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Physics Chapter 15: Refraction
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Refraction In Air: c = 3x108 m/s
In Other Materials Light Travels Slower than c (slightly) The Speed of Light Will Change When Passing From One Material to Another When this Change Occurs The Direction of the Light Also Changes
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Refraction Refraction
The Bending of Light Rays as it Passes from One Medium to Another
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Refraction Refraction
The Direction of Light Changes due to a Change in the Velocity of Light Caused by the Light Rays Passing From One Material to Another
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Refraction Refraction of Light Rays qi qr Air Glass qR
Refraction Toward Normal
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Refraction Refraction of Light Rays qi qr Glass Air qR
Refraction Away From Normal
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Refraction Refraction of Light Waves
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Refraction Frequency Remains Constant through Varying Light Velocities
Wavelength (l) will Fluctuate Smaller when Light is Slowed Larger when Light is Accelerated
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Refraction Refractive Index (n)
Ratio of the Velocity of Light in a Material Compared to The Speed of Light in a Vacuum
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Refraction Refractive Index (n)
The Higher the RI, the Lower the Velocity of Light Through the Media
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Refraction Refractive Index (n) Materials Solids Diamond 2.419
NaCl Crown Glass Ice
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Refraction Refractive Index (n) Materials Liquids CS2 1.628
Benzene 1.501 CCl EtOH H2O
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Refraction Refractive Index (n) Materials Gases CO2 1.00045
Air O H
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Refraction of Light Rays
When a Light Ray Encounters an Interface Between Transparent Materials Some of the Light is Reflected, but Some of the Light Passes Through the Transparent Material
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Refraction Refraction of Light Rays qi qr Air n=1 Glass n=1.523
Lower RI to Higher RI Velocity Decreases Refraction Toward Normal qR
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Refraction Refraction of Light Rays qi qr Glass n=1.523 Air n=1 qR
Higher RI to Lower RI Velocity Increases Refraction Away From Normal
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Refraction Refraction of Light Rays
This Change in the Path of the Light Rays Distorts Our Perception of the Location of an Image Viewed Through an Interface
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Refraction Refraction of Light Rays
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Refraction Example: What angle should the light be aimed to find the sunken treasure chest?
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Refraction Solution:
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Refraction Refraction of Light Rays Snell’s Law of Refraction q1 q2
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Refraction Snell’s Law of Refraction (Example):
Light shining into an aquarium passes from air through glass at an angle of 150 from normal and then into water. What is the final refracted angle into the water? q1 q2
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Refraction Solution: Glass n1 = 1.0 q1 = 150 n2 = 1.5 q2 = ? q1 q2
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Refraction Solution: Water n1 = 1.5 q1 = 9.90 n2 = 1.3 q2 = ? q1 q2
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Refraction Problem: Let’s assume in the last problem that you are looking at your pet piranha in your aquarium. The piranha is 4 inches from the 0.25” thick glass in the aquarium. What is the distance that the image is shifted from its actual position? q1 q2
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Refraction Solution: 0.25” You 4” 150 9.90 11.50 Fish q1 q2
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Refraction Solution: Glass Up 5.10 .25” q1 q2
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Refraction Solution: Water Up 3.50 4” q1 q2
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Refraction Solution: Net 0.022” Up + 0.245” Up = 0.267” Displacement
q1 q2
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Refraction Homework: Page 587 Problems: 10 (30.3o) 11 (26o) 13 ?
14 (q1 = 30.4o, q2 = 22.3o)
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Refraction Lenses Made from Transparent Materials Refract Light
Form an Image of the Light Source
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Refraction Lenses Converging Lens
Paraxial Incident Rays Converge at a Focal Point
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Refraction Lenses Diverging Lens
Paraxial Incident Rays Diverge after Leaving Lens
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Refraction Lenses
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Refraction Ray Diagrams – Converging Lenses Ray 1
Starts At Top of Object Running Parallel to Principal Axis At Lens Center Crosses to Focal Point and Continues
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Refraction Ray Diagrams – Converging Lenses Ray 2
Starts At Top of Object Running Through Focal Point to Lens Center At Lens Center Crosses Continues Parallel to Principal Axis
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Refraction Ray Diagrams – Converging Lenses Ray 3
Starts At Top of Object Running Through Center of Lens
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Refraction Ray Diagrams – Converging Lenses
The Junction of these Rays Indicates the Type, Position, and Size of the Image
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Refraction Image Formation by Converging Lenses
Image Qualities Depend on Object/Lens Position Object Twice the Focal Length (F) from Lens Object Between 2F and F from the Lens Object Within the Focal Length from the Lens
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Refraction Image Formation by Converging Lenses
Object Twice the Focal Length (F) from Lens Image is Real Image is Inverted Image is Reduced
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Refraction Image Formation by Converging Lenses
Object Between 2F and F from the Lens Object is Real Object is Inverted Image is Enlarged
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Refraction Image Formation by Converging Lenses
Object Within the Focal Length from the Lens Image is Virtual Image is Upright Image is Enlarged
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Refraction Image Formation by Diverging Lenses Image is Virtual
Image is Upright Image is Reduced
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Refraction Thin Lens Equation f = Focal Length of the Lens
d0 = Distance Between the Object and the Lens di = Distance Between the Image and the Lens Look Familiar? It should, It’s the Same as the Mirror Equation from Chapter 14
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Refraction Magnification Equation m = Magnification hi = Image Height
ho = Object Height Look Familiar? It should, It’s the Same as the Magnification Equation from Chapter 14
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Refraction Multiple Lenses
Combinations Using Ocular and Objective Lenses The Image Created by the First Lens Acts as the Object for the Second Lens
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Refraction Problem: A ray of sunlight is passing from diamond into crown glass. The angle of incidence is 35.00o. The indices of refraction for the blue and red components of the ray are: blue (nd = 2.444, ncg=1.531), and red (nd = 2.410, ncg=1.520). What is the angle between the refracted blue and red rays in the crown glass?
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Refraction Solution: q1 = 35.00o nd blue = 2.444 ncg blue = 1.531
nd red = 2.410 ncg red = 1.520
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Refraction Solution: q1 = 35.00o nd blue = 2.444 ncg blue = 1.531
nd red = 2.410 ncg red = 1.520
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Refraction Solution: q1 = 35.00o nd blue = 2.444 ncg blue = 1.531
nd red = 2.410 ncg red = 1.520
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Refraction Problem: When a diverging lens is held 13cm above a line of print, the image is 5.0cm beneath the lens. What is the focal length of the lens?
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Refraction Solution: do = -13cm di = -5.0cm
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Refraction Problem: A movie camera has a converging lens with a focal length of 85.0mm. It takes a picture of a 145cm tall person standing 16.0m away. What is the height of the image on the film and is the image upright or inverted?
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Refraction Solution: ho = 145cm f = 85x10-3m do = 16.0cm di = ?
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Refraction Solution: ho = 145cm f = 85x10-3m do = 16.0cm di = m
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Refraction Solution: ho = 145cm f = 85x10-3m do = 16.0cm di = m
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Refraction Homework: Page 588 Problems:
24 (a, -13.3cm, M= b, -10.0cm, M=0.5 c, -6.67cm, M=0.667) 25 (3.40) 26 (a, 40.0cm, M= b, -20.0cm, M=2.0)
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Refraction Total Internal Reflection
Refraction from Larger n to Smaller n (ex. Water to Air) Refraction is AWAY FROM NORMAL q2 = 900 Critical Angle) Any Increase in q1 Causes Total Internal Reflection
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Refraction Total Internal Reflection
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Total Internal Reflection
Refraction Total Internal Reflection Fiber Optics Core (n1) is Larger than Sheath (n2) Design is Such That Any Bend the Cable Can Make is Within qc This Keeps the Light Reflecting Within the Cable!
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Refraction Problem: A ray of light originates in medium “A” and is incident upon medium “B.” What is the critical angle for each pair? A B
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Refraction Total Internal Reflection Solution: Critical Angle A B
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Refraction Total Internal Reflection Solution: Critical Angle A B
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Refraction Total Internal Reflection B must be smaller than A!
Solution: Critical Angle A B B must be smaller than A!
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Refraction Problem: If the Glass in the Illustration is Surrounded by Air, At Which Point (A, B, or C) Will the Laser Beam Exit the Glass?
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Refraction Solution: A = 600 B = 300*** C = 600
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Refraction Problem: What If the Glass in the Illustration is Surrounded by Water, Where Will the Laser Beam Exit the Glass Then?
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Refraction Solution: A = 600*** B = 300 C = 600
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Refraction The Dispersion of Light – Prisms
Remember, Frequency is Constant Wavelength will Change when Light is Refracted When Light Enters a Prism (n1<n2) Refraction Toward Normal When Light Leaves a Prism (n1>n2) Refraction Away From Normal Wavelengths are Dispersed
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Refraction The Dispersion of Light – Prisms
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Refraction The Dispersion of Light – Prisms
Light Must Encounter Two Different Refractive Indices for Refraction to Occur
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Refraction Problem: A beam of sunlight is directed into a prism at the correct incident angle to produce refraction. However, the light passes directly through the prism without being refracted. What could cause this?
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Refraction Solution: Light Must Encounter Two Different Refractive Indices for Refraction to Occur. If the prism were immersed in a fluid with the same refractive index as the glass that the prism is constructed, there would be no refraction and the light would pass through as if the prism were not there.
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Refraction Problem: The prism in the picture is made of ice (n=1.31) and surrounded by oil (n=1.48). The light enters the prism horizontally. What is the angle of the exiting light to normal?
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Refraction Solution: Ice qi = 270 n1 =1.48 n2 =1.31
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Refraction Solution: Oil qi = 240 n1 =1.31 n2 =1.48
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Homework: Page 589 Refraction Problems: 36 (42.8o)
37 (a, 31.3o b, 44.2o c, 49.8o) 38 ?
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