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Copyright © 2009 Pearson Education, Inc. Chapter 31 Maxwell’s Equations and Electromagnetic Waves.

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Presentation on theme: "Copyright © 2009 Pearson Education, Inc. Chapter 31 Maxwell’s Equations and Electromagnetic Waves."— Presentation transcript:

1 Copyright © 2009 Pearson Education, Inc. Chapter 31 Maxwell’s Equations and Electromagnetic Waves

2 Copyright © 2009 Pearson Education, Inc. This figure illustrates the process by which a radio station transmits information. The audio signal is combined with a carrier wave. 31-10 Radio and Television; Wireless Communication

3 Copyright © 2009 Pearson Education, Inc. The mixing of signal and carrier can be done two ways. First, by using the signal to modify the amplitude of the carrier (AM): 31-10 Radio and Television; Wireless Communication

4 Copyright © 2009 Pearson Education, Inc. Second, by using the signal to modify the frequency of the carrier (FM): 31-10 Radio and Television; Wireless Communication

5 Copyright © 2009 Pearson Education, Inc. At the receiving end, the wave is received, demodulated, amplified, and sent to a loudspeaker. 31-10 Radio and Television; Wireless Communication

6 Copyright © 2009 Pearson Education, Inc. The receiving antenna is bathed in waves of many frequencies; a tuner is used to select the desired one. 31-10 Radio and Television; Wireless Communication

7 Copyright © 2009 Pearson Education, Inc. 31-10 Radio and Television; Wireless Communication A straight antenna will have a current induced in it by the varying electric fields of a radio wave; a circular antenna will have a current induced by the changing magnetic flux.

8 ConcepTest 31.3TV Antennas Before the days of cable, televisions often had two antennae on them, one straight and one circular. Which antenna picked up the magnetic oscillations? the circular one 1) the circular one the straight one 2) the straight one both equally; they were straight and circular for different reasons 3) both equally; they were straight and circular for different reasons

9 The varying B field in the loop means the flux is changing and therefore an emf is induced. ConcepTest 31.3TV Antennas Before the days of cable, televisions often had two antennae on them, one straight and one circular. Which antenna picked up the magnetic oscillations? the circular one 1) the circular one the straight one 2) the straight one both equally; they were straight and circular for different reasons 3) both equally; they were straight and circular for different reasons

10 ConcepTest 31.4Radio Antennas If a radio transmitter has a vertical antenna, should a receiver’s antenna be vertical or horizontal to obtain the best reception? vertical 1) vertical horizontal 2) horizontal doesn’t matter 3) doesn’t matter

11 vertical antennaelectric field oscillates up and downreceiver’s antenna should also bevertical If a wave is sent out from a vertical antenna, the electric field oscillates up and down. Thus, the receiver’s antenna should also be vertical so that the arriving electric field can set the charges in motion. ConcepTest 31.4Radio Antennas E field of wave E field of wave If a radio transmitter has a vertical antenna, should a receiver’s antenna be vertical or horizontal to obtain the best reception? vertical 1) vertical horizontal 2) horizontal doesn’t matter 3) doesn’t matter

12 Copyright © 2009 Pearson Education, Inc. Maxwell’s equations are the basic equations of electromagnetism: Summary of Chapter 31

13 Copyright © 2009 Pearson Education, Inc. Electromagnetic waves are produced by accelerating charges; the propagation speed is given by The fields are perpendicular to each other and to the direction of propagation. Summary of Chapter 31

14 Copyright © 2009 Pearson Education, Inc. The wavelength and frequency of EM waves are related: The electromagnetic spectrum includes all wavelengths, from radio waves through visible light to gamma rays. The Poynting vector describes the energy carried by EM waves: Summary of Chapter 31

15 Copyright © 2009 Pearson Education, Inc. Chapter 32 Light: Reflection and Refraction

16 Copyright © 2009 Pearson Education, Inc. The Ray Model of Light Reflection; Image Formation by a Plane Mirror Formation of Images by Spherical Mirrors Index of Refraction Refraction: Snell’s Law Units of Chapter 32

17 Copyright © 2009 Pearson Education, Inc. Visible Spectrum and Dispersion Total Internal Reflection; Fiber Optics Refraction at a Spherical Surface Units of Chapter 32

18 Copyright © 2009 Pearson Education, Inc. Light very often travels in straight lines. We represent light using rays, which are straight lines emanating from an object. This is an idealization, but is very useful for geometric optics. 32-1 The Ray Model of Light

19 Copyright © 2009 Pearson Education, Inc. Law of reflection: the angle of reflection (that the ray makes with the normal to a surface) equals the angle of incidence. 32-2 Reflection; Image Formation by a Plane Mirror

20 Copyright © 2009 Pearson Education, Inc. When light reflects from a rough surface, the law of reflection still holds, but the angle of incidence varies. This is called diffuse reflection. 32-2 Reflection; Image Formation by a Plane Mirror

21 Copyright © 2009 Pearson Education, Inc. With diffuse reflection, your eye sees reflected light at all angles. With specular reflection (from a mirror), your eye must be in the correct position. 32-2 Reflection; Image Formation by a Plane Mirror

22 Copyright © 2009 Pearson Education, Inc. 32-2 Reflection; Image Formation by a Plane Mirror Example 32-1: Reflection from flat mirrors. Two flat mirrors are perpendicular to each other. An incoming beam of light makes an angle of 15° with the first mirror as shown. What angle will the outgoing beam make with the second mirror?

23 Copyright © 2009 Pearson Education, Inc. What you see when you look into a plane (flat) mirror is an image, which appears to be behind the mirror. 32-2 Reflection; Image Formation by a Plane Mirror

24 Copyright © 2009 Pearson Education, Inc. This is called a virtual image, as the light does not go through it. The distance of the image from the mirror is equal to the distance of the object from the mirror. 32-2 Reflection; Image Formation by a Plane Mirror

25 Copyright © 2009 Pearson Education, Inc. 32-2 Reflection; Image Formation by a Plane Mirror Example 32-2: How tall must a full-length mirror be? A woman 1.60 m tall stands in front of a vertical plane mirror. What is the minimum height of the mirror, and how close must its lower edge be to the floor, if she is to be able to see her whole body? Assume her eyes are 10 cm below the top of her head.

26 ConcepTest 32.1Reflection When watching the Moon over the ocean, you often see a long streak of light on the surface of the water. This occurs because: 1) the Moon is very large 2) atmospheric conditions are just right 3) the ocean is calm 4) the ocean is wavy 5) motion of the Moon

27 angle of incidence also changesdifferent spots on the water can reflect the Mooninto your eyes at different times When the water surface changes, the angle of incidence also changes. Thus, different spots on the water can reflect the Moon into your eyes at different times. ConcepTest 32.1Reflection When watching the Moon over the ocean, you often see a long streak of light on the surface of the water. This occurs because: 1) the Moon is very large 2) atmospheric conditions are just right 3) the ocean is calm 4) the ocean is wavy 5) motion of the Moon Follow-up: Where else does this occur?

28 Copyright © 2009 Pearson Education, Inc. Spherical mirrors are shaped like sections of a sphere, and may be reflective on either the inside (concave) or outside (convex). 32-3 Formation of Images by Spherical Mirrors

29 Copyright © 2009 Pearson Education, Inc. Rays coming from a faraway object are effectively parallel. 32-3 Formation of Images by Spherical Mirrors

30 Copyright © 2009 Pearson Education, Inc. Parallel rays striking a spherical mirror do not all converge at exactly the same place if the curvature of the mirror is large; this is called spherical aberration. 32-3 Formation of Images by Spherical Mirrors

31 Copyright © 2009 Pearson Education, Inc. If the curvature is small, the focus is much more precise; the focal point is where the rays converge. 32-3 Formation of Images by Spherical Mirrors

32 Copyright © 2009 Pearson Education, Inc. Using geometry, we find that the focal length is half the radius of curvature: Spherical aberration can be avoided by using a parabolic reflector; these are more difficult and expensive to make, and so are used only when necessary, such as in research telescopes. 32-3 Formation of Images by Spherical Mirrors

33 Copyright © 2009 Pearson Education, Inc. We use ray diagrams to determine where an image will be. For mirrors, we use three key rays, all of which begin on the object: 1.A ray parallel to the axis; after reflection it passes through the focal point. 2.A ray through the focal point; after reflection it is parallel to the axis. 3.A ray perpendicular to the mirror; it reflects back on itself. 32-3 Formation of Images by Spherical Mirrors

34 Copyright © 2009 Pearson Education, Inc. 32-3 Formation of Images by Spherical Mirrors

35 Copyright © 2009 Pearson Education, Inc. The intersection of these three rays gives the position of the image of that point on the object. To get a full image, we can do the same with other points (two points suffice for may purposes). 32-3 Formation of Images by Spherical Mirrors

36 Copyright © 2009 Pearson Education, Inc. Geometrically, we can derive an equation that relates the object distance, image distance, and focal length of the mirror: 32-3 Formation of Images by Spherical Mirrors

37 Copyright © 2009 Pearson Education, Inc. We can also find the magnification (ratio of image height to object height): The negative sign indicates that the image is inverted. This object is between the center of curvature and the focal point, and its image is larger, inverted, and real. 32-3 Formation of Images by Spherical Mirrors

38 Copyright © 2009 Pearson Education, Inc. 32-3 Formation of Images by Spherical Mirrors Example 32-4: Image in a concave mirror. A 1.50-cm-high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.0 cm. Determine (a)the position of the image, and (b) its size.

39 Copyright © 2009 Pearson Education, Inc. 32-3 Formation of Images by Spherical Mirrors Conceptual Example 32-5: Reversible rays. If the object in this figure is placed where the image is, where will the new image be? Figure 32-16 goes here.

40 Copyright © 2009 Pearson Education, Inc. If an object is outside the center of curvature of a concave mirror, its image will be inverted, smaller, and real. 32-3 Formation of Images by Spherical Mirrors

41 Copyright © 2009 Pearson Education, Inc. 32-3 Formation of Images by Spherical Mirrors Example 32-6: Object closer to concave mirror. A 1.00-cm-high object is placed 10.0 cm from a concave mirror whose radius of curvature is 30.0 cm. (a) Draw a ray diagram to locate (approximately) the position of the image. (b) Determine the position of the image and the magnification analytically.

42 Copyright © 2009 Pearson Education, Inc. For a convex mirror, the image is always virtual, upright, and smaller. 32-3 Formation of Images by Spherical Mirrors

43 Copyright © 2009 Pearson Education, Inc. Problem Solving: Spherical Mirrors 1.Draw a ray diagram; the image is where the rays intersect. 2.Apply the mirror and magnification equations. 3.Sign conventions: if the object, image, or focal point is on the reflective side of the mirror, its distance is positive, and negative otherwise. Magnification is positive if image is upright, negative otherwise. 4.Check that your solution agrees with the ray diagram. 32-3 Formation of Images by Spherical Mirrors

44 Copyright © 2009 Pearson Education, Inc. 32-3 Formation of Images by Spherical Mirrors Example 32-7: Convex rearview mirror. An external rearview car mirror is convex with a radius of curvature of 16.0 m. Determine the location of the image and its magnification for an object 10.0 m from the mirror.

45 Copyright © 2009 Pearson Education, Inc. Far Side – Gary Larson

46 Copyright © 2009 Pearson Education, Inc. In general, light slows somewhat when traveling through a medium. The index of refraction of the medium is the ratio of the speed of light in vacuum to the speed of light in the medium: 32-4 Index of Refraction

47 Copyright © 2009 Pearson Education, Inc. Light changes direction when crossing a boundary from one medium to another. This is called refraction, and the angle the outgoing ray makes with the normal is called the angle of refraction. 32-5 Refraction: Snell’s Law

48 Copyright © 2009 Pearson Education, Inc. Refraction is what makes objects half- submerged in water look odd. 32-5 Refraction: Snell’s Law

49 Copyright © 2009 Pearson Education, Inc. The angle of refraction depends on the indices of refraction, and is given by Snell’s law: 32-5 Refraction: Snell’s Law

50 Copyright © 2009 Pearson Education, Inc. 32-5 Refraction: Snell’s Law Example 32-9: Apparent depth of a pool. A swimmer has dropped her goggles to the bottom of a pool at the shallow end, marked as 1.0 m deep. But the goggles don’t look that deep. Why? How deep do the goggles appear to be when you look straight down into the water?

51 ConcepTest 32.4aRefraction I 1 air Parallel light rays cross interfaces from air into two different media, 1 and 2, as shown in the figures below. In which of the media is the light traveling faster? 1) medium 1 2) medium 2 3) both the same 2

52 ConcepTest 32.4aRefraction I 1 air The greater the difference in the speed of light between the two media, the greater the bending of the light rays. Parallel light rays cross interfaces from air into two different media, 1 and 2, as shown in the figures below. In which of the media is the light traveling faster? 1) medium 1 2) medium 2 3) both the same 2 Follow-up: air 12? Follow-up: How does the speed in air compare to that in # 1 or # 2?

53 ConcepTest 32.5aGone Fishin’ I To shoot a fish with a gun, should you aim directly at the image, slightly above, or slightly below? 1) aim directly at the image 2) aim slightly above 3) aim slightly below

54 ConcepTest 32.5aGone Fishin’ I higher aimlower Due to refraction, the image will appear higher than the actual fish, so you have to aim lower to compensate. To shoot a fish with a gun, should you aim directly at the image, slightly above, or slightly below? 1) aim directly at the image 2) aim slightly above 3) aim slightly below

55 Copyright © 2009 Pearson Education, Inc. 32-6 Visible Spectrum and Dispersion The visible spectrum contains the full range of wavelengths of light that are visible to the human eye.

56 Copyright © 2009 Pearson Education, Inc. 32-6 Visible Spectrum and Dispersion The index of refraction of many transparent materials, such as glass and water, varies slightly with wavelength. This is how prisms and water droplets create rainbows from sunlight.

57 Copyright © 2009 Pearson Education, Inc. 32-6 Visible Spectrum and Dispersion This spreading of light into the full spectrum is called dispersion.

58 Copyright © 2009 Pearson Education, Inc. If light passes into a medium with a smaller index of refraction, the angle of refraction is larger. There is an angle of incidence for which the angle of refraction will be 90°; this is called the critical angle: 32-7 Total Internal Reflection; Fiber Optics

59 Copyright © 2009 Pearson Education, Inc. If the angle of incidence is larger than this, no transmission occurs. This is called total internal reflection. 32-7 Total Internal Reflection; Fiber Optics

60 Copyright © 2009 Pearson Education, Inc. 32-7 Total Internal Reflection; Fiber Optics Conceptual Example 32-11: View up from under water. Describe what a person would see who looked up at the world from beneath the perfectly smooth surface of a lake or swimming pool.

61 Copyright © 2009 Pearson Education, Inc. 32-7 Total Internal Reflection; Fiber Optics Optical fibers also depend on total internal reflection; they are therefore able to transmit light signals with very small losses.


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