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Light and Reflection.  The difference among waves is the nature of the vibrating source.  An electromagnetic wave is the coupled vibration of electric.

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Presentation on theme: "Light and Reflection.  The difference among waves is the nature of the vibrating source.  An electromagnetic wave is the coupled vibration of electric."— Presentation transcript:

1 Light and Reflection

2  The difference among waves is the nature of the vibrating source.  An electromagnetic wave is the coupled vibration of electric and magnetic fields free of matter.  An electromagnetic wave is produced by a vibrating (moving) electric field or magnetic field.  All electromagnetic waves travel at the same speed. Electromagnetic Radiation

3 Speed of light (c): 2.997 924 58 x 10 8 m/s in a vacuum 186000 mi/s 2.997 09 x 10 8 m/s in air c = f Dual Nature of Light Light waves Light particles = photons h = 6.626 x 10 -34 J. s E = hf

4 Electromagnetic waves are oscillating electric and magnetic fields.

5 Visible light is electromagnetic radiation with a frequency range of 4.3 x 10 14 to 7 x 10 14 Hz.

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7  The study and description of light in terms of straight line propagation is called optics. In optics, lines, called rays, are drawn perpendicular to the wave front.  Ray optics are used to analyze reflection and refraction of light.  Ray optics do not account for the wave properties of diffraction and interference. Optics: the study of the behavior and properties of light

8  When light waves encounter a barrier (change media) they are absorbed, transmitted, or reflected.  The texture of the reflecting surface affects how it reflects light.  Diffuse reflection is reflection from a rough, texture surface such as paper or unpolished wood.  Specular reflection is reflection from a smooth, shiny surface such as a mirror or a water surface. A good mirror can reflect 90% of the incident light.

9  When light is reflected it obeys the Law of Reflection The angle of incidence and the angle of reflection are always equal.  The angle of incidence is the angle between a ray that strikes a surface and the line perpendicular (the normal) to that surface at the point of contact.  The angle of reflection is the angle formed by the normal and the direction in which a reflected ray moves.  The normal is an imaginary line drawn perpendicular to the reflecting surface used to measure the incident and reflected angles. The Law of Reflection

10  Mirror and Lens Terminology  Object distance is p.  Image distance is q.  Object height is h.  Image height is h’.  Angle of incidence is   Angle of reflection is  ’.

11  Mirror and Lens Terminology  Virtual Image: The image formed by rays that appear to come from the image point behind the mirror—but never really do.  A virtual image is not real.  A virtual image can never be displayed on a physical surface. A dental mirror

12  Mirror and Lens Terminology A dental mirror  Real Image: The image formed by rays of light that actually pass through a point on the image.  A real image can be displayed on a physical surface (a screen).

13  Flat Mirrors A dental mirror  Flat mirrors form virtual images that are the same distance from the mirror’s surface as the object is.

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15  The Law of Reflection  The angle of incidence and the angle of reflection are always equal.   i =  r ii ’i’i ’r’r rr

16  Curved Mirror and Lens Terminology  R is the radius of curvature, which is the same as the radius of the spherical shell of which the mirror is a small part.  C is the center of curvature.  The principal axis (optic axis) is the line that extends infinitely from the center of the mirror’s surface through the center of curvature.

17  Curved Mirror and Lens Terminology  The focal point is the location at which a mirror or a lens focuses rays parallel to the principal axis or from which such rays appear to diverge.  f is the focal length  f = 1/2R.

18  The Mirror Equation  Image location can be predicted by using the mirror equation.

19  The Mirror Equation   Signs and Meanings + p: object is in front of mirror + q: image is real and in front of mirror - q: image is virtual and is located behind mirror + R and +f: concave mirror +h and +h’: object and image are above the principal axis –h’: image is inverted and below the principal axis

20  The Equation for Magnification What does this mean in words?  Signs and Meanings +M: upright and virtual image -M: inverted and real image

21  Ray Diagrams are drawings that use simple geometry to locate an image formed by a mirror or lens. The image will be located at the place where the rays intersect. Only two rays are necessary to locate the image on a ray diagram, but it's useful to add the third as a check. 1. Sketch the object and mirror and relative distance between them. 2. Draw rays 1.The parallel ray is drawn from the tip of the object parallel to the principal axis. It then reflects off the mirror and either passes through the focal point, or can be extended back to pass through the focal point. 2.The second ray (the chief ray) is drawn from the tip of the object to the mirror through the center of curvature. This ray will hit the mirror at a 90° angle, reflecting back the way it came through the center. The chief and parallel rays meet at the tip of the image. 3.The third ray, the focal ray, is a mirror image of the parallel ray. The focal ray is drawn from the tip of the object through (or towards) the focal point, reflecting off the mirror parallel to the principal axis. All three rays should meet at the same point.

22  Ray Diagrams

23  Concave Mirror Image  If the object is outside the focal length, a concave mirror will form a real, inverted, and diminished (smaller) image. o = p i = q positive in front of the mirror.

24  Concave Mirror Image  If an object is placed inside the focal length of a concave mirror, and enlarged virtual and erect image will be formed behind the mirror. o = p i = q positive in front of the mirror.

25 1. A 1.50 m tall child is in a mirror gallery at the amusement park. She is standing in front of a concave mirror with a radius of 4.00 m. She starts walking toward the mirror from a distance of 9.00 m, and she stops every meter to observe her image. a. Find the focal point of this mirror and label it F. b. Mark the child’s locations 9.00 m, 5.00 m, and 1.00 m in front of the mirror, and label them A, B, C. c. Sketch ray diagrams to locate the image formed when the child is at A. Measure the distance from the image to the mirror and record it below. Distance of A’s image = d. Repeat question c for the object at positions B and C. Distance of B’s image = Distance of C’s image = 2. Calculate the image location for the object at A, B, and C in item 1, using the mirror equation. Compare your results with your diagrams. Distance of A’s image = Distance of B’s image = Distance of C’s image =

26 1. A 1.50 m tall child is in a mirror gallery at the amusement park. She is standing in front of a concave mirror with a radius of 4.00 m. She starts walking toward the mirror from a distance of 9.00 m, and she stops every meter to observe her image. a. Find the focal point of this mirror and label it F. b. Mark the child’s locations 9.00 m, 5.00 m, and 1.00 m in front of the mirror, and label them A, B, C. F C A B

27 1. A 1.50 m tall child is in a mirror gallery at the amusement park. She is standing in front of a concave mirror with a radius of 4.00 m. She starts walking toward the mirror from a distance of 9.00 m, and she stops every meter to observe her image. c. Sketch ray diagrams to locate the image formed when the child is at A. Measure the distance from the image to the mirror and record it below. Distance of A’s image = 2.5 m F A Ray 1 Ray 2 Ray 3

28 1. c. Sketch ray diagrams to locate the image formed. d. Repeat question c for the object at positions B (5.0 0 m). Distance of B’s image = 3.3 m F B Ray 1 Ray 2 Ray 3

29 1. c. Sketch ray diagrams to locate the image formed. d. Repeat question c for the object at positions C (1.0 0 m). Distance of C’s image = -1.8 m F C Ray 1 Ray 2 Ray 3 o i

30 2. Calculate the image location for the object at A (9.00 m), B (5.00 m), and C (1.00 m) in item 1, using the mirror equation. Compare your results with your diagrams. Distance of A’s image = 2.57 m 1/f = 1/p + 1/q f = focal point, p = object’s distance to mirror, q = image’s distance to mirror f = 2.00 m,p = 9.00 m,q = ? 1/f – 1/p = 1/q 1/q = 1/f – 1/p 1/q = 1/2.00 m – 1/9.00 m 1/q = 0.500 m -1 – 0.111 m -1 1/q = 0.389 m -1 q = 1/0.389 m= 2.57 m

31 2. Calculate the image location for the object at A (9.00 m), B (5.00 m), and C (1.00 m) in item 1, using the mirror equation. Compare your results with your diagrams. Distance of B’s image = 3.33 m 1/f = 1/p + 1/q f = focal point, p = object’s distance to mirror, q = image’s distance to mirror f = 2.00 m,p = 5.00 m,q = ? 1/f – 1/p = 1/q 1/q = 1/f – 1/p 1/q = 1/2.00 m – 1/5.00 m 1/q = 0.500 m -1 – 0.200 m -1 1/q = 0.300 m -1 q = 1/0.300 m= 3.33 m

32 2. Calculate the image location for the object at A (9.00 m), B (5.00 m), and C (1.00 m) in item 1, using the mirror equation. Compare your results with your diagrams. Distance of C’s image = -2.00 m 1/f = 1/p + 1/q f = focal point, p = object’s distance to mirror, q = image’s distance to mirror f = 2.00 m,p = 1.00 m,q = ? 1/f – 1/p = 1/q 1/q = 1/f – 1/p 1/q = 1/2.00 m – 1/1.00 m 1/q = 0.500 m -1 – 1.00 m -1 1/q = - 0.500 m -1 q = - 1/0.500 m= - 2.00 m

33  A convex spherical mirror is a mirror whose reflecting surface is the outward-curved segment of a sphere.  Diverging mirror  Light rays diverge upon reflection from a convex mirror, forming a virtual image that is always smaller than the object.  Image: virtual, upright, diminished  (-) f, (-) q, (+) p, (+)M < 1

34  Convex or Diverging Mirrors  A convex mirror forms a virtual image. A convex mirror forms a virtual image.

35  The Mirror Equation  Signs and Meanings for Convex Mirrors + p: object is in front of mirror - q: image is virtual and is located behind mirror - R and -f: convex mirror +h and +h’: object and image are above the principal axis –h’: image is inverted or below the principal axis

36 Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Curved Mirrors Chapter 13 Sample Problem Convex Mirrors An upright pencil is placed in front of a convex spherical mirror with a focal length of 8.00 cm. An erect image 2.50 cm tall is formed 4.44 cm behind the mirror. Find the position of the object, the magnification of the image, and the height of the pencil.

37 Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Curved Mirrors Chapter 13 Sample Problem, continued Convex Mirrors Given: Because the mirror is convex, the focal length is negative. The image is behind the mirror, so q is also negative. f = –8.00 cm q = –4.44 cm h’ = 2.50 cm Unknown: p = ? h = ?

38 Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Curved Mirrors Chapter 13 Sample Problem, continued Convex Mirrors Diagram:

39 Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Curved Mirrors Chapter 13 Sample Problem, continued Convex Mirrors 2. Plan Choose an equation or situation: Use the mirror equation and the magnification formula. Rearrange the equation to isolate the unknown:

40 Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Curved Mirrors Chapter 13 Sample Problem, continued Convex Mirrors 3. Calculate Substitute the values into the equation and solve:

41 Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Curved Mirrors Chapter 13 Sample Problem, continued Convex Mirrors 3. Calculate, continued Substitute the values for p and q to find the magnifi- cation of the image. Substitute the values for p, q, and h’ to find the height of the object.

42  Spherical Aberration and Parabolic Mirrors  The blurring of an image that occurs when light from the margin of a mirror or lens with a spherical surface comes to a shorter focus than light from the central portion.

43  Color

44  The retina is lined with a variety of light sensing cells known as rods and cones. The rods on the retina are sensitive to the intensity of light, but cannot distinguish wavelengths (colors). Three types of cone cells in the eye allow humans to see in color.  Light of different wavelengths stimulate a combination of cone cells so that a wide range of colors can be perceived. cyclohexane.tumblr.com

45  Color  The color of an object depends on which wavelengths of light shine on the object and which wavelengths are reflected. Red Orange Yellow Green Cyan Blue violet White Red Green Blue Primary Additive Colors of light produce white light when added.

46  Color  R + G Y  G + B C  B + R M Additive primary colors produce white light when combined. R + G + B W Light of different colors can be produced by adding light consisting of the primary additive colors (red, green, and blue). Adding different colors of light is used in color television, color computer monitors, and on-stage theater lighting. Complementary colors are two colors that add together to produce white light. Two primary colors combine to produce the complement of the third primary color. C + R W Cyan is the complementary of red. Y + B W Yellow is the complementary of blue. M + G W Magenta is the complementary of green.

47  Color Filters are transparent materials which selectively absorb (or block) one or more primary colors of light and allow the remaining colors of light to pass through (or be transmitted). The color of the filter describes which color of light is transmitted by the filter. Opaque objects contain pigments that selectively absorb light and reflect whatever light colors are not absorbed. The color of pigments describes which color(s) of light are reflected. The color of an object is the color of light the object reflects.

48  Color  A pure pigment absorbs a single frequency of light. ▪ Yellow absorbs blue. ▪ Magenta absorbs green. ▪ Cyan absorbs red.  Three primary colors of paint (primary pigments): yellow, cyan, and magenta  An artist can create any color by using varying amounts of the three primary colors of paint.  Each primary color of paint absorbs one primary color of light, its complementary color.

49  Color  What pigment(s) (yellow, cyan, or magenta) would you use to produce the following colors on white canvas that is going to be illuminated with white light?  Yellow  Red  Green  Blue  Cyan  Magenta yellow yellow and magenta yellow and cyan magenta and cyan cyan magenta

50 Chapter 13 Section 4 Color and Polarization

51  Blue Sky and Red Sunset  The sky is blue because the atmosphere (N2 and O2) scatters blue light.  Sunsets and sunrises are red because the sunlight passes through more atmosphere and the longer wavelengths are not scattered as much by the small atmospheric particles.  White clouds?  Brown smog?

52 When a light wave vibrates in a variety of directions, the light is said to be unpolarized. When a light wave's are isolated to a single plane, the light is said to be polarized. A Polaroid filter polarizes light by blocking part of the vibrations while letting through those that are in a specific plane

53 http://www.physicsclassroom.com/Class/light/u12l1e.cfm

54 GLARE

55 http://www.physicsclassroom.com/Class/light/u12l1e.cfm

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