Presentation on theme: "Foundations of Physics"— Presentation transcript:
1 Foundations of Physics CPO ScienceFoundations of PhysicsUnit 6, Chapter 17
2 Chapter 17 Light and Color Unit 6: Light and OpticsChapter 17 Light and Color17.1 Reflection and Refraction17.2 Mirrors, Lenses, and Images17.3 Optical Systems
3 Chapter 17 ObjectivesDescribe the functions of convex and concave lenses, a prism, and a flat mirror.Describe how light rays form an image.Calculate the angles of reflection and refraction for a single light ray.Draw the ray diagram for a lens and a mirror showing the object and image.Explain how a fiber-optic circuit acts like a pipe for light.Describe the difference between a real image and a virtual image and give an example of each.
5 17.1 Reflection and Refraction Key Question:How do we describe the reflection and refraction of light?*Students read Section AFTER Investigation 17.1
6 17.1 Reflection and Refraction The overall study of how light behaves is called optics.The branch of optics that focuses on the creation of images is called geometric optics, because it is based on relationships between angles and lines that describe light rays.
7 17.1 Reflection and Refraction A lens is an optical device that is used to bend light in a specific way.A converging lens bends light so that the light rays come together to a point.A diverging lens bends light so it spreads light apart instead of coming together.
8 17.1 Reflection and Refraction Mirrors reflect light and allow us to see ourselves.A prism is another optical device that can cause light to change directions.A prism is a solid piece of glass with flat polished surfaces.
9 17.1 ReflectionImages appear in mirrors because of how light is reflected by mirrors.The incident ray follows the light falling onto the mirror.The reflected ray follows the light bouncing off the mirror.
10 17.1 ReflectionIn specular reflection each incident ray bounces off in a single direction.A surface that is not shiny creates diffuse reflection.In diffuse reflection, a single ray of light scatters into many directions.
11 Law of Reflection The incident ray strikes the mirror. The reflected ray bounces off.The angle of incidence equals the angle of reflection.
12 17.1 Law of reflection30o30o1) You are asked for a ray diagram and the angle of reflection.2) You are given the angle of incidence.3) The law of reflection states the angle of reflection equals the angle of incidence.4) The angle of reflection is 30°.A light ray is incident on a plane mirror with a 30 degree angle of incidence.Sketch the incident and reflected rays and determine the angle of reflection.
13 17.1 RefractionLight rays may bend as they cross a boundary from one material to another, like from air to water.This bending of light rays is known as refraction.The light rays from the straw are refracted (or bent) when they cross from water back into air before reaching your eyes.
14 17.1 RefractionWhen a ray of light crosses from one material to another, the amount it bends depends on the difference in index of refraction between the two materials.
15 17.1 Index of refractionThe ability of a material to bend rays of light is described by the index of refraction (n).
17 17.1 Snell's law of refraction Snell’s law is the relationship between the angles of incidence and refraction and the index of refraction of both materials.Angle of refraction(degrees)Angle of incidence(degrees)ni sin Qi = nr sin QrIndex of refraction of incident materialIndex of refraction of refractive material
18 17.1 Calculate the angle of refraction A ray of light traveling through air is incident on a smooth surface of water at an angle of 30° to the normal.Calculate the angle of refraction for the ray as it enters the water.1) You are asked for the angle of refraction.2) You are told the ray goes from air into water at 30 degrees.3) Snell’s law:ni sin(θi) = nr sin(θr)ni = 1.00 (air), nr = 1.33 (water)4) Apply Snell’s law to find θr.1.00sin(30°) = 1.33 sin(θr)sin(θr) = 0.5 ÷ 1.33 = 0.376Use the inverse sine function to find the angle that has asine ofθr = sin-1(0.376) = 22°
19 The angle at which light begins reflecting back into a refractive material is called the critical angle, and it depends on the index of refraction.The critical angle for water is about 49 degrees.
20 17.1 Dispersion and prismsWhen white light passes through a glass prism, blue is bent more than red.Colors between blue and red are bent proportional to their position in the spectrum.
21 17.1 Dispersion and prismsThe variation in refractive index with color is called dispersion.A rainbow is an example of dispersion in nature.Tiny rain droplets act as prisms separating the colors in the white light rays from the sun.
22 17.2 Mirrors, Lenses, and Images Key Question:How does a lens or mirror form an image?*Students read Section AFTER Investigation 17.2
23 17.2 Mirrors, Lenses, and Images We see a world of images created on the retina of the eye by the lens in the front of the eye.
24 17.2 Mirrors, Lenses, and Images Objects are real physical things that give off or reflect light rays.Images are “pictures” of objects that are formed in space where light rays meet.
25 17.2 Mirrors, Lenses, and Images The most common image we see every day is our own reflection in a mirror.The image in a mirror is called a virtual image because the light rays do not actually come together.The virtual image in a flat mirror is created by the eye and brain.
26 17.2 Mirrors, Lenses, and Images Light rays that enter a converging lens parallel to its axis bend to meet at a point called the focal point.The distance from the center of the lens to the focal point is called the focal length.The optical axis usually goes through the center of the lens.
27 A converging lens bends an incident light ray parallel to the optical axis toward the focal point. A diverging lens bends an incident light ray parallel to the axis outward, away from the focal point
28 17.2 The image formed by a lens A lens can form a virtual image just as a mirror does.Rays from the same point on an object are bent by the lens so that they appear to come from a much larger object.
29 17.2 The image formed by a lens A converging lens can also form a real image.In a real image, light rays from the object actually come back together.
30 17.2 Drawing ray diagramsA ray diagram is the best way to understand what type of image is formed by a lens, and whether the image is magnified or inverted.These three rays follow the rules for how light rays are bent by the lens:A light ray passing through the center of the lens is not deflected at all (A).A light ray parallel to the axis passes through the far focal point (B).A light ray passing through the near focal point emerges parallel to the axis (C).
33 17.3 Optical Systems Key Question: How are the properties of images determine?*Students read Section AFTER Investigation 17.3
34 17.3 Optical SystemsAn optical system is a collection of mirrors, lenses, prisms, or other optical elements that performs a useful function with light.Characteristics of optical systems are:The location, type, and magnification of the image.The amount of light that is collected.The accuracy of the image in terms of sharpness, color, and distortion.The ability to change the image, like a telephoto lens on a camera.The ability to record the image on film or electronically.
35 17.3 The sharpness of an image Defects in the image are called aberrations and can come from several sources.Chromatic aberration is caused by dispersion, when different colors focus at different distances from the lens.
36 17.3 The sharpness of an image Spherical aberration causes a blurry image because light rays farther from the axis focus to a different point than rays near the axis.
37 17.3 The sharpness of an image Diffraction causes a point on an object to focus as a series of concentric rings around a bright spot.
38 17.3 Thin lens formula 1 + 1 = 1 do di df The thin lens formula is a mathematical way to do ray diagrams with algebra instead of drawing lines on graph paper.= 1do di dfObjectdistance(cm)focallength (cm)Image distance(cm)
39 17.3 Use the thin lens formula 1) You are asked for image distance.2) You are given the focal length and object distance.3) The thin lens formula applies:1/di = 1/f — 1/do4) Solve for di1/di = 1/4 — 1/61/di = 3/12 — 2/12 = 1/12di = 12 cmThe image forms 12 cm to the right of the lens.Calculate the location of the image if the object is 6 cm in front of a converging lens with a focal length of 4 cm.
40 17.3 Image relayA technique known as image relay is used to analyze an optical system made of two or more lenses.