Peterson’s SAT II Success: Physics 120 The law of reflection can be stated as follows: The angle of incidence equals the angle of reflection. A spherical concave mirror reflects light toward a point on the central radius called the focal point. The radius of curvature is the distance from the mirror to its center if the mirror were a full sphere. This is called the principal axis. Concave spherical mirrors are important tools in science because of their ability to focus incident light. The point where incident rays are focused is called the focal point and is one half the radius of curvature. f r = 2 CHAPTER 3 Peterson’s: www.petersons.com 121 Two rules worth remembering about light rays that strike a con- cave mirror are: 1. Any light ray that is moving parallel to the principal axis and is incident to the mirror will be reflected though the focal point. 2. Any light ray that passes through the focal point and is incident to the mirror will be reflected parallel to the principal axis. Light rays from objects (O) placed in front of a concave mirror produce images (I). The images produced may be real (projectable onto a screen) or virtual (appearing to be located on the other side of the mirror); they may be enlarged (magnified) or shrunk (reduced); or they may be right side up (erect) or upside down (inverted). The following diagrams illustrate the position of the image when the object is placed in various locations in front of the mirror. Object distance is considered infinite. Light rays are parallel as they approach and strike the mirror. Image is located at f. It is reduced, inverted, and real. Object is located outside r, but not at infinity. Image is located between f and r. It is reduced, inverted, and real. WAVE PROPERTIES Peterson’s SAT II Success: Physics 122 Object is located at r. Image is located at r. It is the same size as the object, inverted, and real. Object is located between r and f. Image is located beyond r. It is magnified, inverted, and real. Object is located inside f. Image is located behind the mirror. It is reduced, erect, and virtual. CHAPTER 3 Peterson’s: www.petersons.com 123 The letters I and O are used to represent the height of the image (I) and the height of the object (O). The distance from the mirror to the image is labeled q and the distance from the mirror to the object is labeled p. Mirrors are used in a variety of ways throughout the world. Shopkeepers use convex mirrors to keep an eye on the aisles of their stores, and sharp corners on roadways have mirrors set out so drivers can see the road ahead. The magnifying capability of mirrors allows astronomers and laboratory researchers to perform their work. This capability of a mirror is based upon the set of ratios (of the object and image heights) compared to the distance of the object and the image from the mirror. h h q p i 0 = Convex spherical mirrors produce images that are always virtual. The focal length for convex mirrors is always negative. Object is located in front of the mirror. Image is located behind the mirror and appears to be inside it. It is reduced, erect, and virtual. The equation that describes the location of an image in a mirror is called the mirror equation. 111 fpq =+ The focal length (f) is positive for concave mirrors and negative for convex mirrors. If the image (q) or the object (p) are located in front of the mirror, they are positive. If the image (q) and/or the object is/are located behind the mirror, it is negative. WAVE PROPERTIES Peterson’s SAT II Success: Physics 124 Example Let’s do a typical reflection problem. Find the location of the image for an object that is placed 37 cm in front of a mirror having a focal length of 3.6 cm. Describe the image. Solution 111 11 1 36 1 37 1 277 027 fpq fpq q =+ −= −= − rearranges to 1 cm cm cm cm . 1 cm 4cm = == q q 1 025. The image is located 4 cm from the mirror. The object is outside the radius of curvature (2f = 7.2 cm), and the image is located between r and f. It is reduced, inverted, and real. Now we’ll use the original information from the problem above, but we’ll replace the concave mirror with a convex mirror. 111 11 1 36 1 37 1 277 − =+ − −= − −= −− fpq fpq q rearranges to 1 cm cm cm . 0027 1 304 3 29 cm 1 cm cm = =− = − q q CHAPTER 3 Peterson’s: www.petersons.com 125 The image is located 3.29 cm behind the mirror (the negative sign for q). It is reduced, erect, and virtual. REFRACTION Waves that move into a new medium bend as they enter the me- dium. All waves can be refracted, but our discussion here will be limited to light. The light beam shown below passes from air into a cube of plastic. At the air/plastic boundary, the light ray bends (refracts) toward the normal when the light ray passes into the plastic. The ray moves in a straight line while in the plastic until it reenters the air. At the plastic/air boundary the light ray bends (refracts) away from the normal. When light enters an optically more dense material, it refracts toward the normal. When it enters an optically less dense material, it refracts away from the normal. WAVE PROPERTIES Peterson’s SAT II Success: Physics 126 The mathematical relationship between the velocity of light in one material (usually air) compared to another material was determined to be nn 112 2 sin sinθθ= (Snell’s Law) , where n is the index of refraction of the material through which light is passing. Example A typical Snell’s Law problem is one where a scuba diver shines a light upward into the air from under the water. The light beam makes an angle of 30° from the vertical. What is the angle of the light beam as it enters the air? Solution The index of refraction (n) for water is 1.33 and for air is 1. Stating Snell’s Law nn n n 112 2 2 11 2 2 2 133 5 1 655 sin sin sin sin sin ( . )(. ) . sin θθ θ θ θ θ = = == = 440 5. ° The light beam will enter the air at an angle of 40.5° from the normal. Occasionally, a light ray strikes the surface boundary between two materials at an angle (called critical angle) that is too large (from the normal) to pass through the interface between the materials. The light ray is reflected from the surface where it becomes “trapped” inside the material, a condition called total internal reflection. The sparkle of a diamond and optic fiber lights and cables are examples of total internal reflection. Lenses are important because they can focus light. Convex lenses converge light rays passing through them toward the focal point f. This kind of lens is called convergent. Concave lenses separate light rays passing through them as if the separated light rays had originated at the focal point. This kind of lens is called divergent. The two light rays of which to take note are: • A light ray that is parallel to the principal axis and is incident upon the lens will be refracted through the focal point (on the other side of the lens). CHAPTER 3 Peterson’s: www.petersons.com 127 • A light ray that passes through the focal point and is incident upon the lens will be refracted parallel to the principal axis (on the other side). Light rays from object (O) placed in front of a curved lens produce images (I). The diagrams below illustrate the position of the image when the object is placed in various positions. Object distance is considered infinite. Light rays are parallel. Image is located at f. It is reduced, inverted, and real. Object distance is outside 2f but not infinite. Image is located between 2f and f. It is reduced, inverted, and real. WAVE PROPERTIES Peterson’s SAT II Success: Physics 128 Object is located at 2f. Image is located at 2f. It is same size as object, inverted, and real. Object is at f. Image is at infinity. Object inside f. Image same side and outside object; magnified, erect, and virtual. The lens equation 111 fpq =+ is the same as the equation used for mirrors. Remember though, the image distance (q) is positive when on the opposite side of the lens. CHAPTER 3 Peterson’s: www.petersons.com 129 Concave lenses produce virtual images. As with the mirror equation, the lens equations can be used to locate an image. Example An object is placed 20 cm from a convex lens of 8 cm focal length. Locate and describe the image. Solution 111 fpq =+ Rearrange and substitute: 111 1 8 11 125 05 1 1 075 fpq q q q −= −= −= == cm 20cm cm cm cm 13.25cm . The image is located 13.25 cm from the lens on the side opposite from the object. It is reduced, inverted, and real. WAVE PROPERTIES [...]... remember n = 1 in this case nλ = d sin θ and n = 1 λ = d sin θ λ = (1 × 10 6 m )(.5 41) λ = 5. 46 × 10 −7 m or 5 46. 1m If the angle to the fourth order fringe of a blue light is known to be 3.9°, and the distance between slits is 0025 cm, find the wavelength of the light Peterson’s: www.petersons.com 13 3 CHAPTER 3 Solution Remember this is the 4th order fringe, thus n = 4 The sine of 3.9º is 6. 8 × 10 1 nλ =... compose the basis of the kinetic theory: 1 All matter is made of very small particles called atoms and molecules 2 The particles of matter are in constant random motion 3 The particles of matter experience perfectly elastic collisions with one another and with the walls of their containers 14 2 Peterson’s SAT II Success: Physics THERMAL PROPERTIES OF MATTER The particles of a gas in a closed container... 13 3 CHAPTER 3 Solution Remember this is the 4th order fringe, thus n = 4 The sine of 3.9º is 6. 8 × 10 1 nλ = d sin θ and n = 4 ∴ 4λ = d sin θ  0025cm  −2  10 0cm/m  (6. 8 × 10 )  λ= 4 λ = 4.25 × 10 −7 or 425nm 13 4 Peterson’s SAT II Success: Physics CHAPTER SUMMARY CHAPTER SUMMARY • Waves are periodic vibrations that carry energy • Interference can be constructive or destructive • The velocity of... interference (light spots) and destructive interference (dark spots) 13 2 Peterson’s SAT II Success: Physics WAVE PROPERTIES The light traveling from Slit 1 (S1) travels a number of whole wavelengths (nλ) to reach the screen at P1 Additionally, the light from Slit 2 (S2) would travel the same number of whole wavelengths plus one more, nλ + 1 Typically more than one bright spot on the fringe is visible The... Eventually, the gas particles strike one of the sides of the container The impact each particle exerts on the container wall is its momentum, which is a product of the particle’s mass and its velocity The number of times the container wall is struck by the gas particles within depends on the number of gas particles in the container and the velocity at which the particles are moving The faster the particles move,... absolute scale • There are 18 0 degrees between the freezing and boiling points of water on the Fahrenheit scale • There are 10 0 degrees between the freezing and boiling points of water on the Celsius scale • The Kelvin scale places the freezing point of water at 273K and the boiling point of water at 373K, which is also a 10 0-unit difference The Celsius and the Kelvin scales have a 1: 1 relationship Thus to... These are called phase changes • Solids change to liquids by melting • Liquids change to solids by freezing • Liquids change to gases by evaporation • Gases change to liquids by condensation 14 0 Peterson’s SAT II Success: Physics THERMAL PROPERTIES OF MATTER Above is a generalized heat and temperature graph for materials Note the increase in temperature in the substance until a phase change begins to... can take these results a step further and apply them to light Light is considered a transverse wave that only differs from our rope example in that light vibrates 360 ° around the line path of the light ray 13 0 Peterson’s SAT II Success: Physics WAVE PROPERTIES The polarized eyeglasses many people wear restrict the intensity of the light that reaches their eyes by using a device (polarizer) that only... passed through a focal point on the other side of the mirror 1 1 1 • The mirror equation f = p + q is the same for both concave and convex mirrors: f is positive for concave mirrors, and f is negative for convex mirrors • Light passing between two transparent materials is refracted at the surface boundary of the materials • Snell’s Law is n1 sin 1 = n2 sinθ2 where n is the index of refraction for the materials,... to the container wall and the more often the particles strike the container wall The enclosed gas particles have constant mass, which means the momentum of the gas only changes when the velocity of the particle changes If the particle moves more slowly, then its kinetic energy decreases; if it moves faster, its kinetic energy increases A higher number of particles within the container will collide with . describe the image. Solution 11 1 fpq =+ Rearrange and substitute: 11 1 1 8 11 12 5 05 1 1 075 fpq q q q −= −= −= == cm 20cm cm cm cm 13 .25cm . The image is located 13 .25 cm from the lens on the. convex mirror. 11 1 11 1 36 1 37 1 277 − =+ − −= − −= −− fpq fpq q rearranges to 1 cm cm cm . 0027 1 304 3 29 cm 1 cm cm = =− = − q q CHAPTER 3 Peterson’s: www.petersons.com 12 5 The image. length of 3 .6 cm. Describe the image. Solution 11 1 11 1 36 1 37 1 277 027 fpq fpq q =+ −= −= − rearranges to 1 cm cm cm cm . 1 cm 4cm = == q q 1 025. The image is located 4 cm from the mirror.