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All About Light. Light is a part of the electromagnetic spectrum.

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Presentation on theme: "All About Light. Light is a part of the electromagnetic spectrum."— Presentation transcript:

1 All About Light

2 Light is a part of the electromagnetic spectrum.

3 Light can travel through empty space as well as air and water.

4 Light is a transverse wave.

5 Light comes in discreet packages called photons.

6 Light has properties of a particle and wave, known as the dual nature of light.

7 Similarities of Light to Sound Source determines the frequency. Speed depends on what it is traveling through. Light travels fastest in a vacuum, slowest through water. v = f λ still applies. The speed of light in a vacuum is constant and represented by c. c = f λc = 3.00 ˣ 10 8 m/s

8 The primary colors of light are red, green, and blue.

9 The secondary colors of light are yellow, cyan, and magenta.

10 All of the colors combined make white light.

11 The mixing of all of these colors makes every color we see.

12 Polarization of Light

13 Incident light is unpolarized and is at all angles.

14 The first filter removes the horizontal light.

15 The second filter removes the vertical light.

16 No light gets past the second filter!

17 This is how sunglasses work!

18 Glare is horizontal light reflected off of the ground.

19 Vertical Polarizers eliminates the horizontal light leaving only vertical.

20 More on Reflection and Refraction.

21 The Law of Reflection always applies.

22 A Smooth surface gives a regular reflection where light comes back in parallel beams.

23 A rough surface gives a diffuse reflection because light is scattered.

24 Keep in mind we see nothing unless it reflects light!!

25 Refraction goes along with reflection!!

26 Notice the refracted ray is smaller than the incident ray.

27 This is because light is slower in water than in air!

28 Facts about Refraction When light goes from a faster to slower medium, it will bend towards the normal. The Angle of Refraction is always dependent on the Angle of Incidence!!! The higher the Index of Refraction the slower light travels through the medium!!

29 The Light travels towards water at 60° Normal

30 So the Angle of Incidence is 60°. (Θ i = 60°) Normal ΘiΘi

31 So the Angle of Refraction is 40.6°. (Θ r = 40.6°) Normal ΘiΘi ΘrΘr

32 Since light travels through glass even slower it will bend more towards the normal!! Normal

33 So the Angle of Incidence is still 60°. (Θ i = 60°) Normal ΘiΘi

34 Since light moves slower through glass than water it bends more as it crosses the boundary!! Normal ΘiΘi

35 So the Angle of Refraction is 34.5°. (Θ r = 34.5°) Normal ΘiΘi ΘrΘr

36 Using this information we can calculate the Index of Refraction (n). n = sin Θ r sin Θ i

37 Finding the Index of Refraction for water… n = sin 40.6° sin 60° n =1.33

38 The Index of Refraction can also be done in terms of the speed of light! n = v c 1.33 = 3.00 × 10 8 m/s 2.25 × 10 8 m/s

39 Use the speed of light in a vacuum over the speed of light in water. n = v c 1.33 = 3.00 × 10 8 m/s 2.25 × 10 8 m/s

40 Finding the Index of Refraction for glass… n = sin 34.5° sin 60° n =1.53

41 Remember Index of Refraction can never be less than 1 (vacuum) !! n = sin 34.5° sin 60° n =1.53

42 Using Snell’s Law we can predict how light will bend if we know the indexes of refraction!! ΘiΘi

43 ΘiΘi nini sin Θ i =nrnr sin Θ r If Θ i = 30° and the light is traveling from water to air…

44 ΘiΘi nini sin Θ i =nrnr sin Θ r 1.33sin 30° =1.0sin Θ r

45 ΘiΘi 1.33 sin 30°= 1.0 sin Θ r

46 ΘiΘi = 1.0 sin Θ r

47 ΘiΘi = 0.665sin Θ r

48 ΘiΘi = ΘrΘr sin -1

49 ΘiΘi =41.7°ΘrΘr ΘiΘi

50 ΘiΘi ΘiΘi When light went from air to water it bent towards the normal, but water to air it bends away from the normal!!

51 ΘiΘi ΘiΘi Why???

52 ΘiΘi ΘiΘi Because light travels faster in air than water!!!

53 ΘiΘi ΘiΘi When light travels from a slower to faster medium it bends away from the normal

54 Common Indexes of Refraction Vacuum Air Water Ethanol Crown Glass Quartz Flint Glass Diamond

55 Mirrors and Lenses

56 First we will look at curved mirrors, which are just a section of a circle.

57 The focal point of a concave mirror is the point halfway between the center of the circle and the surface of the mirror.

58 Following three rules we can find the location of any image. FC

59 #1 Draw a line from the top of the object reflected through the focal point. FC

60 #2 Draw a ray through the focal point and reflected parallel. FC

61 #3 draw a line through the center and straight back to the object. FC

62 We now know the location of the image. Any two of our lines will tell us. FC

63 It is an inverted image because it is upside down. FC

64 It is a real image because the beams of light actually pass through. FC

65 We can locate this image with the same rules. FC

66 What do we do? FC

67 Extend the lines behind the mirror!!! F C

68 Is this image real? F C

69 No, the light does not actually travel there! F C

70 This image is said to be virtual ! F C

71 Is the image inverted or erect? F C

72 Erect, because it is not upside down! F C

73 Convex mirrors produce small images. The focal point is behind the mirror.

74 The images are never real.

75 These mirrors are valued for their wide angle views!!

76 We apply the rules we learned before. FC

77 We treat the reflected ray as if it were coming from the focus. FC

78 The second line is to the focus and straight outward. FC

79 The image is smaller giving us our “wide angle view” FC

80 Now on to Lenses!!

81 Locating an image with a lens FF

82 #1 Draw a line from the top of the object and then through a focal point. FF

83 #2 Draw a line to the top of the object straight through the center. FF

84 The image is inverted and real. FF

85 If the object is in front of the focal point…. FF

86 What do we do???? FF

87 Extend the lines backwards…… FF

88 The image is erect and virtual…… FF

89 Convex Lens Image Zones FF2F Zone 1: Image is on the same side, erect, magnified, and virtual.

90 Convex Lens Image Zones FF2F Zone 2: Image is on the opposite side, inverted, magnified, and real.

91 Convex Lens Image Zones FF2F Zone 3: Image is on the opposite side, inverted, diminished, and real.

92 Zone I

93 At F image at infinity and does not exist.

94 Zone II

95 At 2F image at 2F

96 Zone III

97 A convex lens focuses light on a point…..

98 A concave lens spreads light out….

99 A concave lens is used to help people with myopia.

100 Optics The Math of Mirrors and Lenses I See You!!

101 The Mirror Equation 111 += soso sisi f s o is the object’s distance from the mirror. s i is the image’s distance from the mirror. f is the focal length of the mirror.

102 The Mirror Equation 111 += soso sisi f If the answer is positive the image is real !! If the answer is negative the image is virtual !!

103 The Magnification Equation = ˗ soso sisi If the answer is positive the image is upright !! If the answer is negative the image is inverted !! m m in this equation means magnification!!

104 More on Magnification ×hoho hihi ∣m ∣ Use the absolute value for magnification!! = h o is the height of the object!! h i is the height of the image!!

105 20 cm focal point soso 30 cm Where is the image? sisi = sisi =0.05 sisi =59 cm The positive result means a real image.

106 20 cm focal point soso 30 cm What is the size of the image?? sisi = 59 soso = 20 -m 59 cm -m sisi m=-2.95 The negative result means an inverted image.

107 20 cm focal point hoho 30 cm What is the size of the image?? 59 cm hihi 2.95 ∣m ∣× × hoho 5 = = hihi hihi hihi =14.75 cm

108 Light can be used to send digital signals through Total Internal Reflection…

109 This is used in fiber optic cables…

110 Total Internal Reflection This occurs because an incident beam enters a medium with a lower index of refraction, such as from water to air. (Light is refracted away from the normal) The critical angle (θ c ) is reached when the light is refracted at 90° and travels along the surface. At angles greater than the critical angle (θ c ) total internal reflection occurs.

111 Critical Angle ( θ c )

112 Notice Total Internal Reflection only occurs when the angle of the Incident beam is greater than the critical angle. θ i ˃ θ c

113 Summary of Requirements for Total Internal Reflection Index of Refraction must be lower for medium incident beam in entering. n i ˂ n r or n 1 ˂ n 2 The angle of incidence must be greater than the critical angle! θ i ˃ θ c or θ 1 ˃ θ c

114 Calculating Critical Angle n1n1 n2n2 sin θ c = n 1 is the index of refraction for the medium where the incident beam originated n 2 is the index of refraction for the medium the incident beam enters

115 From Water to Air n1n1 n2n2 sin θ c == sin θ c = sin -1 θc θc =(0.7518) θc θc = 48.8°

116 If a light source is 5 m underwater, how far away on the surface will total internal reflection occur?? Air n = 1.00 Water n = 1.33 y = 5 m x = ? m

117 Air n = 1.00 Water n = 1.33 y = 5 m x = ? m tan 48.8° = y x = 5 m x x= tan 48.8° x=5.71 m

118 Dispersion of Light Different colors of light have different indexes of refraction. This causes the “rainbow effect” under the proper conditions. For some media, such as air, this difference is miniscule and rarely noticed. For glass the difference in indexes of refraction can be profound for each color of light.

119 Indexes of Refraction for a piece of glass. Red1.502 Orange1.506 Yellow1.511 Green1.517 Blue1.523 Violet1.530

120 θ 1 = 60° We make use that the sum of the angles in a triangle are always 180° θ2 θ2 θ3 θ3 θ4 θ4

121 θ 1 = 60° (90° - θ 2 ) + 60° + (90° - θ 3 ) = 180° θ2 θ2 θ3 θ3 θ4 θ4

122 θ 1 = 60° (90° - θ 2 ) + (90° - θ 3 ) = 120° θ2 θ2 θ3 θ3 θ4 θ4 60°

123 θ 1 = 60° θ 2 + θ 3 = 60° θ2 θ2 θ3 θ3 θ4 θ4 60°

124 θ 1 = 60° θ 3 = 60° - θ 2 θ2 θ2 θ3 θ3 θ4 θ4

125 θ 1 = 60° Now we can use Snell’s Law to solve for the rest!!! θ2 θ2 θ3 θ3 θ4 θ4

126 θ 1 = 60° 1 sin 60° = n sin θ 2 n sin θ 3 = 1 sin θ 4 θ2 θ2 θ3 θ3 θ4 θ4

127 θ 1 = 60° We will do this for red n = θ2 θ2 θ3 θ3 θ4 θ4

128 θ 1 = 60° 1 sin 60° = sin θ 2 θ2 θ2 θ3 θ3 θ4 θ4

129 θ 1 = 60° sin / = θ 2 θ2 θ2 θ3 θ3 θ4 θ4

130 θ 1 = 60° 35.21° = θ 2 θ2 θ2 θ3 θ3 θ4 θ4

131 θ 1 = 60° θ 3 = 60° - θ 2 θ 3 = 60° ° θ2 θ2 θ3 θ3 θ4 θ4

132 θ 1 = 60° θ 3 = 24.79° θ2 θ2 θ3 θ3 θ4 θ4

133 θ 1 = 60° n sin θ 3 = 1 sin θ 4 θ2 θ2 θ3 θ3 θ4 θ4

134 θ 1 = 60° sin 24.79° = 1 sin θ 4 θ2 θ2 θ3 θ3 θ4 θ4

135 θ 1 = 60° sin 24.79° = 1 sin θ 4 θ2 θ2 θ3 θ3 θ4 θ4

136 θ 1 = 60° 39.05° = θ 4 θ2 θ2 θ3 θ3 θ4 θ4

137 Now calculate the rest!! Color n θ 2 θ 3 θ 4 Red ° 24.79° 39.05° Orange1.506 Yellow1.511 Green1.517 Blue1.523 Violet1.530

138 Indexes of Refraction for a piece of glass. Color n θ 2 θ 3 θ 4 Red ° 24.79° 39.05° Orange ° 24.90° 39.35° Yellow ° 25.03° 39.74° Green ° 25.19° 40.22° Blue ° 25.35° 40.70° Violet ° 25.53° 41.25°

139 The End

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