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1. 2 Today’s agendum: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave.

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Presentation on theme: "1. 2 Today’s agendum: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave."— Presentation transcript:

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2 2 Today’s agendum: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave.

3 3 We began this course by studying fields that didn’t vary with time—the electric field due to static charges, and the magnetic field due to a constant current. In case you didn’t notice—about five lectures ago things started moving! We found that changing magnetic field gives rise to an electric field. Also a changing electric field gives rise to a magnetic field. These time-varying electric and magnetic fields can propagate through space.

4 4 Electromagnetic Waves These four equations provide a complete description of electromagnetism. Maxwell’s Equations

5 5 Production of Electromagnetic Waves Apply a sinusoidal voltage to an antenna. Charged particles in the antenna oscillate sinusoidally. The accelerated charges produce sinusoidally varying electric and magnetic fields, which extend throughout space. The fields do not instantaneously permeate all space, but propagate at the speed of light. direction of propagation y z x

6 6 This static image doesn’t show how the wave propagates. Here are a couple of animations, available on-line: direction of propagation y z x

7 7 Electromagnetic waves are transverse waves, but are not mechanical waves (they need no medium to vibrate in). direction of propagation Therefore, electromagnetic waves can propagate in free space. At any point, the magnitudes of E and B (of the wave shown) depend only upon x and t, and not on y or z. A collection of such waves is called a plane wave. y z x

8 8 Manipulation of Maxwell’s equations leads to the following plane wave equations for E and B: These equations have solutions: You can verify this by direct substitution. where E max and B max in these notes are sometimes written by others as E 0 and B 0.

9 9 You can also show that At every instant, the ratio of the magnitude of the electric field to the magnitude of the magnetic field in an electromagnetic wave equals the speed of light.

10 10 Summary of Important Properties of Electromagnetic Waves The solutions of Maxwell’s equations are wave-like with both E and B satisfying a wave equation. Electromagnetic waves travel through empty space with the speed of light c = 1/(  0  0 ) ½. E max and B max are the electric and magnetic field amplitudes.

11 11 Summary of Important Properties of Electromagnetic Waves The components of the electric and magnetic fields of plane EM waves are perpendicular to each other and perpendicular to the direction of wave propagation. The latter property says that EM waves are transverse waves. The magnitudes of E and B in empty space are related by E/B = c. direction of propagation y z x

12 12 Today’s agendum: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave.

13 13 The magnitude S represents the rate at which energy flows through a unit surface area perpendicular to the direction of wave propagation. Energy Carried by Electromagnetic Waves Electromagnetic waves carry energy, and as they propagate through space they can transfer energy to objects in their path. The rate of flow of energy in an electromagnetic wave is described by a vector S, called the Poynting vector.* Thus, S represents power per unit area. The direction of S is along the direction of wave propagation. The units of S are J/(s·m 2 ) =W/m 2. *J. H. Poynting, 1884.

14 14 z x y c S E B Because B = E/c we can write These equations for S apply at any instant of time and represent the instantaneous rate at which energy is passing through a unit area. For an EM wave so

15 15 The time average of sin 2 (kx -  t) is ½, so EM waves are sinusoidal. The average of S over one or more cycles is called the wave intensity I.

16 16 The energy densities (energy per unit volume) associated with electric field and magnetic fields are: Using B = E/c and c = 1/(  0  0 ) ½ we can write Energy Density

17 17 For an electromagnetic wave, the instantaneous energy density associated with the magnetic field equals the instantaneous energy density associated with the electric field. Hence, in a given volume the energy is equally shared by the two fields. The total energy density is equal to the sum of the energy densities associated with the electric and magnetic fields:

18 18 When we average this instantaneous energy density over one or more cycles of an electromagnetic wave, we again get a factor of ½ from the time average of sin 2 (kx -  t). so we see that Recall The intensity of an electromagnetic wave equals the average energy density multiplied by the speed of light. and

19 19 Example: a radio station on the surface of the earth radiates a sinusoidal wave with an average total power of 50 kW. Assuming the wave is radiated equally in all directions above the ground, find the amplitude of the electric and magnetic fields detected by a satellite 100 km from the antenna. R Station Satellite All the radiated power passes through the hemispherical surface* so the average power per unit area (the intensity) is *In problems like this you need to ask whether the power is radiated into all space or into just part of space.

20 20 R Station Satellite

21 21 Example: for the radio station in the example on the previous two slides, calculate the average energy densities associated with the electric and magnetic field.

22 22 Today’s agendum: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave.

23 23 Momentum and Radiation Pressure EM waves carry linear momentum as well as energy. When this momentum is absorbed at a surface pressure is exerted on that surface. If we assume that EM radiation is incident on an object for a time  t and that the radiation is entirely absorbed by the object, then the object gains energy  U in time  t. Maxwell showed that the momentum change of the object is then: The direction of the momentum change of the object is in the direction of the incident radiation. incident

24 24 If instead of being totally absorbed the radiation is totally reflected by the object, and the reflection is along the incident path, then the magnitude of the momentum change of the object is twice that for total absorption. The direction of the momentum change of the object is again in the direction of the incident radiation. incident reflected

25 25 The radiation pressure on the object is defined as the force per unit area: From Newton’s 2 nd Law (F = dp/dt) we have: For total absorption, So Radiation Pressure incident (Equations on this slide involve magnitudes of vector quantities.)

26 26 This is the instantaneous radiation pressure in the case of total absorption: For the average radiation pressure, replace S by =S avg =I:

27 27 Following similar arguments it can be shown that: incident reflected

28 28 Example: a satellite orbiting the earth has solar energy collection panels with a total area of 4.0 m2. If the sun’s radiation is incident perpendicular to the panels and is completely absorbed find the average solar power absorbed and the average force associated with the radiation pressure. The intensity (I or Saverage) of sunlight prior to passing through the earth’s atmosphere is 1.4 kW/m2. Assuming total absorption of the radiation: Caution! The letter P (or p) has been used in this lecture for power, pressure, and momentum!

29 29 Today’s agendum: Introduction to Light. You must develop a general understanding of what light is and how it behaves. Reflection and Refraction (Snell’s “Law”). You must be able to determine the path of light rays using the laws of reflection and refraction. Total Internal Reflection and Fiber Optics. You must be able to determine the conditions under which total internal reflection occurs, and apply total internal reflection to fiber optic and similar materials. Dispersion. You must understand that the index of refraction of a material is wavelength-dependent.

30 30 Light Normally, “light” refers to the type of electromagnetic wave that stimulates the retina of our eyes.light Light acts like a wave except when it acts like particles.

31 31 *Light—Waves or Particles? (quantum007.jpg) *Both! Take Physics 203 for further enlightenment!

32 32 Light is a type of electromagnetic wave and travels with the speed c = x10 8 m/s in a vacuum. (Just use 3x10 8 !) The Speed of Light How many physicists does it take to change a light bulb? Eleven. One to do it and ten to co-author the paper.

33 33 Visible light is a small part of the electromagnetic spectrum.

34 34 Although light is actually an electromagnetic wave, it generally travels in straight lines (like particles do!). We can describe many properties of light by assuming that it travels in straight-line paths in the form of rays. A ray is a straight line along which light is propagated. In other contexts, the definition of ray might be extended to include bent or curved lines. Geometric Optics (ray.jpg)

35 35 A light ray is an infinitely thin beam of light. Of course, there really isn’t such a thing, but the concept helps us visualize properties of light. (rays.jpg) there really isn’t such a thing

36 36 Light rays from some external source strike an object and reflect off it in all directions. We only see those light rays that reflect in the direction of our eyes. If you can see something, it must be reflecting light! Zillions* of rays are simultaneously reflected in all directions from any point of an object. Later, when we study mirrors and lenses, we won’t try do draw them all! Just enough representative ones to understand what the light is doing. *one zillion = 10 a big number

37 37 Today’s agendum: Introduction to Light. You must develop a general understanding of what light is and how it behaves. Reflection and Refraction (Snell’s “Law”). You must be able to determine the path of light rays using the laws of reflection and refraction. Total Internal Reflection and Fiber Optics. You must be able to determine the conditions under which total internal reflection occurs, and apply total internal reflection to fiber optic and similar materials. Dispersion. You must understand that the index of refraction of a material is wavelength-dependent.

38 38 Light striking a surface may be reflected, transmitted, or absorbed. Reflected light leaves the surface at the same angle it was incident on the surface: ii rr Reflection Real Important Note: the angles are measured relative to the surface normal.

39 39 Reflection from a smooth surface is specular (mirror- like). Reflection from a rough surface is diffuse (not mirror-like).

40 40 Light travels in a straight line except when it is reflected or when it moves from one medium to another. Refraction—the “bending” of light rays when light moves from one medium to a different one—takes place because light travels with different speeds in different media. Refraction

41 41 The speed of light in a vacuum is c = 3x10 8 m/s. The index of refraction of a material is defined by where c is the speed of light in a vacuum and v is the speed of light in the material. The speed and wavelength of light change when it passes from one medium to another, but not the frequency, so If you study light in advanced classes, you’ll find it is more complex than this.

42 42 Because light never travels faster than c, n  1. For water, n = 1.33 and for glass, n  1.5. Indices of refraction for several materials are listed in your text. Example: calculate the speed of light in diamond (n = 2.42).

43 43 When light moves from one medium into another, some is reflected at the boundary, and some is transmitted. The transmitted light is refracted (“bent”).  a is the angle of incidence, and  b is the angle of refraction. Snell’s “Law” air (n a ) water (n b ) aa bb n b >n a incident ray refracted ray air (n b ) water (n a ) bb aa n a >n b incident ray refracted ray

44 44 air (n a ) water (n b ) aa bb n b >n a incident ray refracted ray Light passing from air (n  1) into water (n  1.33). Light “bends” towards the normal to the surface as it slows down in water.

45 45 air (n b ) water (n a ) bb aa n a >n b incident ray refracted ray Light passing from water (n  1.33) into air (n  1). Light “bends” away from the normal to the surface as it speeds up in air.

46 46 Snell’s “Law”, also called the law of refraction, gives the relationship between angles and indices of refraction: aa bb water (n b ) air (n a ) aa bb water (n b ) air (n a )  is the angle the ray makes with the normal! You are free to choose which is “a” and which is “b.”

47 47 Today’s agendum: Introduction to Light. You must develop a general understanding of what light is and how it behaves. Reflection and Refraction (Snell’s “Law”). You must be able to determine the path of light rays using the laws of reflection and refraction. Total Internal Reflection and Fiber Optics. You must be able to determine the conditions under which total internal reflection occurs, and apply total internal reflection to fiber optic and similar materials. Dispersion. You must understand that the index of refraction of a material is wavelength-dependent.

48 48 Suppose n 2

49 49  1 <  C  1 close to  C  1 >  C 11 Another visualization here. here

50 50 n2n2 n 1 >n 2 Ray incident normal to surface is not “bent.”  Ray incident normal to surface is not “bent.” Some is reflected,Ray incident normal to surface is not “bent.” Some is reflected, some is transmitted.

51 51 n2n2 n 1 >n 2  Increasing angle of incidence…

52 52 n2n2 n 1 >n 2  Increasing angle of incidence…more…

53 53 n2n2 n 1 >n 2  Increasing angle of incidence…more…critical angle reached…Increasing angle of incidence…more…critical angle reached… some of incident energy is reflected, some is “transmitted along the boundary layer.

54 54 n2n2 n 1 >n 2  Light incident at any angle beyond  C is totally internally reflected.

55 55 application: fiber optics ea/report.html

56 56 ii ff n i =1 (air) n f >1 Light is incident at an angle  i on a transparent fiber. The light refracts at an angle  f. Example: determine the incident angle  i for which light strikes the inner surface of a fiber optic cable at the critical angle.

57 57 ii ff 90  -  f n i =1 (air) n f >1 Light strikes the fiber wall an an angle of 90-  f normal to the surface. At the critical angle, instead of exiting the fiber, the refracted light travels along the fiber-air boundary. In this case, 90  -  f is the critical angle. 90  Solve the above for  f and use to solve for  i.

58 58 application: swimming underwaterswimming underwater If you are looking up from underwater, if your angle of sight (relative to the normal to the surface) is too large, you see an underwater reflection instead of what’s above the water.

59 59 application: perfect mirrorsperfect mirrors (used in binoculars) application: diamondsdiamonds

60 60 Today’s agendum: Introduction to Light. You must develop a general understanding of what light is and how it behaves. Reflection and Refraction (Snell’s “Law”). You must be able to determine the path of light rays using the laws of reflection and refraction. Total Internal Reflection and Fiber Optics. You must be able to determine the conditions under which total internal reflection occurs, and apply total internal reflection to fiber optic and similar materials. Dispersion. You must understand that the index of refraction of a material is wavelength-dependent.

61 61 Dispersion We’ve treated the index of refraction of a material as if it had a single value for all wavelengths of light. In fact, the index of refraction is generally wavelength- (or color-) dependent. When white light passes from air into glass, the different colours are refracted by different angles, and therefore spread out, or are dispersed. It is observed that the shorter the wavelength of the light, the greater is the refraction.

62 62 Picture from the Exploratorium (http://www.exploratorium.edu/).

63 63 Today’s agendum: Plane Mirrors. You must be able to draw ray diagrams for plane mirrors, and be able to calculate image and object heights, distances, and magnifications. Spherical Mirrors: concave and convex mirrors. You must understand the differences between these two kinds of mirrors, be able to draw ray diagrams for both kinds of mirrors, and be able to solve the mirror equation for both kinds of mirrors.

64 64 Mirrors Plane mirrors form virtual images; no light actually comes from the image. The solid red rays show the actual light path after reflection; the dashed black rays show the perceived light path. Images Formed by Plane Mirrors

65 65 *The object distance and image distance are equal: s=-s’. The object height and image height are equal: y=y’. The magnification of a plane mirror is therefore one. ss’ y’y The image is upright and virtual. The image is reversed front-to-back relative to the object. *The – sign is needed because of sign conventions—see later.

66 66 ss’ Example: how tall must a full-length mirror be? A light ray from the top of your head reflects directly back from the top of the mirror.

67 67 ss’ y/2    To reach your eye, a light ray from your foot must reflect halfway up the mirror (because  I =  R =  ).

68 68 ss’ y/2    The mirror needs to be only half as tall as you. This calculation assumed your eyes are at the top of your head.

69 69 Today’s agendum: Plane Mirrors. You must be able to draw ray diagrams for plane mirrors, and be able to calculate image and object heights, distances, and magnifications. Spherical Mirrors: concave and convex mirrors. You must understand the differences between these two kinds of mirrors, be able to draw ray diagrams for both kinds of mirrors, and be able to solve the mirror equation for both kinds of mirrors.

70 70 Images Formed by Spherical Mirrors Spherical mirrors are made from polished sections cut from a spherical surface. The center of curvature, C, is the center of the sphere, of which the mirror is a section. Of course, you don’t really make these mirrors by cutting out part of a sphere of glass. C

71 71 The radius of curvature, R, is the radius of the sphere, or the distance from V to C. CV R

72 72 The principal axis (or optical axis) is the line that passes through the center of curvature and the center of the mirror. CV R Principal or Optical Axis The center of the mirror is often called the vertex of the mirror.

73 73 Paraxial rays are parallel to the principal axis of the mirror (from an object infinitely far away). Reflected paraxial rays pass through a common point known as the focal point F. CV F

74 74 The focal length f is the distance from P to F. Your text shows that f = R/2. CP F f R

75 75 Reality check: paraxial rays don’t really pass exactly through the focal point of a spherical mirror (“spherical aberration”). CV F

76 76 If the mirror is small compared to its radius of curvature, or the object being imaged is close to the principal axis, then the rays essentially all focus at a single point. CV F We will assume mirrors with large radii of curvature and objects close to the principal axis.

77 77 In “real life” you would minimize aberration by using a parabolic mirror. CV F

78 78 Today’s agendum: Plane Mirrors. You must be able to draw ray diagrams for plane mirrors, and be able to calculate image and object heights, distances, and magnifications. Spherical Mirrors: concave and convex mirrors. You must understand the differences between these two kinds of mirrors, be able to draw ray diagrams for both kinds of mirrors, and be able to solve the lens equation for both kinds of mirrors.

79 79 Concave and Convex Mirrors There are two kinds of spherical mirrors: concave and convex. F concaveconvex

80 80 Ray Diagrams for Mirrors We can use three “principal rays” to construct images. In this example, the object is “outside” of F. Ray 1 is parallel to the axis and reflects through F. Ray 2 passes through F before reflecting parallel to the axis. Ray 3 passes through C and reflects back on itself. F C We’ll also use three for convex mirrors, but there will be a different version of ray 2.

81 81 We use three “principal rays” to construct images. C An image is formed where the rays converge. The image from a concave mirror, object outside the focal point, is real, inverted, and smaller than the object. “Real” image: you could put a camera there and detect the image. F Ray Diagrams for Concave Mirrors Two rays would be enough to show us where the image is. We include the third ray for “safety.” You don’t have to use principal rays, but they are easiest to trace.

82 82 C The image from a concave mirror, object inside the focal point, is virtual, upright, and larger than the object. Ray 1: parallel to the axis then through F. Ray 2: “through” F then parallel to the axis. Ray 3: “through” C. F With this size object, there was a bit of spherical aberration present, and I had to “cheat” my C a bit to the left to make the diagram look “nice.”

83 83 C You could show that if an object is placed at the focal point, reflected rays all emerge parallel, and *no image is formed. Ray 1: parallel to the axis then through F. Ray 2: “through” F then parallel to the axis. Can’t do! Ray 3: through C. F no image *Actually, the image is formed at infinity. Worth thinking about: what if the object is placed between F and C?

84 84 With a bit of geometry, you can show that C The magnification is the ratio of the image to the object height: F The Mirror Equation f s s’ y y’

85 85 Sign conventions for the mirror equation: C F f s s’ y y’ When the object, image, or focal point is on the reflecting side of the mirror, the distance is positive. When the object, image, or focal point is “behind” the mirror, the distance is negative. The image height is positive if the image is upright, and negative if the image is inverted relative to the object.

86 86 Example: a dime (height is 1.8 cm) is placed 100 cm away from a concave mirror. The image height is 0.9 cm and the image is inverted. What is the focal length of the mirror. C F f s s’ y y’ s’, s, or f on reflecting side are + y is – if image is inverted

87 87 Applications of concave mirrors. Shaving mirrors. Makeup mirrors. Solar cookers. Flashlights, headlamps, stove reflectors. Satellite dishes (when used with electromagnetic radiation).

88 88 Today’s agendum: Plane Mirrors. You must be able to draw ray diagrams for plane mirrors, and be able to calculate image and object heights, distances, and magnifications. Spherical Mirrors: concave and convex mirrors. You must understand the differences between these two kinds of mirrors, be able to draw ray diagrams for both kinds of mirrors, and be able to solve the lens equation for both kinds of mirrors.

89 89 C F Ray Diagrams for Convex Mirrors Ray 1: parallel to the axis then through F. Ray 2: “through” Vertex. Ray 3: “through” C. The image is virtual, upright, and smaller than the object.

90 90 C F Instead of sending ray 2 “through” V, we could have sent it “through” F. The ray is reflected parallel to the principal axis. Your text talks about all four of the “principal rays” we have used.

91 91 C F Because they are on the “other” side of the mirror from the object, s’ and f are negative. The mirror equation still works for convex mirrors. s f s’ y y’

92 92 The ray diagram looks like the one on the previous slide, but with the object much further away (difficult to draw). On reflecting side  positive. Not on reflecting side  negative. Example: a convex rearview car mirror has a radius of curvature of 40 cm. Determine the location of the image and its magnification for an object 10 m from the mirror.

93 93 …algebra… Remind me… what does it say on passenger side rear view mirrors?

94 94 Applications of convex mirrors. Passenger side rear-view mirrors. Grocery store aisle mirrors. Anti-shoplifting (surveillance) mirrors. Christmas tree ornaments. Railroad crossing mirrors.

95 95 Sign Conventions Introduced Today When the object, image, or focal point is on the reflecting side of the mirror, the distance is positive. When the object, image, or focal point is “behind” the mirror, the distance is negative. The image height is positive if the image is upright, and negative if the image is inverted relative to the object.

96 96 Summary of Sign Conventions Object Distance. When the object is on the same side as the incoming light, the object distance is positive (otherwise is negative). Here’s a compact way of expressing mirror and lens (coming soon) sign conventions all at once. Image Distance. When the image is on the same side as the outgoing light, the image distance is positive (otherwise is negative). Radius of Curvature. When the center of curvature C is on the same side as the outgoing light, R is positive (otherwise is negative).

97 97 Today’s agendum: Refraction at Spherical Surfaces. You must be able to calculate properties of images formed by refraction at spherical surfaces. Thin Lenses: Concave and Convex Lenses, Ray Diagrams, Solving the Lens Equation. You must understand the differences between these two kinds of lenses, be able to draw ray diagrams for both kinds of lenses, and be able to solve the lens equation for both kinds of lenses. If Time Allows: Lens Combinations, Optical Instruments. You should be aware of this useful information.

98 98 Refraction at Spherical Surfaces Convex surface: CF axis R f 11 22 nana n b >n a Geometry: a light ray parallel to the axis passes through F.

99 99 An extended object will form an image inside the n b medium. CF axis R f 11 22 nana n b >n a This image is real and inverted. s s’ Ray 1: parallel to the axis, through F. Ray 3: through C.

100 100 Concave surface: CF R n b >n a Geometry: a light ray parallel to the axis seems to have come from F. axis nana f

101 101 CF R n b >n a The image is virtual and upright. axis nana f An extended object will form an image inside the n a medium. Ray 1: parallel to the axis, through F. Ray 3: through C. There are three different places to put the object. The different images formed are always virtual and upright.

102 102 We can use geometry to derive an equation relating the image and source positions, and an equation for the magnification. CF axis R f nana nbnb s s’

103 103 CF R nbnb axis nana f The same equations work for concave surfaces. s s’

104 104 Approximations Were Used! The equations in this section are excellent approximations if both the angles of incidence and refraction are small.

105 105 Sign Conventions  R is positive when it is in the medium into which the light propagates. R is negative when it is in the medium from which the light radiates.  The image distance is positive when the image is in the medium into which the light propagates, and negative if it is in the medium from which the light radiates (virtual image).  The object distance is positive when the object is in the medium from which the light radiates (the usual case—a real object), and negative if on the side opposite to the light source (a virtual object). These are really “the same” as for mirrors.

106 106 Example: a Jurassic mosquito is discovered embedded in an amber sphere which has an index of refraction of 1.6. The radius of curvature of the sphere is 3.0 mm. The mosquito is located on the principal axis and appears to be imbedded 5.0 mm into the amber. How deep is the mosquito really? R s s’ n a =1.6n b =1 The object is in the amber, so n a =1.6 and n b =1. The image is in the medium from which the light radiates so s’=-5.0 mm. Notice the reversed orientation...

107 107 R s s’ n a =1.6n b =1 R is negative because it is in the medium from which the light radiates. R=-3.0 mm.

108 108 Today’s agendum: Refraction at Spherical Surfaces. You must be able to calculate properties of images formed by refraction at spherical surfaces. Thin Lenses: Concave and Convex Lenses, Ray Diagrams, Solving the Lens Equation. You must understand the differences between these two kinds of lenses, be able to draw ray diagrams for both kinds of lenses, and be able to solve the lens equation for both kinds of lenses. If Time Allows: Lens Combinations, Optical Instruments. You should be aware of this useful information.

109 109 Thin Lenses A lens in this section is taken to be a single object made of transparent material of refractive material n>1. There are two surface boundaries. Light from an object incident on the first surface forms an image, which becomes the object for the second surface. A thin lens is one for which the distance from the object to each of the two surfaces is the “same” (and the distance from the image to each surface is the “same”). This would NOT qualify as a thin lens.

110 110 Until I figure out how to use Powerpoint to fill in the lens color, I will make my lenses look “hollow,” like this. There are several surface combinations from which we can make lenses. Here are three (there are more).

111 111 Thin lenses can be converging or diverging. Converging and Diverging Lenses The converging lens is thicker in the center. The diverging lens is thicker at the edges. There are focal points on both sides of each lens. The focal length is the same whether light passes from left to right or right to left.

112 112 There are two surfaces at which light refracts. Our equations (provided later) “automatically” take care of this. In your diagrams, simply draw the incident ray up to the center of the lens, then draw the refracted ray in its final direction.

113 113 Today’s agendum: Refraction at Spherical Surfaces. You must be able to calculate properties of images formed by refraction at spherical surfaces. Thin Lenses: Concave and Convex Lenses, Ray Diagrams, Solving the Lens Equation. You must understand the differences between these two kinds of lenses, be able to draw ray diagrams for both kinds of lenses, and be able to solve the lens equation for both kinds of lenses. If Time Allows: Lens Combinations, Optical Instruments. You should be aware of this useful information.

114 114 Ray Diagrams for Converging Lenses Ray 1 is parallel to the axis and refracts through F. Ray 2 passes through F’ before refracting parallel to the axis. Ray 3 passes straight through the center of the lens. The image is real and inverted. In this case, it is larger than the object. O F’F I

115 115 Ray Diagrams for Diverging Lenses Ray 1 is parallel to the axis and refracts as if through F. Ray 2 heads towards F’ before refracting parallel to the axis. Ray 3 passes straight through the center of the lens. The image is virtual and upright. It is smaller than the object. O FF’I

116 116 Converging and Diverging Lenses The image formed by a diverging lens is always virtual, upright, and smaller than the object. See this web page.this The image formed by a converging lens may be real, inverted, and either smaller or larger than the object. It may also be virtual, upright, and larger than the object. See this web page.this Do these lens properties remind you of anything you’ve studied recently?

117 117 Today’s agendum: Refraction at Spherical Surfaces. You must be able to calculate properties of images formed by refraction at spherical surfaces. Thin Lenses: Concave and Convex Lenses, Ray Diagrams, Solving the Lens Equation. You must understand the differences between these two kinds of lenses, be able to draw ray diagrams for both kinds of lenses, and be able to solve the lens equation for both kinds of lenses. If Time Allows: Lens Combinations, Optical Instruments. You should be aware of this useful information.

118 118 The Lensmaker’s Equation s s s’

119 119 Sign Conventions for The Lens Equation The object distance s is positive if the object is on the side of the lens from which the light is coming; otherwise s is negative. The image distance s’ is positive if the image is on the opposite side of the lens from where the light is coming; otherwise s’ is negative. The focal length f is positive for converging lenses and negative for diverging lenses. The image height y’ is positive if the image is upright and negative if the image is inverted relative to the object.

120 120 Example: an object is located 5 cm in front of a converging lens of 10 cm focal length. Find the image distance and magnification. Is the image real or virtual? O FF’ Image distance is 10 cm, image is on side of lens light is coming from, so image is virtual. M=2 so image is upright. It’s just a coincidence that the image is located at F’.

121 121 Today’s agendum: Refraction at Spherical Surfaces. You must be able to calculate properties of images formed by refraction at spherical surfaces. Thin Lenses: Concave and Convex Lenses, Ray Diagrams, Solving the Lens Equation. You must understand the differences between these two kinds of lenses, be able to draw ray diagrams for both kinds of lenses, and be able to solve the lens equation for both kinds of lenses. If Time Allows: Lens Combinations, Optical Instruments. You should be aware of this useful information.

122 122 Lens Combinations To determine the image formed by a combination of two lenses, simply... …calculate the image formed by the first lens… …then use the first lens image as the source (object) for the second lens. There is no homework on lens combinations.

123 123 Optical Instruments A Simple Magnifier h  O h ´´ Magnifier F I q p  25 cm (near point) O

124 124 Refracting Telescope For viewing very far objects. Object distance taken as infinity.

125 125 Terrestrial Telescopes For producing upright images: Galilean telescope Field-lens telescope

126 126 Reflecting Telescope Newtonian-focus reflecting telescope

127 127 Compound Microscope Again has objective and eyepiece, but because it is for viewing very near objects it is very different from the telescope. Eyepiece magnification: Objective magnification: Overall magnification:

128 128 Summary of Sign Conventions When the object, image, or focal point is on the reflecting side of the mirror, the distance is positive. When the object, image, or focal point is “behind” the mirror, the distance is negative. The image height is positive if the image is upright, and negative if the image is inverted relative to the object. Mirrors The object distance s is positive if the object is on the side of the lens from which the light is coming; otherwise s is negative (and the object is virtual). The image distance s’ and radius of curvature R are positive if the image is on the side of the lens into which the light is going; otherwise negative. The focal length f is positive for converging lenses and negative for diverging lenses. The image height is positive if the image is upright, and negative if the image is inverted relative to the object. Lenses

129 129 Summary of Sign Conventions Object Distance. When the object is on the same side as the incoming light, the object distance is positive (otherwise is negative). Here’s a more compact way of expressing the sign conventions all at once. Image Distance. When the image is on the same side as the outgoing light, the image distance is positive (otherwise is negative). Radius of Curvature. When the center of curvature C is on the same side as the outgoing light, R is positive (otherwise is negative).

130 130 Today’s agendum: Review of Waves. You are expected to recall facts about waves from Physics 103. Young’s Double Slit Experiment. You must understand how the double slit experiment produces an interference pattern. Conditions for Interference in the Double Slit Experiment. You must be able to calculate the conditions for constructive and destructive interference in the double slit experiment. Intensity in the Double Slit Experiment. You must be able to calculate intensities in the double slit experiment.

131 131 Interference Review of Waves This section is a review of material you learned in your previous physics course (perhaps Physics 103). Consider a wave described by The phase of this wave is Also y x

132 132 If  is constant with time (i.e., d  /dt=0), then we are moving with the wave, and The phase velocity, v p, is given by Imagine yourself riding on any point on this wave. The point you are riding moves to the right. The velocity it moves at is vp. If the wave is moving from left to right then  /k must be positive. y x

133 133 When waves of the same nature travel past some point at the same time, the amplitude at that point is the sum of the amplitudes of all the waves The amplitude of the electric field at a point is found by adding the instantaneous amplitudes, including the phase, of all electric waves at that point. In Physics 103 you may have learned that power (or intensity) is proportional to amplitude squared. The intensity of the superposed waves is proportional to the square of the amplitude of the resulting sum of waves. Superposition—a Characteristic of All Waves

134 134 Constructive Interference: If the waves are in phase, they reinforce to produce a wave of greater amplitude. Destructive Interference: If the waves are out of phase, they reinforce to produce a wave of reduced amplitude. Interference—a Result of the Superposition of Waves

135 135 Today’s agendum: Review of Waves. You are expected to recall facts about waves from Physics 103. Young’s Double Slit Experiment. You must understand how the double slit experiment produces an interference pattern. Conditions for Interference in the Double Slit Experiment. You must be able to calculate the conditions for constructive and destructive interference in the double slit experiment. Intensity in the Double Slit Experiment. You must be able to calculate intensities in the double slit experiment.

136 136 This experiment demonstrates the wave nature of light. Consider a single light source, and two slits. Each slit acts as a secondary source of light. Light waves from secondary slits interfere to produce alternating maxima and minima in the intensity. Reference and “toys:” fsu magnet lab, colorado.fsucolorado Young’s Double Slit Experiment Interesting reading: the double slit experiment and quantum mechanics.the double slit experiment and quantum mechanics

137 137 How does this work? Light waves from the two slits arriving at the detection screen in phase will interfere constructively and light waves arriving out of phase will interfere destructively. In phase— constructive. Out of phase— destructive.

138 138 Today’s agendum: Review of Waves. You are expected to recall facts about waves from Physics 103. Young’s Double Slit Experiment. You must understand how the double slit experiment produces an interference pattern. Conditions for Interference in the Double Slit Experiment. You must be able to calculate the conditions for constructive and destructive interference in the double slit experiment. Intensity in the Double Slit Experiment. You must be able to calculate intensities in the double slit experiment.

139 139 Conditions for Interference Sources must be monochromatic- of a single wavelength. Sources must be coherent-- must maintain a constant phase with respect to each other. Here’s the geometry I will use in succeeding diagrams.

140 140  d L2L2 L1L1  L = L 2 –L 1 = d sin  For an infinitely distant* screen:  P S2S2 S1S1 LL R y L1L1 L2L2  d  *so that all the angles labeled  are approximately equal

141 141 Destructive Interference: Constructive Interference: The parameter m is called the order of the interference fringe. The central bright fringe at  = 0 (m = 0) is known as the zeroth-order maximum. The first maximum on either side (m = ±1) is called the first-order maximum.  d L2L2 L1L1  L = L 2 –L 1 = d sin  

142 142  P S2S2 S1S1 LL R y L1L1 L2L2  d Bright fringes: Do not use the small-angle approximation unless it is valid!

143 143  P S2S2 S1S1 LL R y L1L1 L2L2  d Dark fringes: Do not use the small-angle approximation unless it is valid!

144 144 Example: a viewing screen is separated from the double-slit source by 1.2 m. The distance between the two slits is mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. Determine the wavelength of the light. Bright fringes:  P S2S2 S1S1 LL R y L1L1 L2L2 

145 145 Example: a viewing screen is separated from the double-slit source by 1.2 m. The distance between the two slits is mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. Find the distance between adjacent bright fringes. Bright fringes:  P S2S2 S1S1 LL R y L1L1 L2L2 

146 146 Example: a viewing screen is separated from the double-slit source by 1.2 m. The distance between the two slits is mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. Find the width of the bright fringes. Define the bright fringe width to be the distance between two adjacent destructive minima.  P S2S2 S1S1 LL R y L1L1 L2L2 

147 147 Today’s agendum: Review of Waves. You are expected to recall facts about waves from Physics 103. Young’s Double Slit Experiment. You must understand how the double slit experiment produces an interference pattern. Conditions for Interference in the Double Slit Experiment. You must be able to calculate the conditions for constructive and destructive interference in the double slit experiment. Intensity in the Double Slit Experiment. You must be able to calculate intensities in the double slit experiment.

148 148 Intensity in the Double Slit Experiment Our equations for the minima and maxima intensity positions are for the centers of the fringes. In this section, we calculate distribution of light intensity in the double-slit interference pattern.

149 149 The derivation of the double-slit intensity equation is not particularly difficult, so study it if you find derivations helpful for your understanding. A path length difference  L= corresponds to a phase difference of  =2 . A path length difference  L=m corresponds to a phase difference of  =2  m. In general, for non-integral m, the phase difference at P between the waves from S 1 and S 2 is

150 150 Your text writes the equation for the intensity distribution in the double-slit experiment in terms of the phase difference on the previous slide. Your starting equation for the intensity is where I 0 is 4 times the peak intensity of either of the two interfering waves: Why did my previous diagrams show this?

151 151 Today’s agendum: Interference Due to Reflection. Phase Change Due to Reflection. You must be able to determine whether or not a phase change occurs when a wave is reflected. Thin Film Interference. You must be able to calculate thin film thicknesses for constructive or destructive interference. Examples. You must be able to solve problems similar to these examples.

152 152 Light undergoes a phase change of 180° (  radians) upon reflection from a medium that has a higher index of refraction than the one in which the wave is traveling. Interference from Reflection Phase Change Due to Reflection

153 153 Today’s agendum: Interference Due to Reflection. Phase Change Due to Reflection. You must be able to determine whether or not a phase change occurs when a wave is reflected. Thin Film Interference. You must be able to calculate thin film thicknesses for constructive or destructive interference. Examples. You must be able to solve problems similar to these examples.

154 154 Thin film interference is caused by… Thin Film Interference …phase difference of reflected waves due to reflection off a higher-n material, and… …phase difference of reflected waves due to path length differences.

155 155 No phase change Air Film t n Air < n Film Do the reflected rays interfere destructively or constructively? Caution! The wavelength in the film is different than in air. 180° phase change Thin Film Interference Ray  undergoes a phase change on reflection.   Ray  has a phase change due to the path difference. Dark lines in drawings are there to help you see the boundaries, and are not a separate medium.

156 156 We will get destructive interference when the path difference is an integral number of wavelengths: Assume the incident light is nearly perpendicular to the film surface. No phase change Air Film t n Air < n Film 180° phase change The path length difference is approximately 2t. There is a 180  phase difference (½ of a wavelength) due to the first reflection.

157 157 We will get constructive interference when the path difference is a half-integral number of wavelengths: Assume the incident light is nearly perpendicular to the film surface. No phase change Air Film t n Air < n Film 180° phase change We get constructive interference when the path difference is film /2, 3 film /2, 5 film /2, etc.

158 158 No phase change Air Film t n Air < n Film 180° phase change You need to apply the reasoning used in deriving them to each of your thin film interference problems. These are only true when the film is surrounded by a medium with lower index of refraction than the film! The equations below are not on your starting equation sheet.

159 159 No phase change Air Film t n Air < n Film 180° phase change These are valid when the light is incident almost perpendicular to the film: Caution! The incident ray in the diagram clearly does not qualify visually as “almost perpendicular.” That’s because the angle relative to the normal is exaggerated for viewing convenience.

160 160 No phase change Air Film t n Air < n Film 180° phase change Caution! For truly non-perpendicular incidence, you have to take into account the extra path length of the ray reflected at the air-film interface (as well as the extra path length inside the film).

161 161 Thin Film Interference Problem Solving Tips  Identify the thin film causing the interference.  Determine the phase relationship between the portion of the wave reflected at the upper surface and the portion reflected at the lower surface.  Phase differences have two causes: (1) path differences and (2) phase changes upon reflection.  When the total phase change is an integer multiple of the wavelength (, 2, 3, etc.) the interference is constructive, and when it is a half-integer multiple of the wavelength ( /2, 3 /2, 5 /2, etc.) it is destructive.

162 162 Today’s agendum: Interference Due to Reflection. Phase Change Due to Reflection. You must be able to determine whether or not a phase change occurs when a wave is reflected. Thin Film Interference. You must be able to calculate thin film thicknesses for constructive or destructive interference. Examples. You must be able to solve problems similar to these examples.

163 163 Example: a glass lens is coated on one side with a thin film of MgF2 to reduce reflection from the lens surface. The index of refraction for MgF2 is 1.38 and for glass is What is the minimum thickness of MgF2 that eliminates reflection of light of wavelength λ = 550 nm? Assume approximately perpendicular angle of incidence for the light. Both rays  and  experience a 180  phase shift on reflection so the total phase difference is due to the path difference of the two rays. Air MgF 2 t n= 1.38 n Air = ° phase change 180° phase change   glass, n g =1.50

164 164 The minimum thickness is for m=0. The reflected light is minimum when the two light rays meet the condition for destructive interference: the path length difference is a half-integral multiple of the light wavelength in MgF 2. Air MgF 2 t n= 1.38 n Air = ° phase change 180° phase change   glass, n g =1.50

165 165 Example: two glass plates 10 cm long are in contact on one side and separated by a piece of paper 0.02 mm thick on the other side. What is the spacing between the interference fringes? Assume monochromatic light with a wavelength in air of λ = 500 nm incident perpendicular to the slides. Ray  is not phase shifted on reflection. Ray  is shifted 180  on reflection. H   t x L = 10 cm H = 2x10 -5 m For destructive interference The light that is partly reflected at the bottom of the first glass surface and partly transmitted is responsible for the interference fringes.* *This reference explains why there is no visible interference due to the relatively thick glass plates themselves.This

166 166 H   t x L = 10 cm H = 2x10 -5 m Successive dark fringes are separated by 1.25 mm. x is the distance from the contact point to where destructive interference takes place.

167 167 H   t x L = 10 cm H = 2x10 -5 m Successive bright fringes occur for m+½ and (m+1)+½. For constructive interference

168 168 H   t x L = 10 cm H = 2x10 -5 m Successive bright fringes are separated by 1.25 mm.

169 169 Example: suppose the glass plates have ng = 1.50 and the space between them contains water (nw = 1.33). What happens now? Ray  is not phase shifted on reflection. Ray  is shifted 180  on reflection. Both are the same as before. H   t x L = 10 cm H = 2x10 -5 m For destructive interference But the path difference now occurs in water, where the light will have a wavelength Repeat the calculation, using water.

170 170 H   t x L = 10 cm H = 2x10 -5 m Successive dark fringes are separated by 0.94 mm. For destructive interference, we now have

171 171 Two lectures ago I showed you these two plots of the intensity distribution in the double-slit experiment: Which is correct? Peak intensity varies with angle.Peak intensity independent of angle.

172 172 Diffraction Light is an electromagnetic wave, and like all waves, “bends” around obstacles. d <>d This bending, which is most noticeable when the dimension of the obstacle is close to the wavelength of the light, is called “diffraction.” Only waves diffract.

173 173 Diffraction pattern from a penny positioned halfway between a light source and a screen. The shadow of the penny is the circular dark spot. Notice the circular bright and dark fringes. The central bright spot is not a defect in the picture. It is a result of light “bending” around the edges of the penny and interfering constructively in the exact center of the shadow.

174 174 Single Slit Diffraction In the previous chapter we calculated the interference pattern from a pair of slits. One of the assumptions in the calculation was that the slit width was very small compared with the wavelength of the light. Now we consider the effect of finite slit width. We start with a single slit.  a      Each part of the slit acts as a source of light rays, and these different light rays interfere.

175 175      a/2 a  Divide the slit in half. Ray  travels farther* than ray  by (a/2)sin . Likewise for rays  and . If this path difference is exactly half a wavelength (corresponding to a phase difference of 180°) then the two waves will cancel each other and destructive interference results. Destructive interference: *All rays from the slit are converging at a point P very far to the right and out of the picture.

176 176      a/2 a  Destructive interference: If you divide the slit into 4 equal parts, destructive interference occurs when If you divide the slit into 6 equal parts, destructive interference occurs when

177 177      a/2 a  In general, destructive interference occurs when The above equation gives the positions of the dark fringes. The bright fringes are approximately halfway in between.

178 178 a  O x y Use this geometry for tomorrow’s single-slit homework problems. If  is small,* then it is valid to use the approximation sin   . (  must be expressed in radians.) *The approximation is quite good for angles of 10  or less, and not bad for even larger angles.

179 179 I won’t derive the intensity distribution for the single slit. The general features of that distribution are shown below. Most of the intensity is in the central maximum. It is twice the width of the other (secondary) maxima. Single Slit Diffraction Intensity

180 180 New starting equations for single-slit intensity: “Toy”

181 181 Example: 633 nm laser light is passed through a narrow slit and a diffraction pattern is observed on a screen 6.0 m away. The distance on the screen between the centers of the first minima outside the central bright fringe is 32 mm. What is the slit width? y 1 = (32 mm)/2tan  = y 1 /L tan   sin    for small 

182 182 The ability of optical systems to distinguish closely spaced objects is limited because of the wave nature of light. If the sources are far enough apart so that their central maxima do not overlap, their images can be distinguished and they are said to be resolved. Resolution of Single Slit (and Circular Aperture)

183 183 When the central maximum of one image falls on the first minimum of the other image the images are said to be just resolved. This limiting condition of resolution is called Rayleigh’s criterion.

184 184 From Rayleigh’s criterion we can determine the minimum angular separation of the sources at the slit for which the images are resolved. For a slit of width a: For a circular aperture of diameter D: Resolution is wavelength limited! These come from the small angle approximation, and geometry.

185 185 If a single slit diffracts, what about a double slit? Remember the double-slit interference pattern from the chapter on interference? If the slit width (not the spacing between slits) is small (i.e., comparable to the wavelength of the light), you must account for diffraction. interference only

186 186 S2S2 S1S1 d a  P L y r1r1 r2r2 Double Slit Diffraction

187 187 A diffraction grating consists of a large number of equally spaced parallel slits.  d  = d sin  The path difference between rays from any two adjacent slits is  = dsin . Interference maxima occur for Diffraction Gratings If  is equal to some integer multiple of the wavelength then waves from all slits will arrive in phase at a point on a distant screen.

188 188 Ok what’s with this equation monkey business? double-slit interference constructive single-slit diffraction destructive! diffraction grating constructive

189 189  d  = d sin  The intensity maxima are brighter and sharper than for the two slit case. Interference Maxima: Diffraction Grating Intensity Distribution

190 190

191 191 Application: spectroscopy You can view the atomic spectra for each of the elements here.here visible light hydrogen helium mercury

192 192 Interference Maxima: First-order violet: Example: the wavelengths of visible light are from approximately 400 nm (violet) to 700 nm (red). Find the angular width of the first-order visible spectrum produced by a plane grating with 600 slits per millimeter when white light falls normally on the grating.

193 193 First-order red: visible light 10.9 

194 194 No matter what the grating spacing, d, the largest angle for the 2 nd order spectrum (for the red end) is always greater than the smallest angle for the 3 rd order spectrum (for the violet end), so 2 nd and 3 rd orders always overlap. Example: for this diffraction grating show that the violet end of the third-order spectrum overlaps the red end of the second- order spectrum. Third-order violet: Second-order red:

195 195 Diffraction gratings let us measure wavelengths by spreading apart the diffraction maxima associated with different wavelengths. In order to distinguish two nearly equal wavelengths the diffraction must have sufficient resolving power, R. Consider two wavelengths λ 1 and λ 2 that are nearly equal. The average wavelength is and the difference is The resolving power is defined as Diffraction Grating Resolving Power

196 196 For a grating with N lines illuminated it can be shown that the resolving power in the m th order diffraction is Dispersion mercury Spectroscopic instruments need to resolve spectral lines of nearly the same wavelength. The greater the angular dispersion, the better a spectrometer is at resolving nearby lines.

197 197 Example: Light from mercury vapor lamps contain several wavelengths in the visible region of the spectrum including two yellow lines at 577 and 579 nm. What must be the resolving power of a grating to distinguish these two lines? mercury

198 198 Example: how many lines of the grating must be illuminated if these two wavelengths are to be resolved in the first-order spectrum? mercury

199 199


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