes became smaller. This coul happen if the wavelength became smaller, if the slit with became bigger, or if the screen istance were shortene. 0. (a) Given: f.5 GHz.5 0 Hz; w.9 cm.9 0 m; L 0.45 m Require: length of B; length of C Analysis: To preict the values for B an C, I will calculate them. If I ha access to the equipment, I coul check my preicte answers by oing the experiment an measuring the results. Rearrange the universal wave equation, c λf, to calculate wavelength; c λ λ. Then use Δ y L to calculate B. To calculate C, use trigonometry C. f w Solution:!y L w (.3330$ m) ( 0.45 m ).9 0 $ m m (two extra igits carrie)!y 0.3 m Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-30

22 A, B, an C form a right-angle triangle. Use the Pythagorean theorem: ( ) + ( m) C 0.45 m C 0.55 m Statement: The values for B an C are 0.3 m an 0.55 m, respectively.. Given: w 0.0 cm.00! 0 m; 36 Require: λ Analysis: Use the formula for minima an n ; sin! n n w ;! wsin n n Solution:! wsin n n (.00 0$ m)sin $ m! cm Statement: The wavelength of the microwaves is cm. 3. (a) By making the aperture of a telescope larger, we can reuce the amount of iffraction. (b) An Airy isk consists of a bright central spot surroune by concentric rings of light. (c) A reflecting telescope uses multiple mirrors to form an image. A refracting telescope uses lenses with no obstructions (mirrors) to form an image. With no obstructions in its optical path, a refracting telescope shoul give the clearest image. 4. For a iffraction grating to isplay only first-orer maxima, it must have a wie ispersion an cause the secon-orer maxima to appear at an angle greater than 90. For any given wavelength, the ispersion of a iffraction grating is inversely proportional to the separation of the lines, so by aing more lines per centimetre to the grating, the ispersion is increase. 5. Given: w. 0 6 m; L.96 m; λ 4 nm m; λ 664 nm m; m Require: Δy ; θ Analysis: Rearrange the formula m! wsin m to solve for θ for both wavelengths; sin! m m ; then use trigonometry to calculate the with of the light spectrum. w Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-3

23 Solution: For λ : For λ : sin! m m w () () ( 4.0$7 m ) ( m ) sin! sin!.0 $6.0 6 m m!!.9 (two extra igits carrie) 8.43 (two extra igits carrie) By trigonometry, the positions of the colour bans are (.96 m)sin m an (.96 m)sin m. Statement: For the 4 nm wavelength, the angle an position are an 0.58 m; for the 664 nm wavelength, the angle an position are 8 an 0.94 m. 6. Given: N 500 lines/cm; θ 33 ; m Require: λ Analysis: Use the equation w to calculate the slit separation. Then rearrange the N equation mλ w sinθ to calculate the wavelength;! wsin m m m. Solution: w N 500 lines/cm.96!0 4 cm w.96!0 6 m! wsin m m (.96 0$5 m)sin33! 530 nm Statement: The wavelength of the light is 530 nm. 7. Given: N 8600 lines/cm; θ 6.6 ; θ 4. Require: λ, λ Analysis: Use w to calculate the slit separation. Rearrange the equation N mλ w sinθ to calculate the wavelengths;! wsin m m m. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-3

24 Solution: w N 8600 lines/cm.6!0 4 cm w.6!0 6 m For m an θ 6.6 : For m an θ 4. :! wsin m m (.6 0$6 m)sin6.6! 5. 0 $7 m! wsin m m (.6 0$6 m)sin4.! $7 m Statement: The wavelengths of the observe light are m an m. 8. (a) (b) The smallest possible value of sinθ is the number of lines per metre times the wavelength, an this value must be less than to prouce a maximum. Thus, the maximum wavelength is the inverse of the number of lines per metre. At lines/cm, only wavelengths less than m, or 380 nm, will prouce a maximum, an light with wavelengths in this range are not in the visible spectrum. Evaluation 9. Answers may vary. Sample answer: Water waves are visible an we can measure them without sophisticate technology. To observe interference an iffraction of light, we nee more sophisticate equipment. 0. No, I am not guarantee better reception if my cellphone provier as a secon tower because the signal from the two towers coul interfere estructively an reuce the quality of the reception.. (a) Given: raio wave transmitter an raio wave receiver; reflecte signal has halfwavelength phase shift on reflection; Δ 0.40 km 400 m; h 6.0 m Require: longest λ for constructive interference Analysis: Calculate the real path ifference between the irect signal an the reflecte signal. Then, incluing the extra half-wavelength phase shift for reflection, use the fact that the effective path ifference must be a whole number of wavelengths for constructive interference. Solution: The reflecte signal reflects off the groun at the mipoint of the towers, so ivie 400 m by : 00 m. The istance the signal travels is the hypotenuse of a rightangle triangle on the way own to the groun an the same again on the way up to the receiver: (00 m) + (6.0 m) m (two extra igits carrie) The irect signal travels 400 m. Calculate the real path ifference: m 400 m 8.78 m Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-33

25 The effective path ifference is to a whole number of wavelengths: 8.78 m + λ mλ 8.78 m + λ. For constructive interference, this is equal The smallest possible value for m is m : 8.78 m + λ λ 8.78 m λ λ 38 m Statement: The longest possible wavelength for constructive interference is 38 m. (b) Given: raio wave transmitter an raio wave receiver; reflecte signal has halfwavelength phase shift on reflection; Δ 0.40 km 400 m; h 6.0 m Require: longest λ for estructive interference Analysis: The problem is similar to part (a) except that the effective path ifference must be n λ. Solution: 8.78 m + λ n λ The smallest possible value for n, is n : m + λ λ 8.78 m λ λ 9 m Statement: The longest possible wavelength for estructive interference is 9 m.. (a) Yes, the image of the slit in the mirror is coherent with the slit itself. The two light rays come from a single source an share a fixe phase relationship. (b) The two branches of the light wave will interfere with each other at the screen, so the screen shows a ouble-slit interference pattern. (c) The fringe closest to the mirror surface is bright. The path istance will be about the same for both paths, an the light oes not have a phase change when reflecting from the mirror. The waves shoul be in phase with each other an will interfere constructively. 3. Given:! 583 nm m; there are 3 cycles of alternating light patterns in L Require: t Analysis:!x L 3 L 3!x Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-34

26 !x L t 3!x t!x t 6 Solution: t 6! 6( m) t m Statement: The thickness varies from 0 m to 9.3! 0 6 m. 4. Given:! 685 nm m; 6.5 mm m Require: θ.λ Analysis: sinθ Solution: sin!..(6.850$7 m ) 6.50 $3 m! Statement: The angle between the central an first ark fringes is (a) Telescope B has a smaller aperture than telescope A, so telescope B shoul experience more iffraction. (b) Magnification for telescope A is 00 times ( 00 00). Magnification for telescope B is 00 times ( 50 00). (c) The rule suggests that the maximum magnification for the telescope in the avertisement is 50 times ( 75 50). I o not believe the avertisement; the magnification appears to be greatly exaggerate. 6. Given:! 30 ; f khz. 0 4 Hz; v m/s Require: w Analysis: Rearrange the universal wave equation, v λf; to calculate wavelength, %! v m + f. Then rearrange the equation sin! $ & ' ( to calculate the with of the m w! m + $ % & ' vibrating surface of the transmitter, w. Use m. sin( m Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-35

27 Solution:!! v m + $ % & ' w f sin( m m/ s! + $. 0 4 Hz % & (.7 )0* m)!.7 0 m sin30 w 0.35 m Statement: The with of the vibrating surface of the transmitter is 0.35 m. 7. Sample answer: When we want to see objects using ifferent wavelengths of electromagnetic raiation, we want the best resolution possible. The best resolution is achieve by reucing iffraction. For a fixe aperture, a smaller wavelength means less iffraction. X-rays have much smaller wavelengths than raio waves, so X-rays woul allow for better resolution. 8. Given: λ 400 nm to 750 nm m to m Require: N Analysis: mλ w sinθ ; m N. For exactly two spectral orers to be visible, use λ θ 90 an m. Solution: m! wsin m! wsin90! w() w! N w N! Longer wavelengths have maxima farther from the centre. Use λ m: N (7.50!0 7 m) 6.67!0 5 lines/m 6.67!0 3 lines/cm N 6700 lines/cm Statement: The maximum number of lines for this grating is 6700 lines/cm. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-36

28 9. Given: nsoap film.33; N 50 lines/mm; θ 7 ; n ; m Require: t Analysis: Calculate the slit with using the equation w. Then rearrange the equation N mλ w sinθ to etermine the wavelength;! wsin m. The white light changes phase m m at the air soap bounary but not at the soap air bounary. A ark ban is create, so use the estructive interference formula, t nλ n! ; t. n n soap film soap film Solution: w N 50 mm! w.88 0!3 mm (two extra igits carrie)! wsin m m (.88 0$3 mm)sin $4 mm! $7 m (one extra igit carrie) t ()! n soap film m ()(.33) t.0 7 m 7 Statement: The minimum possible thickness of the soap film is. 0 m. 30. Lightning prouces strong changing electric an magnetic fiels, which propagate together as electromagnetic waves at ifferent frequencies. The waves at raio frequencies can therefore be picke up by a receiver, an are hear as pops or whistles. 3. Answers may vary. Sample answer: Holograms were originally ifficult to uplicate, but hologram printers have come own in price, so it is possible to counterfeit holograms more easily now. NOtES is a new technology that creates structures that are easy to recognize visually but ifficult to reprouce because they involve the creation of perforations on a nanoscale. I think the NOtES technology is more effective because the equipment necessary is too involve an too expensive for a counterfeiter to invest in. Stuents responses shoul also inclue an example of the stuent s preferre technology being use in an application not mentione in the text, for example, a shimmering penant or other piece of jewellery. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-37

29 3. Answers may vary. Sample answer: Some pros of GPS technology are that GPS units are easy to use, they can be use for navigation anywhere in the worl, an they use the most efficient route when navigating. A con of GPS technology is that GPS units in cars can be a istraction, which can lea to car accients. GPS units can also sometimes provie incorrect information. Reflect on Your Learning 33. Answers may vary. Sample answer: Learning about the interference an iffraction of water waves gave me a way to visualize light waves an unerstan how light waves work because they are very similar to water waves, except in wavelength. 34. Answers may vary. Sample answer: I woul like to learn more about the history of fibre optics in telecommunications. I want to know how this iea was create an how it has improve over the years. I woul also like to learn about future applications of fibre optics in telecommunications, an how future technology can benefit society. Perhaps my interest in fibre optics technology will lea to a career in that fiel. 35. Answers may vary. Sample answer: The information that I learne in this unit gave me a basic unerstaning about iffraction gratings. 36. Answers may vary. Sample answer: I woul like to know why astronomical observatories use ifferent regions of the electromagnetic spectrum, the ifferent properties of the ifferent regions, an what types of information astronomers learn from each region of the electromagnetic spectrum. 37. Answers may vary. Sample answer: Base on what I know about GPS technology, it oes change the way I think about privacy: with all the evices that use this technology, it means we can potentially be tracke an locate by anyone. 38. Answers may vary. Sample answer: The unit on light in grae 0 was mainly about optical images an the geometry of light. The grae unit goes into much more mathematical etail about light, especially in the area of light interference, an is more about the physics of light. Research 39. Answers may vary. This report shoul escribe fibre optic cables as strans of optically pure glass, as thin as a human hair, that carry igital information over long istances. The stuents may iscuss avantages compare to wire cables, such as being less expensive, having higher carrying capacity an less signal egraation, being nonflammable, lightweight, an flexible, an being able to carry light signals as well as igital signals. Some isavantages inclue price an fragility; they can be affecte by chemicals; they can become opaque when expose to raiation, an they require special technology to connect pieces together. 40. Answers may vary. Sample answers coul inclue points such as the following: Diffraction gratings were first prouce over 00 years ago. Early manufacturing of iffraction gratings use mechanical ruling engines. Later methos use photolithographic techniques. A more current metho uses a photosensitive gel sanwiche between two substrates. A holographic interference pattern exposes the gel, which is later evelope. Some applications use crystals as gratings for short wavelength rays, such as X-rays. Stuents reports shoul also escribe the size of gratings possible, which may have as few as 4000 lines/mm. The smallest gratings currently mae are CDs. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-38

30 4. Answers may vary. Stuents reports shoul escribe how rainbows are create using the principles of refraction stuie in this unit. They shoul also explain the polarization of the light of a rainbow cause by reflection insie rainrops, the angle of which is close to Brewster s angle. A seconary rainbow can be explaine as ouble reflections insie the rainrops at angles slightly ifferent than those for the primary rainbow. Stuents reports coul also inclue benefit by incluing images of rainbows. 4. Answers may vary. Stuents reports shoul highlight some of the methos use in astronomy to overcome the istortion effects of iffraction an shoul inclue the following information: The point sprea function is a mathematical means of escribing the ieal or minimal istortion pattern for a telescope. Distortions can be correcte by computer software, calibration using lasers, or by situating the telescope in an ieal Earth location that has minimal light interference. The reports shoul also escribe some of the techniques use for image processing in astronomy. Visuals coul also be inclue. 43. (a) Answers may vary. Sample answer: Many types of lasers are use in surgeries, incluing a variety of ermatological applications, removal of varicose veins, corrective eye surgery, removal of cancerous tissue, an treatment of prostate conitions. Examples of the ifferent lasers use are carbon ioxie, erbium (yttrium aluminum garnet), green light, re light, an pulse ye lasers. Stuents shoul give examples of which lasers are use in specific applications. (b) Answers may vary. Sample answer: There are many avantages of laser surgery over traitional invasive scalpel surgery: cutting precision, improve therapeutic results, reuce risk of infection, blooless surgery with most lasers, less scarring, limite injury to normal tissue, quick recovery, an short or no hospital stay. (c) Answers may vary. Sample answer: Disavantages of laser surgery: the possibility of burns; some meications (incluing tanning lotions) are a problem because they make the skin more susceptible to burning an scarring, thus interfering with laser proceures; consierable skill by the surgeon is neee because small errors in placement can amage healthy tissue. Other isavantages of laser surgery are that the equipment neee is costly, multiple lasers coul be require for treatment, any type of local infection coul complicate laser treatment, an the possibility of scarring or altere skin texture. 44. (a) Answers may vary. Sample answer: X-rays are prouce uring a process of raioactive ecay when high-energy electrons strike a target mae of a heavy metal. As electrons collie with this material, their paths are eflecte by the nucleus of the metal atoms. This eflection results in the prouction of X-rays as the electrons lose energy. (b) Answers may vary. Sample answer: X-ray machines consist of a long tube with an electron emitter, which is usually a tungsten filament at one en an a metal electroe at the other. The tungsten filament emits electrons when it is heate to 000 C. The electrons are accelerate along the tube an strike the metal, creating X-rays. (c) Answers may vary. Sample answer: One risk of using X-rays to treat tumours is that X-rays can amage non-tumour tissue. When the amage tissue repairs itself, it may not be able to repair property. There coul also be amage to major organs such as the heart. In some cases, there is postoperative bleeing or infection. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-39

31 () Answers may vary. Sample answer: Researchers at the University of North Carolina (UNC) have evelope carbon nanotubes to replace conventional X-ray proucing machines, which require high temperatures to operate. Instea of a single tungsten emitter, the UNC team uses an array of vertical carbon nanotubes that serve as hunres of tiny electron guns. The avantage of this new technology is that X-rays are prouce almost instantly when a voltage is applie, avoiing the warm-up time of traitional machines. This new multibeam X-ray source makes taking clear, high-resolution X-ray images of tumours an boy organs much easier. This new machine can turn multiple nanotube emitters on an off in sequence. Therefore, it is able to take rapi pictures from ifferent angles without moving the machine. With computer help, this can lea to threeimensional images of a tumour with limite istortion or amage to surrouning tissue. Copyright 0 Nelson Eucation Lt. Unit 4: The Wave Nature of Light U4-40