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Chapter 26 Geometrical Optics. The Reflection of Light. If a stone is dropped into a pond, circular waves emanate from the point where it landed. Rays, perpendicular to the wave fronts, give the direction in which the waves propagate. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
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1. Chapter 26 Geometrical Optics
2. The Reflection of Light If a stone is dropped into a pond, circular waves emanate from the point where it landed. Rays, perpendicular to the wave fronts, give the direction in which the waves propagate.
3. As one moves farther from a point wave source, the wave fronts become more nearly flat.
4. Reflection Reflection occurs when light bounces off a surface. There are two types of reflection:
5. 1. Specular reflection: Off a shiny surface 2. Diffuse reflection: Off a rough surface .
6. Mirrors are great reflectors.
7. The Law of “REFLECTION” The Law of Reflection states that- " the angle of incidence (incoming ray) equals the angle of reflection (outgoing ray)" The law works for FLAT, PLANE surfaces only. The angles are measured from a perpendicular line to the surface called a NORMAL. NORMAL
8. A ray of light reflects from a plane mirrorwith an angle of incidence of 37o. If themirror is rotated by an angle of 5o, throughwhat angle is the reflected ray rotated?
9. Sample Problem: Standing 2.0 m in front of a small vertical mirror, you see the reflection of your belt buckle, which is 0.70 m below your eyes What is the vertical location of the mirror relative to the level of your eyes?
10. Sample Problem: Standing 2.0 m in front of a small vertical mirror, you see the reflection of your belt buckle, which is 0.70 m below your eyes What is the vertical location of the mirror relative to the level of your eyes?
11. Forming Images with a Plane Mirror Light reflected from the flower and vase hits the mirror. Obeying the law of reflection, it enters the eye. The eye interprets the ray as having had a straight-line path, and sees the image behind the mirror.
12. Plane Mirror Suppose we had a flat , plane mirror mounted vertically. A candle is placed 10 cm in front of the mirror. WHERE IS THE IMAGE OF THE CANDLE LOCATED? mirror On the surface of the mirror? Behind the mirror? Object Distance, Do = 10 cm Same side as the object?
13. Plane Mirror Suppose we had a flat , plane mirror mounted vertically. A candle is placed 10 cm in front of the mirror. WHERE IS THE IMAGE OF THE CANDLE LOCATED? mirror Virtual Image Object Distance, Do = 10 cm Image Distance, Di = 10 cm Do = Di, and the heights are equal as well
14. Properties of Mirror Images Produced by Plane Mirrors: • A mirror image is upright, but appears reversed right to left. • A mirror image appears to be the same distance behind the mirror that the object is in front of the mirror. • A mirror image is the same size as the object.
15. Virtual Images Virtual Images are basically images which cannot be visually projected on a screen. If this box gave off light, we could project an image of this box on to a screen provided the screen was on the SAME SIDE as the box. You would not be able to project the image of the vase or your face in a mirror on a screen, therefore it is a virtual image. CONCLUSION: VIRTUAL IMAGES are ALWAYS on the OPPOSITE side of the mirror relative to the object.
16. Ray tracing: plane mirror Construct the image using two rays.
17. Ray tracing: plane mirror Construct the image using two rays.
18. A corner reflector reflects light parallel to the incident ray, no matter the incident angle.
19. 26-3 Spherical Mirrors A spherical mirror has the shape of a section of a sphere. If the outside is mirrored, it is convex; if the inside is mirrored, it is concave.
20. Spherical mirrors have a central axis (a radius of the sphere) and a center of curvature (the center of the sphere).
21. Spherical Mirrors – Concave & Convex Also called DIVERGING mirror Also called CONVERGING mirror
22. Parallel rays hitting a spherical mirror come together at the focal point (or appear to have come from the focal point, if the mirror is convex).
23. Parts of a Spherical Concave Mirror
24. Converging (Concave) Mirror A converging mirror is one that is spherical in nature by which it can FOCUS parallel light rays to a point directly in front of its surface. Every spherical mirror can do this and this special point is at a “fixed” position for every mirror. We call this point the FOCAL POINT. To find this point you MUST use light from “infinity” Light from an “infinite” distance, most likely the sun.
25. Converging (Concave) Mirror Since the mirror is spherical it technically has a CENTER OF CURVATURE, C.The focal point happens to be HALF this distance. Note ‘C’ is the radius We also draw a line through the center of the mirror and call it the PRINCIPAL AXIS.
26. 26-4 Ray Tracing and the Mirror Equation We use three principal rays in finding the image produced by a concave mirror. The parallel ray (P ray) reflects through the focal point. The focal ray (F ray) reflects parallel to the axis. The center-of-curvature ray (C ray) reflects back along its incoming path.
27. 26-4 Ray Tracing and the Mirror Equation These three rays are illustrated here.
28. Ray Diagram A ray diagram is a pictorial representation of how the light travels to form an image and can tell you the characteristics of the image. C object f Principal axis Rule One: Draw a ray, starting from the top of the object, parallel to the principal axis and then through “f” after reflection.
29. Ray Diagram A ray diagram is a pictorial representation of how the light travels to form an image and can tell you the characteristics of the image. C object f Principal axis Rule One: Draw a ray, starting from the top of the object, parallel to the principal axis and then through “f” after reflection.
30. Ray Diagrams C object f Principal axis Rule Two: Draw a ray, starting from the top of the object, through the focal point, then parallel to the principal axis after reflection.
31. Ray Diagrams C object f Principal axis Rule Three: Draw a ray, starting from the top of the object, through C, then back upon itself. THEY INTERSECT What do you notice about the three lines? The intersection is the location of the image.
32. Ray Diagram – Image Characteristics C object f Principal axis After getting the intersection, draw an arrow down from the principal axis to the point of intersection. Then ask yourself these questions: Is the image on the SAME or OPPOSITE side of the mirror as the object? Same, therefore it is a REAL IMAGE. Is the image ENLARGED or REDUCED? Is the image INVERTED or RIGHT SIDE UP?
33. 26-3 Spherical Mirrors For a convex mirror, the focal length is negative, as the rays do not go through the focal point. The opposite is true for a concave mirror.
34. 26-3 Spherical Mirrors We have made the assumption here that the rays do not hit the mirror very far from the principal axis. If they do, the image is blurred; this is called spherical aberration, and can be remedied by using a parabolic mirror instead.
35. 26-3 Spherical Mirrors When the Hubble Space Telescope was first launched, its optics were marred by spherical aberration. This was fixed with corrective optics.
36. 26-4 Ray Tracing and the Mirror Equation The process is similar for a concave mirror, although there are different results depending on where the object is placed.
37. 26-4 Ray Tracing and the Mirror Equation This image shows how these three rays are used to find the image formed by a convex mirror. The image is located where the projections of the three rays cross. The size of the image can also be determined.
38. 26-4 Ray Tracing and the Mirror Equation We derive the mirror equation using the ray diagrams:
39. 26-4 Ray Tracing and the Mirror Equation Using the similar triangles and the fact that f = ½ R, we get the mirror equation: Here, do is the distance from the mirror to the object, di is the distance from the mirror to the image, and f is the focal length.
40. Mirror/Lens Equation Assume that a certain concave spherical mirror has a focal length of 10.0 cm. Locate the image for an object distance of 25 cm and describe the image’s characteristics. 16.67 cm What does this tell us? First we know the image is BETWEEN “C” & “f”. Since the image distance is POSITIVE the image is a REAL IMAGE. Real image = positive image distance Virtual image = negative image distance What about the size and orientation?
41. 26-4 Ray Tracing and the Mirror Equation
42. We can also find the magnification:
43. Sample Problem: Calculate the magnification for the previous problem.
44. Here are the sign conventions for concave and convex mirrors:
45. Sample Problem Assume that a certain concave spherical mirror has a focal length of 10.0 cm. Locate the image for an object distance of 5 cm and describe the image’s characteristics. -10 cm Characteristics? VIRTUAL (opposite side) Enlarged Upright 2x
46. Refraction Refraction is based on the idea that LIGHT is passing through one MEDIUM into another. The question is, WHAT HAPPENS? Suppose you are running on the beach with a certain velocity when you suddenly need to run into the water. What happens to your velocity? IT CHANGES!
47. Refraction Refraction is the movement of light from one medium into another medium. Refraction cause a change in speed of light as it moves from one medium to another. Refraction can cause bending of the light at the interface between media.
48. 26-5 The Refraction of Light Light moves at different speeds through different media. When it travels from one medium into another, the change in speed causes the ray to bend.
49. Refraction What EXACTLY is light doing when it reaches a new medium? We don’t want you to think ALL of the light refracts. Some of the light REFLECTS off the boundary and some of the light REFRACTS through the boundary. Angle of incidence = Angle of Reflection Angle of Incidence > or < the Angle of refraction depending on the direction of the light
50. 26-5 The Refraction of Light The angle of refraction is related to the different speeds: The speed of light in a medium is given by the index of refraction of that medium:
51. Refraction – Going from Air to Water The index of refraction, n, for air (vacumm) is equal to 1. The index of refraction for water is 1.33. If you are going from a LOW “n” to a HIGH “n”, your speed DECREASES and the angle BENDS TOWARDS the normal
52. Refraction – Going from Water into Air The index of refraction, n, for air (vacumm) is equal to 1. The index of refraction for water is 1.33. If you are going from a HIGH “n” to a LOW “n”, your speed INCREASES and the angle BENDS AWAY the normal Note: If the angles are EQUAL, then the “n” must be equal for each. The ray will pass straight through.
53. The Refraction of Light Here are some typical indices of refraction:
54. Snell’s Law A scientist by the name of Snell discovered that the ratios of the index’s and the ratio of the sine of the angles are the same value!
55. Refraction can make objects immersed in water appear broken, and can create mirages.
56. If light enters a medium of lower index of refraction, it will be bent away from the normal. If the angle of incidence is large enough, the angle of refraction is 90°; at larger incident angles the light will be totally reflected.
57. This is called total internal reflection, and the incident angle at which the angle of refraction is 90° is called the critical angle, C. Total internal reflection is used in some binoculars and in optical fibers.
58. Sample Problem: Light enters an oil from the air at an angle of 50o with the normal, and the refracted beam makes an angle of 33o with the normal. a) Draw this situation. b) Calculate the index of refraction of the oil. c) Calculate the speed of light in the oil.
59. Sample Problem: Light enters an oil from the air at an angle of 50o with the normal, and the refracted beam makes an angle of 33o with the normal. a) Draw this situation. b) Calculate the index of refraction of the oil. c) Calculate the speed of light in the oil.
60. Sample Problem: Light enters water from a layer of oil at an angle of 50o with the normal. The oil has a refractive index of 1.65, and the water has a refractive index of 1.33. a) Draw this situation. b) Calculate the angle of refraction. c) Calculate the speed of light in the oil, and in the water.
61. Sample Problem: Light enters water from a layer of oil at an angle of 50o with the normal. The oil has a refractive index of 1.65, and the water has a refractive index of 1.33. a) Draw this situation. b) Calculate the angle of refraction. c) Calculate the speed of light in the oil, and in the water.
62. There is a special angle called Brewster’s angle; light reflected at this angle is totally polarized. Reflected light is completely polarized when the reflected and refracted beams are at right angles to one another. The direction of polarization is parallel to the reflecting surface.
63. Brewster’s angle can be calculated using the appropriate geometry:
64. 26-6 Ray Tracing for Lenses Lenses are used to focus light and form images. There are a variety of possible types; we will consider only the symmetric ones, the double concave and the double convex.
65. If we think of a convex lens as consisting of prisms, we can see how light going through it converges at a focal point (assuming the lens is properly shaped).
66. A concave lens can also be modeled by prisms:
67. Ray Tracing for Lenses The three principal rays for lenses are similar to those for mirrors: The P ray—or parallel ray—approaches the lens parallel to its axis. The F ray is drawn toward (concave) or through (convex) the focal point. The midpoint ray (M ray) goes through the middle of the lens. Assuming the lens is thin enough, it will not be deflected. This is the thin-lens approximation.
68. 26-6 Ray Tracing for Lenses These diagrams show the principal rays for both types of lenses:
69. 26-6 Ray Tracing for Lenses As with mirrors, we use these principal rays to locate the image:
70. The convex lens forms different image types depending on where the object is located with respect to the focal point:
71. Applications of Converging Lenses Obviously, converging lenses play an important role in our lives as our eyes are these types of lenses. Often times we need additional corrective lenses to fix our vision. In figure A, we see an eye which converges what we see on the retina. In figure B, we see an eye which converges too LATE. The eye itself is often too short and results in the person being far sighted. In figure C, we see an eye which converges too SOON. The eye itself is often too long and results in the person being near sighted In the later 2 cases, a convex or concave lens is necessary to ensure the image is on the retina.
72. 26-7 The Thin-Lens Equation We derive the thin-lens equation in the same way we did the mirror equation, using these diagrams:
73. The Thin-Lens Equation
74. Sign conventions for thin lenses:
75. Sample Problem: A converging lens, focal length 20 cm, has a 5-cm high object placed 30 cm from it. a) Draw a ray diagram and construct the image. b) Use the lens equations to calculate i. the position of image ii. the magnification iii. the size of image c) Name the image
76. Solution b) i. 1/di + 1/do = 1/f 1/di + 1/30 = 1/20 1/di = 1/20 - 1/30 = 3/60 –2/60 = 1/60 di = 60 cm ii. M = - di/do = -60/30 = -2 iii. M = hi/ ho hi = M ho= (-2)5 = -10 cm
77. Sample Problem: A converging lens, focal length 10 cm, has a 2-cm high object placed 5 cm from it. a) Draw a ray diagram and construct the image. b) Use the lens equations to calculate i. the position of image ii. the magnification iii. the size of image c) Name the image
78. Solution b) i. 1/di+ 1/do= 1/f 1/di+ 1/5 = 1/10 1/di= 1/10 - 1/5 = 1/10 –2/10 = -1/10 di= -10 cm ii. M = - di/do= -(-10)/5 = 2 iii. M = hi/ ho hi = M ho= (2)2 = 4 cm
79. For diverging lenses: f is negative dois positive diis negative M is positive and < 1 hi is positive and < ho
80. Sample Problem: A diverging lens, focal length -15 cm, has a 4-cm high object placed 10 cm from it. a) Draw a ray diagram and construct the image. b) Use the lens equations to calculate i. the position of image ii. the magnification iii. the size of image c) Name the image
81. Solution b) i. 1/di+ 1/do= 1/f 1/di+ 1/10 = 1/(-15) 1/di= -1/10 - 1/15 = -3/30 – 2/30 = -5/30 = -1/6 di= -6 cm ii. M = - di/do= -(-6)/10 = 0.6 iii. M = hi/ ho hi= M ho= (0.6)(4) = 2.4 cm
82. Sample Problem: Two lenses that are 35 cm apart are used to form an image as shown below. Lens 1 is converging and has a focal length f1 = 14 cm; Lens 2 is diverging and has a focal length f2 = -7.0 cm. The object is placed 24 cm to the left of lens 1. a) Use a ray diagram to find the approximate location of the image. b) Is the image upright or inverted? Real or virtual? Explain.
83. 26-8 Dispersion and the Rainbow The index of refraction varies slightly with the frequency of light; in general, the higher the frequency, the higher the index of refraction. This means that refracted light is “spread out” in a rainbow of colors; this phenomenon is known as dispersion.
84. 26-8 Dispersion and the Rainbow Rainbows are created by the dispersion of light as it refracts in a rain drop.
85. 26-8 Dispersion and the Rainbow As the drop falls, all the colors of the rainbow arrive at the eye.
86. 26-8 Dispersion and the Rainbow Sometimes a faint secondary arc can be seen.
87. Summary of Chapter 26 A wave front is a surface along which the wave phase is constant. Rays, perpendicular to the wave fronts, indicate the direction of propagation. The angle of incidence equals the angle of reflection. The image formed by a plane mirror is upright, but appears reversed left to right; appears to be the same distance behind the mirror as the object is in front of it; and is the same size as the object.
88. Summary of Chapter 26 Spherical mirrors have spherical reflecting surfaces. A concave mirror is curved inward, and a convex one outward. Focal length of a convex mirror: Focal length of a concave mirror: An image is real if light passes through it, virtual if it does not. Mirror equation:
89. Summary of Chapter 26 Magnification: Refraction is the change in direction of light due to a change in speed. The index of refraction gives the speed of light in a medium:
90. Summary of Chapter 26 Snell’s law: Light entering a medium of higher n is bent towards the normal; light entering a medium of lower n is bent away from the normal. When light enters a medium of lower n, there is a critical angle beyond which the light will be totally reflected.
91. Summary of Chapter 26 At Brewster’s angle, the reflected light is totally polarized: A lens uses refraction to bend light and form images. Thin-lens equation:
92. Summary of Chapter 26 Magnification: The index of refraction varies with frequency; different frequencies of light are bent different amounts. This is called dispersion.