Pearson Physics Level 30 Unit VII Electromagnetic Radiation: Chapter 13 Solutions

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1 Pearson Physics Level 30 Unit VII Electromagnetic Raation: Chapter 13 Solutions Student Book page 636 Concept Check Infrared raation is observed by the human body through the sensing of temperature. In a way similar to visible light, infrared raation can be reflected; so it can be observed both rectly and inrectly. Student Book page 641 Concept Check 1. In the context of EMR, the wave model of light and the particle model of light are used to simplify a complex entity or process. In both cases, a model is used to explain many of the observed behaviours of EMR. For example, a particle and a wave can both reflect in a similar fashion to that of light and EMR in general.. In the particle model, a particle is considered to be analogous to a photon of light, inferring that EMR will have the behavioural characteristics of a particle. In the wave model, a wave is considered to be analogous to a photon of light, inferring that EMR will behave similarly to mechanical waves. Other examples of models are the Rutherford, Dalton, and Bohr models of the atom. 3. The EMR spectrum is based on wavelength and frequency. As the frequency of the EMR increases, so does the associated energy. Student Book page 647 Concept Check 1. Drawing a path to connect two opposite electric field vectors reveals that there must be a change in rection of the electric field along a square path. The changing electric field will produce a magnetic field accorng to Faraday s Law: The circulation of the electric field around a closed path determines the rate of change of the magnetic field that passes through this closed path.. A conductor is not required for EMR to propagate since a changing electric field exists in the region near an accelerating charge. In turn the changing electric field induces a changing magnetic field that also propagates as a region of influence in empty space. 3. A changing electric field exists in the region near any accelerating charge. Therefore, if a charge accelerates, a changing electric field is present, which induces a changing magnetic field, propagating outwards from the source at the speed of light. 1

2 13.1 Check and Reflect Knowledge 1. EMR can be detected rectly from the source, or inrectly after being reflected.. EMR spectrum should be similar to Figure The spectrum is organized by wavelength and/or frequency. 3. Definitions of EMR a) particle model describes EMR as a stream of tiny particles raating out from a source b) wave model - describes EMR as a stream of transverse waves raating out from a source c) quantum model describes EMR as screte bundles of energy, each of which is a particle that has wavelike characteristics. 4. Young s two slit experiment produced a pattern of light with varying intensity after light passed through two small slits. This pattern is similar to two-source interference patterns in water waves, as seen in section 8.3. Therefore, Young s experiment presented strong evidence for the wave model of light. 5. Hertz used his apparatus to produce a rapidly changing electric field (observed as a spark) which induced a current in a nearby antenna. This observation supports Maxwell s theory that a changing electric field will induce a changing magnetic field, both varying over time and traveling outward from the source. When the changing electric and magnetic field reached the antenna, an induced current was observed. 6. Maxwell s study of capacitors helped him understand that an electric field could exist in the absence of meum or conductor. This led to an explanation of why EMR can travel in empty space. 7. If the electric field were constant, like one existing between two stationary point charges, then no electromagnetic raation would be produced. Applications 8. The frequency of the changing electric field at the spark gap was Hz. Therefore, the frequency of the EMR observed at an antenna located nearby would also be Hz. 9. EMR that is incident on a rao receiver will induce a current flow, therefore, the antenna must be a good conductor. 10. A changing magnetic field is required to produce a changing electric field. If the magnetic field lines held within a closed path are constant and unchanging, there will not be an electric field along the closed path. 11. If an electric charge is not moving, there is no acceleration or changing electric field, therefore it will not produce a magnetic field. 1. X rays are at a high enough frequency and energy to cause biological damage. Repeated exposure could cause cancer and other health side effects, therefore technicians who work in this environment must either wear protective clothing (to block the X rays) or leave the room (to avoid the X rays). 13. When an electromagnetic wave contacts a conductor, such as an antenna, the changing electric and magnetic fields will induce a current flow in the conductor or antenna.

3 14. EMR could be produced by an oscillating voltage or potential, which accelerates electrons back and forth at a known frequency along a conductor. Other methods could involve an electron collision, causing a rapidly changing electric field to produce EMR. Extensions 15. In a stanng wave pattern for EMR their will be regions of maximum change (antinodes) and regions of no change (nodes) in the magnetic field. At the antinodes the maximum change in magnetic field will induce the maximum current in the ring antenna, which could be measured with an ammeter. In a similar way, at the nodes, there will be no change in the magnetic field and therefore no current induced in the ring antenna. By moving the ring antenna along the stanng wave, the stance separating adjacent nodes or antinodes can be determined. Doubling the stance between the antinodes or nodes reveals the wavelength. 16. Schematics will vary, but should include the following connections: Encoded data is sent to the satellite from a ground based station using EMR (in gital form). The satellite amplifies the signal and transmits it to ground based receivers (sh). The ground based receiver decodes the gital signal and transmits it via conduction to the television. Example 13.1 Practice Problems 1. Given 1 sides / 1 rotation d m v /s Student Book page 650 3

4 frequency of rotation (f) (rot/s Hz) rot m 1 rot Hz side s m s 8 1 rot.88 1 rot side 1s 6.00 s f 400 Hz and Verify If the speed of light is /s and it has to travel a total of 60.0 km from the rotating mirror to the fixed mirror and back, the rotating mirror must spin at 400 Hz in order for the light to successfully travel from the source to the observer.. Given 8 sides / 1 rotation d m v /s frequency of rotation (f)(rot /s Hz) rot m 1 rot Hz side s m s 8 1 rot rot side 1s 3.00 s 3 f Hz and Verify If the speed of light is /s and it has to travel a total of 30.0 km, the rotating mirror must spin at Hz in order for the light to successfully travel from the source to the observer. 3. Given 8 sides / 1 rotation f 500 rot /s (Hz) d m speed of light (c) The round trip stance is m. The travel time Δt is one eighth of the period of rotation. 1 1 Δd T Δ t T c f 8 Δt T Δ t ( s) c 500 Hz s s 8 T s Δ t c.88 / s The speed of light in air was found to be /s. 4

5 Student Book page Check and Reflect Knowledge 1. In Galileo s original experiment, the light travelled between two relatively close observers so quickly that accurate measurements of the time taken could not be made.. The orientation of the mirrors in a Michelson-type experiment has to be very precise in order for the light to be reflected rectly toward the observation point. If the rotating mirror is not turning such that the light is reflected at the precise angle needed, the light will be scattered at various angles and will not reach the observer. 3. When the moon Io emerges from behind Jupiter it can be observed when the light reflected from its surface reaches Earth. The light travels at a constant, finite speed and the time it takes to reach Earth depends on the stance separating Io and Earth. Since the stance between Io and Earth is constantly changing as they rotate around the Sun, the moon Io appeared to be eclipsed later when the separation stance is great and earlier when the separation stance is small. Applications 4. Given 11 d s t 4 min 1min speed of light, (c) Δd c Δ t c 60 s 4 min 1 min c s.1 10 s The speed of light in this experiment is.1 10 /s. 5. Given d.00 c s time for signal travel (t) 7 5

6 Δd c Δ t 7 Δ d.00 Δ t c s Δ t s The signal reaches the satellite in s. 6. Given 1 rotation 64 sides c s 4 d 1.60 frequency of rotation (f) rot m 1 rot f side s m s 1 rot side 1 s rot s f 93 Hz The speed of light can be measured when the rotating mirror spins at a frequency of 93 Hz. 7. Given 1 rotation 16 sides c s 3 d 7.00 frequency of rotation (f) rot m 1 rot f side s m s 8 1 rot side 1 s rot s 3 f.7 10 Hz 6

7 The speed of light can be measured when the rotating mirror spins at a frequency of Hz. 8. Given 1 rotation 8 sides c s f 545 Hz stance (d) rot m s d side s rot 8 1 rot s dtotal 8side 1s 545rot dtotal m m d m The fixed mirror should be place m from the rotating mirror. 9. Given 1 rotation 16 sides c s 4 d 1.00 frequency of rotation (f) rot m 1 rot f side s m s 8 1 rot side 1 s rot s 3 f Hz The mirror was rotating at Hz. 10. Given 1 rot 8 side 3 f Hz 4 d

8 speed of light (c) 1 Δd T c f Δ t 1 3 T s Hz 1 Δ t T ( Δ t s ) 8 4 Δ t s c s c s Based on the student s observations, the speed of light is /s. 11. Given 1 rot 64 side f Hz 4 d 1.31 speed of light (c) 1 Δd T c f Δ t 1 3 T s Hz 1 Δ t T ( Δ t s ) 64 5 Δ t s c s c s The speed of light is /s. 8

9 Extensions 1. a) The exact stance between the satellite and receiver can be determined by comparing the time values on the receiver to the time the satellite sent the signal. The fference in time equals the time the signal took to travel between the satellite and receiver. Using this time and the speed of light, the stance can be determined. b) Once the receiver determines the stance to three or more satellites, it can calculate its exact position using a process called trilateration. In this process, three spheres are produced, each with a raus equal to the stance between one satellite and the receiver. The point where all three spheres intersect is the exact location of the receiver. Concept Check 1. If 1 d o Student Book page 66 is zero, the mirror equation becomes 1 1 f d ; therefore, the image stance is equal to the focal length, so the image will appear at the focal point.. If 1 0 f, the mirror equation becomes: 1 d i 1 d incating that the image and object stances are equal in magnitude. o The negative image stance incates that the image is virtual and on the opposite side of the mirror. Student Book page 664 Example 13. Practice Problems 1. Given d o +0.0 cm f 10.0 cm (verging mirror has a negative focal length) image stance (d i ) magnification (m) Location of the image: + f do f do 10.0 cm cm d 6.67 cm i i 9

10 Magnification of the image: hi m h o d o ( 6.67 cm) 1 m d o cm The image forms 6.67 cm from the mirror. It is magnified by times, and it is a virtual image, incated by the negative image stance.. a) Given d o +6.0 cm f +1.0 cm image stance (d i ) magnification (m) image attributes Location of the image: + f do f do cm cm d 1 cm i Magnification of the image: h d i i m h o d o d ( 1cm) cm i m d o + The image forms 1 cm from the mirror. It is magnified.0 times and it is virtual, erect, and enlarged. b) Given d o cm f 5.00 cm image stance(d i ) magnification (m) image attributes Location of the image: 10

11 + f do f do 5.00 cm cm d 3.33 cm i Magnification of the image: hi m h o d 0 ( 3.33cm) m d o cm and Verify The image forms 3.33 cm from the mirror. It is magnified times and it is virtual, upright, and minished. c) Given d o +.0 cm f 10.0 cm image stance(d i ) magnification (m) image attributes Location of the image: + f do f do 10.0 cm +.0 cm d 1.7 cm i Magnification of the image: hi m h o d o ( 1.667cm) m 0.83 d o +.0 cm and Verify The image forms 1.7 cm from the mirror. It is magnified 0.83 times and it is virtual, upright, and minished. 3. Given h o 5.0 cm d o.0 cm m 4 11

12 image stance (d i ) image focal length (f) Location of the image: m d o mdo ( 4)(.0cm) 8.0cm d 8cm focal length: + f do + f.0 cm 8.0 cm f 1.6 cm and Verify The image forms 8 cm from the mirror and the focal length of the mirror is 1.6 cm. 4. Given h o 4.0 cm d o 15.0 cm f 5.0 cm image stance (d i ) image height (h i ) magnification (m) image attributes Location of the image: i Magnification and height of the image: hi m h d m i + f do f do 5.0 cm 15.0 cm d 7.5 cm o 0 d 7.5cm i 0.50 d o 15.0cm 1

13 h d i i m h o d o hi mho hi ( 0.5)(4.0cm).0cm and Verify The image forms 7.5 cm from the mirror, it is magnified 0.50 times and it is.0 cm high. It is real, inverted, and minished. 5. The object stance is assumed to be infinite; therefore the image stance is equal to the focal length of the mirror (45.0 cm). The image is real, inverted, and minished Check and Reflect Student Book page 665 Knowledge 1. In Figure 13.7 the boundary is not flat and the normal line is always drawn perpencular to the surface of the boundary.. Concept maps vary but should show similar connections: principal axis (PA) an imaginary line drawn through the vertex, perpencular to the surface of the curved mirror at this point focal point (F) the point where light rays that are parallel to the principal axis converge, or appear to verge from, after being reflected centre of curvature (C) the point on the principal axis that would represent the centre of the sphere from which the curved mirror was cut 3. In specular or regular reflection, parallel rays that are incident on a flat, smooth, reflecting surface, such as a mirror, are reflected as parallel rays. In ffuse reflection parallel rays that are incident on a rough, irregular shaped reflecting surface are scattered in many fferent rections. 4. A virtual image is one that forms when light rays appear to be coming from an image which does not exist. A real image is one that can be formed on a ffusely reflecting surface, such as a piece of paper, where rays actually converge. 5. A converging mirror is a concave reflecting surface that causes parallel light rays to converge after being reflected. A verging mirror is a convex reflecting surface that causes parallel light rays to spread out or verge after being reflected. 6. Incident Ray 1 is parallel to the principal axis and will reflect through the focal point on a converging mirror, or appear to have originated from the focal point on a verging mirror. Incident Ray travels through the focal point on a converging mirror, or is rected at the focal point on a verging mirror and will be reflected back parallel to the principal axis. Incident Ray 3 travels through the centre of curvature on a converging mirror, or is rected at the centre of curvature on a verging mirror, and is reflected rectly back along the incident path. 7. A verging mirror causes parallel light rays to spread out, or verge after being reflected. To the observer, the reflected rays that verge appear to originate from a focal point that is behind the mirror. This focal point does not exist, but only appears to exist based on the rection of the reflected rays. 13

14 Applications 8. Given hi 5.0 cm do 10.0 cm h 4.0 cm o focal length (f) hi + ho do f do hd i o h o (5.0 cm)(10.0 cm) 1.5 cm 4.0 cm f 10.0 cm 1.5 cm f 50 cm : The focal length of the mirror is 50 cm. 9. a) object at.0 f: image is real, inverted and same size as object b) object at 0.50 f: image is virtual, erect and enlarged. 14

15 c) object at 3.0 f: image is real, inverted and minished. 10. a) object at 0.50 f: image is virtual, erect and minished. b) object at 1.0 f: image is virtual, erect and minished. 15

16 c) object at 1.5 f: image is virtual, erect and minished. 11. Any incident ray which originates at the focal point will be reflected parallel to the principal axis. In the case of a flashlight or headlight, most of the rays originate or pass through the focal point and are reflected parallel to the principal axis. Extension 1. For assessment of students performance in this research activity, you may wish to use Generic Rubric GR 11. Research Report. You might hand out this rubric at the time that you assign the exercise, so that students know what is expected of them. Example 13.3 Practice Problems 1. Given θ 5 n 1.4 n 1.00 Student Book page

17 angle of incidence (θ 1 ) sinθ1 n sinθ n1 1 nsinθ θ1 sin n1 1 (1.00)(sin 5) θ1 sin.4 θ1 10 The angle of incidence as the light exits the amond is 10. Given θ 1 0 θ 17 n index of refraction of transparent material (n ) sinθ1 n sinθ n1 n1sinθ1 n sinθ n sinθ n 1 1 sinθ (1.00)(sin 0) n (sin17) n 1. The transparent material has an index of refraction equal to Given θ 1 15 o n n 1.54 angle of refraction (θ ) sinθ1 n sinθ n1 1 n1sinθ1 θ sin n 17

18 1 n sinθ θ sin 1 1 n 1 (1.00)(sin15) θ sin 1.54 θ 9.7 The angle of refraction is 9.7 o. 4. Given θ 1 0 θ 15 n index of refraction (n ) sinθ1 n sinθ n1 n1sinθ n 1 sinθ n1sinθ1 n sinθ (1.00)(sin 0) n (sin15) n 1.3 and Verify The unknown material has an index of refraction equal to 1.3. Student Book page 670 Example 13.4 Practice Problems 1. Given The refractive index of each material: water: 1.33 ethanol: 1.37 ruby: 1.54 crown glass: 1.5 The speed of light in air /s speed of light (c) in each material: c n v c v n 18

19 v v v water ethanol v ruby crownglass c s.6 10 n 1.33 s c s n 1.37 s c s n 1.54 s c s n 1.5 s and Verify The speed of light is fferent in each material, travelling slower in materials with a higher index of refraction.. Given θ n n 1.47 λ nm angle of refraction (θ ) wavelength (λ ) sinθ1 n sinθ n1 1 n1sinθ1 θ sin n λ1 n λ n1 λ1n1 λ n θ 1 1 sin λ n n n sinθ 1 (1.00)sin(5) (737nm)(1.00) θ sin λ θ 16.7 λ 501nm λ n The angle of refraction is 16.7 o and the wavelength in the quartz glass is 501 nm. 3. Given λ nm λ 450 nm n

20 refractive index (n ) λ1 n λ n1 λ1n 1 n λ λ1n1 n λ n n (500nm)(1.00) (450nm) 1.11 The refractive index of the unknown material is Student Book page 67 Concept Check 1. Light is refracted toward the normal line when passing from a meum with a low refractive index like air, into a meum with a high refractive index like water.. Total internal reflection cannot occur when light travels from a low index meum like air, into a high index meum like water since the light will always be bent toward the normal line and away from the boundary. Student Book page 673 Example 13.5 Practice Problems 1. Given Refractive index of: water 1.33 amond.4 air 1.00 critical angle for water/air interface, (θ critical ) critical angle for amond /air interface, (θ critical ) critical angle for amond/water interface, (θ critical ) n sinθcritical n1 sin sin 90 critical sin n θ n 0

21 (a) θ (b) θ (c) θ θ θ θ critical water/air critical water/air critical amond/air critical amond/air critical amond/water critical amond/water sin 90 sin sin 90 sin sin 90 sin The critical angle for a water/air boundary is 48.8 o : for a amond/air boundary it is 4.4 o : for a amond water boundary it is 33.3 o. Student Book page 679 Example 13.6 Practice Problems 1. a) Given an object located between C and F relative to a converging lens image attributes Image is beyond C, real, inverted, and enlarged. b) Given an object located between C and F relative to a verging lens image attributes 1

22 Image is between the lens and F, virtual, erect, and minished. c) Given an object located between F and a converging lens image attributes Image is between F and C, virtual, erect, and enlarged. d) Given an object located between F and the lens of a verging lens image attributes

23 Image is between the lens and F, virtual, erect, and minished. Example 13.7 Practice Problems 1. Given h o 3.00 cm d o 15.0 cm f 10.0 cm image stance (d i ) image height ( h i ) image attributes hi + m d o d i f h o d o dh i o hi d f d d i o o Student Book page 681 (30.0cm)(3.00 cm) hi 10.0 cm 15.0 cm 15.0 cm 30.0 cm hi 6.00 cm The image forms 30.0 cm from the lens with a height of 6.00 cm. The image is real, inverted, and enlarged.. Given h o 10.0 cm d o cm f 5.0 cm 3

24 image stance, (d i ) image height (h i ) image type hi + m do f ho dh i o hi d f d d i o o d d i o ( 0.0cm)(10.0 cm) hi 5.0 cm cm cm 0.0 cm hi.00 cm The image forms 0.0 cm from the lens with a height of.00 cm. Since the image stance is negative it is a virtual image. 3. Given d o 7.50 cm d i 500 cm focal length (f) magnification (m) + m f d d d i o o (500 cm) + m f 500cm 7.50cm 7.50 cm f 7.39 cm m 66.7 The projector lens has a focal length of 7.39 cm and the image is magnified 66.7 times Check and Reflect Student Book page 683 Knowledge 1. Light is bent towards the normal line when it passes from a low index meum to a high index meum.. The index of refraction can be measured by observing the change in speed, rection, or wavelength of the EMR as it enters the new meum. 3. Concept maps may vary, but should include the following connections: The angle of incidence is the angle between the incident ray and the normal line. The angle of refraction is the angle between the refracted ray and the normal line. 4. The simple form of Snell s Law assumes that the first meum has an index of refraction equal to one. Therefore, Snell s constant and the index of refraction for the 4

25 second meum are one and the same thing. In the general form of Snell s law, the first meum can have an index of refraction greater than or equal to one. 5. Place a semi-circular block of Lucite on the centre of a sheet of polar coornate paper. Direct a light ray towards the curved surface of the block, (rectly at the centre of the polar coornate paper). Slowly increase the angle of incidence until the light ray internally reflects. The smallest possible angle of reflection will be equal to the critical angle. 6. economically less expensive than copper wire of equivalent length thinner, more flexible, and made of non-flammable materials able to handle a higher data-carrying capacity based on the fibre bundle ameter less signal degradation and interference between multiple signals on the same fibre as compared to copper networks the glass fibres are highly transparent so that repeaters (amplifiers) can be many kilometres apart, as opposed to coaxial cable repeaters that must be less than 1 km apart 7. Incident light with a smaller, shorter wavelength will slow down and refract to a greater extent than light with a larger, longer wavelength. As each wavelength of light refracts at a slightly fferent angle, the wavelengths will separate, producing a continuous spectrum. 8. When light enters a new meum, like glass, the wavelength is reduced but the frequency is unchanged. Accorng to the universal wave equation, (v fλ) if the speed of the wave is reduced, so too is the wavelength. Therefore, the index of refraction is related to both the speed of a wave, and its wavelength. Dispersion is caused by the fact that the index of refraction is fferent for each wavelength in the visible spectrum, causing each wavelength to refract at a slightly fferent angle. Applications 9. Given nwater 1.33 c s speed of light in water (v water ) c c n v v n v s 1.33 v.6 10 s The speed of light in water is.6 10 /s. 10. Given n quartz glass 1.47 θ incident

26 angle of refraction (θ refraction ) sinθincident n sinθrefraction n1 1 ( sinθincident )( n1 ) θrefraction sin n 1 ( sin35.0 )( 1.00) θrefraction sin 1.47 θrefraction 3.0 The angle of refraction is 3.0 when the light enters the quartz glass at Given nwater 1.33 nlucite 1.5 θrefraction 8.0 angle of incidence (θ incidence ) sinθincidence n sinθrefraction n1 1 ( sinθrefraction )( n ) θincidence sin n 1 1 ( sin8.0 )( 1.5) θincidence sin 1.33 θincidence 3.4 The light ray is incident on the Lucite block at an angle of Given nair 1.00 nlucite 1.5 θrefraction 90 critical angle (θ critical ) 6

27 sinθcritical n sin90 n1 1 ( sin 90 )( n ) θcritical sin n 1 1 ( sin 90 )( 1.00) θcritical sin 1.5 θcritical 41.1 The critical angle for an air-lucite interface is 41.1 o. If the angle of incidence is larger than the critical angle the light will internally reflect. 13. Given nair 1.00 nwater 1.33 θrefraction 90 critical angle (θ critical ) sinθcritical n sin90 n1 1 ( sin90 )( n ) θcritical sin n 1 1 ( sin90 )( 1.00) θcritical sin 1.33 θcritical 48.8 Since the angle of incidence is larger than the critical angle, the light will internally reflect at the water-air interface and remain in the water. 14. Given nruby 1.54 nair 1.00 θincident 5.0 λ1 540 nm angle of refraction (θ refraction ) wavelength of light in ruby (λ ) 7

28 sinθincident λ1 n sinθrefraction λ n1 1 ( sinθincident )( n1 ) θrefraction sin n n1λ 1 λ n 1 ( sin5.0 )( 1.00) θrefraction sin 1.54 θrefraction 15.9 ( 1.00)( 540nm) λ 1.54 λ 351nm The angle of refraction is 15.9 o and the light has a wavelength of 351 nm in the ruby. 15. a) image is real, inverted, and the same size as the object. b) image is real, inverted, and enlarged 8

29 c) no image forms since the refracted rays are parallel. 16. Given f 4.50cm 4.50 ho 5.0m do 50.0m image height (h i ) image attitude + f do hi m h d i o o m d 4.50 ( 4.50 )( 5.0m) hi 50.0m hi.5 The image is.5 10 m high and the negative height incates that it is inverted. 17. Incident light with a smaller, shorter wavelength will slow down and refract to a greater extent than light with a larger, longer wavelength. Therefore, the blue light is refracted at a larger angle than the red light. 18. a) Given f 8.00cm d 8.10cm o image stance (d i ) 9

30 + f do 8.00cm 8.10cm 648cm The screen must be placed 648 cm from the lens in order to produce a clear image. b) Given ho 1.75cm do 8.10cm 648cm Image height (h i ) hi do ho dh i o hi d i o ( 648cm)( 1.75cm) hi 8.10cm h 140cm The image on the screen is 140 cm. The negative height incates that it is inverted relative to the object. c) Given f 8.00cm 648cm 300cm 348cm object stance (d o ) + f do do 8.00cm 348cm do 8.19cm If the screen is moved 3.0 m closer to the projector then the object must be positioned 8.19 cm from the lens. Extensions 19. The sparkle and glitter associated with amonds is caused by total internal reflection. The amond s high index of refraction in combination with the shape of the surface both reflects and focusses the light. We observe this effect as sparkles. 30

31 0. A magnifying glass is a converging lens. When multiple rays of EMR energy converge at the focal point of the lens, there can be sufficient energy to ignite combustible material. 1. If a ver is underwater and looks up, a circular hole appears on the surface of the water, which otherwise appears to be a mirror. This is caused by total internal reflection. Light rays which are in the water and are incident on the surface at an angle of 49 o (critical angle for a water-air interface), or larger, are internally reflected. Concept Check Student Book page 685 As the secondary wavelets emerge from the opening(s), they form a new wavefront travelling outward with curved edges. An interference pattern is produced as multiple wavelets overlap and interfere with one another. 31

32 Student Book page 689 Example 13.8 Practice Problems 1. Given 4 d.0 n θ 0.6 wavelength (λ) dsinθ λ n ( m)( sin0.6 ) λ 7 λ 4.5 The light has a wavelength of m.. Given 4 d 5.5 n 3 7 λ 4.60 angle of ffraction (θ) d sinθ λ n 1 λn θ sin d 7 ( 4.60 )( 3) 1 θ sin θ 0.14 The angle of ffraction to the third antinode is 0.14 o. 3. Given 4 d 4.0 n θ wavelength (λ) 3

33 d sinθ λ 1 n ( m)( sin ) λ 1 7 λ 4.4 The wavelength of light producing this pattern is m and this corresponds to the colour violet. Example 13.9 Practice Problems 1. Given 4 d 1.5 L 5.0m x 4.50 n 3 wavelength ( λ) xd λ 1 n L 4 ( 4.50 )( 1.5 ) λ 1 3 ( 5.0m) 7 λ 5.4 The wavelength of light is m.. Given 5 d 3.00 L 1.50m x 3.10 n1 wavelength (λ) Student Book page

34 λ λ xd nl 5 ( 3.10 )( 3.00 ) ()( m) 7 λ 6.0 The wavelength of light producing this pattern m and this corresponds to the colour red. 3. Given 1 x 1.00 L 1.0m 7 λ 5.00 n 3 slit stance (d) xd λ 1 n L 1 λ n L d x d 7 1 ( 5.00 ) 3 ( 1.0m) 1 ( 1.00 ) 5 d 1.5 The two slits are separated by m. Example Practice Problems 1. Given 1cm 4 d cm 500 lines 6 d n 3 1 x.00 1 L 5.00 Student Book page

35 wavelength (λ) xd λ 1 n L λ 1 6 (.00 )( ) ( 5.00 ) 7 λ 6.40 The wavelength of the light is m.. Given n 3 x L λ 5.00 number of lines per cm. xd λ 1 n L 1 λ n L d x d 1 d ( 5.00 ) 3 ( 6.0 ) ( 5.00 ) 5 d 1.5 lines 1 4 lines metre 1.5 metre lines 1m lines m 100 cm cm There are lines/cm in the ffraction grating. 35

36 3. Given 1cm d cm 1000 lines 5 d λ 7.50 n 1 L 3.0m separation stance (x) xd λ nl λnl x d ( m)()( m) x x.5 x.5 cm x 3 cm The bright antinodes are separated by a stance of 3 cm Check and Reflect Student Book page 697 Knowledge 1. Accorng to Huygens Principle, a straight wavefront will take on a semi-circular shape as it emerges from the opening.. When a straight wavefront is incident on two narrow openings, each openings acts as a point source (wavelet), with a circular wavefront emerging. Since the wavelets or point sources have the same wavelength, the circular waves emerge in phase and interfere depenng on the length of the path fference to any point p. 3. Incandescent white lights emit light waves in random bursts, out of phase with any second light source which could be placed nearby. In adtion, the emitted waves have a variety of wavelengths making up all the colours in the spectrum. Therefore, the interference pattern would not be constant, but would vary rapidly over time, making any observation of the pattern impossible. 4. Young needed to have two sources of light that were coherent (in phase) and separated by a very small stance so that an interference pattern could be observed. He d this by splitting a small sunbeam using a card that was slightly smaller than the ameter of the sunbeam. 5. Concept maps will vary but should include the following connections: When the path length of two identical waves is fferent by a whole number of ½ wavelengths they 36

37 are out of phase and destructive interference occurs, producing a nodal fringe. When the path lengths are fferent by a whole number of wavelengths they are in phase and constructive interference will occur to produce an antinodal fringe. Applications 6. Given n 3 7 λ 6.50 θ 0 slit separation (d) d sinθ λ n nλ d sinθ ( 3)( m) d sin 0 6 d 5.7 The two slits are m apart. 7. Given 6 d 6.00 n 7 λ 4.5 angle of ffraction (θ) d sinθ λ 1 n 1 λ n 1 θ sin d 7 1 ( 4.5 ) 1 θ sin θ 6.10 The angle of ffraction to the nd node is 6.10 o. 37

38 8. Given 1cm 4 d cm lines 6 d.00 7 λ 7.00 n 1 L 1.00m stance between1st and central antinodes (x) xd λ nl λnl x d ( m)()( m) x 6.00 x m x 35.0 cm The stance between the 1 st and central antinodes is 35.0 cm. 9. Given 1cm d cm 60lines 4 d f Hz v λ s 14 f Hz 7 λ n 3 1 L 1.0m stance between nd and 3 rd dark fringes (x) xd λ nl λnl x d ( m)()( 1 1.0m) x x

39 The stance between the nd and 3 rd dark fringes is m. 10. Given 1 m d lines n 1 x 5.50 L 1.50 m wavelength (λ) frequency (f) xd λ nl v fλ 1 m ( m ) lines λ ()( m) 7 λ v f s 7 λ f Hz The unknown light source has a wavelength of m and a frequency of Hz. 11. Given 0.01 m d lines n 1 x 3.10 L 1.50 m wavelength (λ) frequency (f) 39

40 xd λ nl v fλ 0.01 m ( m ) lines λ ()( m) 7 λ v f s 7 λ f Hz The unknown light source has a wavelength of m and a frequency of Hz. Extensions 1. Experimental designs will very but should be based on the following: 1. the unknown light source is incident on a ffraction grating which produces an interference pattern on a nearby screen.. The stances between each visible antinode and the central antinode (x) are measured and recorded. 3. Slit separation (d) is determined from the grating, and the stance between the grating and the screen (L) is measured and recorded. 4. For each antinode, the wavelength is calculated using λ xd/nl. An average value for the wavelength is then determined. Materials may include a ffraction grating, screen, ruler, metre-stick and light source. 13. X rays have a much shorter wavelength than visible light rays. Comparing the ffraction pattern produced by visible light and X rays that are incident on the same ffraction grating, one would notice that the separation between antinodes (x) would be smaller for X rays, based on: x λ(nl/d), where (nl/d) is constant. 14. Crystal lattices and structures too small to be seen with visible light have openings which can be thought of as small slits. Using X rays or other, shorter, wavelengths which can ffract in these small openings, the size and shape of the crystal lattice can be inferred. This inference is based on the interference pattern produced on photographic film, which is sensitive to the type of EMR that is incident on the crystal. Chapter 13 Review Student Book pages Knowledge 1. Table should be similar to Table 13.1 with adtional notes related to common uses for each type of raation. 40

41 . The particle model describes EMR as a stream of tiny particles raating out from a source. The wave model describes EMR as a stream of transverse waves raating out from a source. 3. Two critical insights understood by Maxwell were: A moving electric charge (or a changing electric field) produces a magnetic field; and, a changing magnetic field produces an electric field that can cause electric charges to move. 4. Drawing a path to connect these two electric field vectors reveals that there must be a change in rection of the electric field along this square path. Accorng to Faraday s Law, there must be a perpencular magnetic field that exists within the closed path of the changing electric field. 5. The changing electric field produced by a capacitor will produce a changing magnetic field in the same manner that a moving charge in a conductor does. Therefore, a conductor is not necessary for the production and propagation of perpencular magnetic and electric fields. 6. a) electromagnetic waves are produced whenever an electric charge is accelerating. Therefore, as an electric charge oscillates, electrical energy will be lost, and an equivalent amount of energy will be raated outward in the form of oscillating electric and magnetic fields. b) When the electric charge is accelerated in perioc motion, the frequency of oscillation of the charge will correspond exactly to the frequency of the electromagnetic wave that is produced. c) All electromagnetic waves, inclung rao, infrared, visible light, and X rays, will travel at a speed of /s in a vacuum and obey the universal wave equation (c f λ ) relating speed, frequency, and wavelength d) The oscillating electric and magnetic fields will always be perpencular to each other and perpencular to the rection of propagation of the wave. e) Electromagnetic waves should show all the phenomena associated with transverse waves: interference, ffraction, refraction, and polarization. 7. When a changing electromagnetic field encounters a metal antenna, it will induce current flow, which can be measured. The induced current in the antenna will fluctuate in such a way as to match the frequency of the changing electric and magnetic fields in the wave. In Hertz s experiment, when the receiver was tuned to the same frequency as the raator, the induced current flow in the antenna oscillated at a frequency identical to that of the changing electric field in the raator. This was conclusive evidence that Hertz s device had indeed produced the electromagnetic raation that was being observed at his antenna. 8. The eclipse of Io occurs at regular intervals based on its constant period of rotation around Jupiter. Astronomical observations showed that the timing of the eclipse appeared to vary by as much as min depenng on the relative positions of Jupiter and Earth. Christian Huygens reasoned that the variance in time was due to the extra stance (Earth s orbital ameter, estimated as m) that the light needed to travel when Jupiter and Earth were at their maximum stance of separation. His calculations produced a value of.3 10 /s when he vided the estimated orbital ameter of Earth ( m) by the min (130 s) time variance. 9. In Fizeau s water experiment, light travelled through two water tubes, which had water flowing in opposite rections. The speed of the light was indeed influenced by 41

42 the motion of the water, and the light arrived out of phase at an exit point, as would be expected if light has wavelike characteristics. An interferometer was used to observe the resulting interference pattern, which provided strong support for the wave model of light centre of curvature (C) the point in space that would represent the centre of the sphere from which the curved mirror was cut raus of curvature (r) the stance from the centre of curvature to the mirror surface vertex (V) the geometric centre of the curved mirror surface principal axis (PA) an imaginary line drawn through the vertex, perpencular to the surface of the curved mirror at this point principal focal point (F) the point where light rays that are parallel to and close to the principal axis converge, or appear to verge from, after being reflected 4

43 focal length (f) the stance from the vertex to the focal point, measured along the principal axis. 1. A real image forms where multiple rays converge or cross. In the case of a verging mirror, all the reflected rays verge and only appear to have originated from a single point, thus proving a virtual image. A real image can never form since the reflected rays never converge or cross. 13. [Angle measurements are not exact here, but incate the relationship between angles of incidence and refraction.] 14. When light passes from a meum with a high refractive index to one with a low refractive index the light is bent away from the normal line. sinθair 15. The simple form of Snell s Law, n applies whenever the first meum has sinθ r an index of refraction equal to 1.0 (i.e. the meum is air or a vacuum). It can be used to determine the refractive index of any material when light passes from air or a vacuum into that material. The general form of Snell s Law, n1sinθ1 nsinθ, can be applied in all cases, regardless of the indexes of refraction of the mea and/or the rection in which the light travels. 16. Dispersion is the separation of white light into a continuous spectrum of colours - each defined by its wavelength. Two fferent methods of separating all the colours in white light are as follows. In a prism, the refractive index varies for each wavelength (or colour) and so the amount of refraction is fferent for each colour. This causes the colours to be separated as they exit the prism. In ffraction, spersion can be 43

44 observed in the interference pattern because the path lengths which create destructive and constructive interference are defined by the wavelength of each colour in the visible spectrum. Since the wavelengths are fferent, the location of constructive and destructive interference for each wavelength is fferent causing the colours to be spersed or separated in the interference pattern Accorng to Huygens Principle the straight wave front is composed of many small wavelets, with the line tangent to all wavelets forming the wave front. The wavelets travel faster in the less refractive meum, causing the wavefront to travel a greater stance during the t 1 t time interval. This is illustrated by the larger raus of the wavelets in the less refractive meum. Therefore, the process of refraction can be explained using Huygens Principle. This supports the wave model of light. Interference will occur as the wavelets overlap, causing areas of constructive interference and areas of destructive interference. 19. Incandescent lights emit light in random bursts, not necessarily in phase with one another, and with a variety of wavelengths making up all the colours in the spectrum. In this situation, the interference pattern would not be constant, but would vary rapidly over time, making any observation of a pattern impossible. 44

45 0. Electromagnetic waves can be blocked by two polarizing filters held at right angles to one another. This evidence can be explained by considering electromagnetic raation as a transverse wave with perpencular magnetic and electric fields. 1. Each antinode appears as a full spectrum when white light is incident on a ffraction grating because the fferences in path lengths which create destructive and constructive interference are defined by the wavelength of each colour in the visible spectrum. Since the wavelengths are fferent, location of constructive and destructive interference for each wavelength is fferent causing the colours to be spersed or separated in the interference pattern.. Incident light which contacts the edge of a circular sc is ffracted inwards. A tracing of the path of all the ffracted light around the edge of the sc forms a geometric cone. Since the path length from the sc edge to the tip of the cone is identical for all the ffracted light rays, they arrive at the tip of the cone in phase. When all of the light rays arrive in phase, constructive interference is observed as a bright spot. Applications 3. Alternating current means that the charges are constantly accelerating, creating a changing electric field that produces an electromagnetic wave with a frequency equal to that of the alternating rection of the current in the high voltage wires. When these EMR waves encounter rao waves, they may interfere constructively and destructively, thus altering the form of the rao waves. This is observed as interference in the rao signal. 4. (a) Water Given n 1.33 from Table 13.4 water /s c speed of light in water (v) The refractive index of water for 589 nm light is This index is a ratio of the speed of light in a vacuum to that of the light in water. This ratio can be used to determine the speed of the light in water. c n v c v n v 1.33 v m s m s The speed of 589 nm yellow light in water is.6 10 /s. (b) ethanol Given n c 1.37 from Table 13.4 ethanol /s 45

46 speed of light in ethanol (v) The refractive index of ethanol for 589 nm light is This index is a ratio of the speed of light in a vacuum to that of the light in ethanol. This ratio can be used to determine the speed of the light in ethanol. c n v c v n v s v.19 /s The speed of 589 nm yellow light in ethanol is /s. (c) Lucite Given n 1.5 from Table 13.4 Lucite /s c speed of light in Lucite (v) The refractive index of Lucite for 589 nm light is 1.5. This index is a ratio of the speed of light in a vacuum to that of the light in Lucite. This ratio can be used to determine the speed of the light in Lucite. c n v c v n v s v 1.97 /s The speed of 589 nm yellow light in Lucite is /s. (d) quartz glass Given nquartz glass 1.47 from Table c 3.00 /s speed of light in quartz glass (v) The refractive index of quartz glass for 589 nm light is This index is a ratio of the speed of light in a vacuum to that of the light in quartz glass. This ratio can be used to determine the speed of the light in quartz glass. c n v 46

47 c v n v 1.47 v s s The speed of 589 nm yellow light in quartz glass is /s. (e) amond Given namond.4 from Table c 3.00 /s speed of light in amond (v) The refractive index of amond for 589 nm light is.4. This index is a ratio of the speed of light in a vacuum to that of the light in amond. This ratio can be used to determine the speed of the light in amond. c n v c v n v.4 v s s The speed of 589 nm yellow light in amond is /s. 5. Given 1 rotation 8 sides 8 c 3.00 f Hz stance (d) Unit analysis can be used to determine the stance that the light can travel during 1/8th of a rotation. The given information must be arranged as follows rot m s d side s rot 8 1 rot s dtotal 8side 1s rot dtotal d m 47

48 The fixed mirror should be place m from the rotating mirror. 6. Given 8 sides / 1 rotation f 600 rot /s (Hz) d m speed of light (c) The round trip stance is m. The light must travel this stance in time Δt, which is equal to one-eighth of the period of rotation. Using this time and the total stance, the speed of light can be calculated using c Δd t 1 1 Δd T Δ t T c f Δt T Δ t c 8 Δ ( ) s Hz s T s Δ t s c.88 / s The speed of light in air was found to be /s. 7. An image that forms in a plane mirror is erect, virtual, and the same size as the object. 8. Using a ray agram and measuring the stance (shown in green below) between the upper ray and lower ray point of incidence reveals that the portion of the mirror needed to see the entire face is equal to 1.5 cm or half the height of the object. 9. Incident Ray 1 is parallel to the principal axis. Any ray that is parallel to the principal axis will reflect through the focal point on a converging mirror. Incident Ray travels through the focal point. Any ray that passes through the focal point on a converging mirror will be reflected back parallel to the principal axis. Incident Ray 3 travels towards the vertex. The vertex acts as a plane reflector, where the angle of incidence is equal to the angle of reflection. 48

49 (a) the image is located at 37.4 cm from the vertex and it is real (b) the image is inverted (c) the image is magnified 1.5X A virtual image is produced by a converging mirror whenever an object is placed between the focal point and the vertex of the mirror. If the object is beyond the focal point, the image is real and if the object is located at the focal point, no image will form. 3. Given h o cm d o 0.0 cm f 10.0 cm image height (h i ) image stance (d i ) The location of the image can be found using the mirror equation as follows: 49

50 + f do f do 10.0 cm cm 6.67 cm The height of the image can be found by comparing the ratio of the image and object height to the image and object stances as follows: hi d i ho do dh h i o i do ( 6.67 cm)(15.0 cm) hi 0.0 cm hi 5.00 cm The virtual image forms 6.67 cm from the mirror and it is 5.00 cm high. 33. Given h i 0.0 cm d o 1.0 cm f 11.0 cm image height (h i ) The image stance is required to solve for the object height, but it is not given. It can be determined using the thin lens equation as follows: + f do f do 11.0 cm 1.0 cm 13 cm The height of the object can be found by comparing the ratio of the image and object height to the image and object stances as follows: hi d i ho do hd h i o o ( 0.0 cm)(1.0 cm) ho 13 cm ho 1.8 cm The object is 1.8 cm high. 50

51 34. Given θ n n 1.47 λ nm The angle of refraction (θ ) The wavelength in the quartz glass (λ ) The angle of refraction and the wavelength in the quartz glass are calculated using the general form of Snell s Law. sinθ1 n sinθ n1 1 n1sinθ θ sin 1 θ sin θ 3 λ λ λ 1 1 n n1 λ1n 1 n n (1.00)sin(35 ) 1.47 (610 nm)(1.00) nm λ λ The angle of refraction is 3 o and the wavelength in the quartz glass is 415 nm. 35. Total internal reflection can only occur when the light ray is bent away from the normal line as it enters a new meum. This will only occur when light travels from a high index meum into a low index meum. (a) no total internal reflection cannot occur since light is travelling from a low index meum (n air 1.00) into a high index meum (n water 1.33) (b) yes total internal reflection can occur since light is travelling from a high index meum (n water 1.33) into a low index meum (n air 1.00) (c) yes - total internal reflection can occur since light is travelling from a high index meum (n Lucite 1.5) into a low index meum (n water 1.33) (d) no total internal reflection cannot occur since light is travelling from a low index meum (n water 1.33) into a high index meum (n amond.4) 36. Given v unkown /s n water 1.33 from Table 13.4 c /s The refractive index of the unknown material (n unknown ) The index of refraction is a ratio of the speed of light in the meum to the speed of light in a vacuum. 51

52 c n v c nunknown vunknown /s nunknown 8.67 /s nunknown 1.1 and Verify The unknown material has a refractive index of 1.1, which is less than that of water at Given From Table 13.4: n water 1.33 n amond.4 n air 1.00 n Lucite 1.5 The critical angle for each boundary (θ critical ) The critical angle can be found using Snell s Law and assuming that the angle of refraction is equal to 90 0 when total internal reflection occurs. n sinθcritical n1 sin90 1 n1sin 90 θcritical sin n sin90 θcritical water/air sin 1.33 θcritical water/air sin90 θcritical amond/air sin.4 θcritical amond/air sin90 θcritical amond/water sin.4 θcritical amond/water sin90 θcritical Lucite/air sin 1.5 θcritical Lucite/air 41.1 The critical angle for a water air boundary is 48.8º; for a amond air boundary it is 4.4 o ; for a amond-water boundary it is 33.3º; and for a Lucite air boundary it is 41.1º. 38. (a) Incident Ray 1 is parallel to the principal axis. Any ray that is parallel to the principal axis will refract through the focal point on a converging lens. Incident Ray travels through the focal point. Any ray that passes through the focal point on a converging lens will be refracted parallel to the principal axis. 5