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Plane Mirror •A Plane Mirror Is a Flat Mirror That Reflects Light Specularly General Physics II Ray Optics 1 Waves and Wavefronts •From a point source, spherical waves emanate outward. All points on a sphere concentric with the source are in phase (same point in oscillation cycle: crest, trough, etc.). Such a sphere is called a wavefront. Electron waves (IBM Almaden) 2 Rays • The direction of the outward radiating waves is along radial lines emanating outward from the source. These lines are perpendicular everywhere to the wavefronts and are called rays. • At very large distances from the source, the wavefronts have very large radii and are essentially flat planes. The rays are then (nearly) parallel lines. • It is sometimes convenient to analyze the behavior of light in terms of the ray model. In this model, light travels in straight lines (i.e., rays) from a source. 3 Seeing Objects • To see an object, light rays from it must enter your eye. You see a self-luminous object like a flame because light rays emitted by it enter your eye (and is then processed by your brain). • To see a non-luminous object, light rays must be reflected from it and then enter your eye. • The sky appears bright in the daytime because sunlight is scattered in all directions by atoms in the atmosphere. Some of this scattered light enters your eye, which allows you to see the bright sky, even though the particles themselves are not visible. 4 Workbook: Chapter 18, Question 1 5 Reflection • When light is reflected from a smooth flat surface, a parallel bundle of rays, or, parallel beam, that is incident on the surface is reflected as a parallel beam. This is called specular reflection. Mirrors produce specular reflection. • Reflection from a rough surface produces diffuse reflection. An incident parallel beam produces a non-parallel beam on reflection. 6 Law of Reflection • Consider a ray that is incident on a interface. The normal is an imaginary line drawn perpendicular to the interface at the point at which the ray strikes the interface. • The angle of incidence (θi ) is the angle between the incident ray and the normal. The angle of reflection (θr ) is the angle between the reflected ray and the normal. •The law of reflection states that: (i) The angle of incidence is equal to the angle of reflection, i.e., θi = θr . (ii) The incident ray, reflected ray, and the normal lie in one plane. 7 Light enters horizontally into the combination of two perpendicular mirrors as shown below. After reflecting off of both mirrors the direction of the incident light 1. bounces back and forth many times, until it hits the corner. 2. depends on the mirror angle θ. 3. is reflected back and upwards. 4. is turned around by 180o. 5. is reflected back and downwards. 8 Image Formation •An object is anything from which light rays diverge (spread out). The object may be self-luminous or it may reflect light originating from a self-luminous source. Each point on an object acts as a source of diverging rays (even if the object is simply reflecting light). • Let us consider a single point on an object. An image of this point is formed by an optical device (e.g., mirror or lens) when light rays from the object point, after being reflected or refracted by the device, either converge to a point or appear to diverge from a point. This point is called the image point. • If the light rays actually converge to and pass through the image point, the image is a real image. • If the light rays appear to diverge from the image point, the image is a virtual image. No light rays actually pass through a virtual image. 9 Image Formation by a Plane Mirror •A plane mirror is a flat mirror that reflects light specularly. • Due to reflection of light originating from the point object P, an image is formed by the mirror at P’. Two rays from the object are all that are needed to locate the image exactly. • Note that light rays only appear to be diverging from the image point P’. Thus, the image is virtual. • From the geometry, the object distance (s) has the same magnitude as the image distance (s’). Hence, the image is as far behind the mirror as 10 the object is in front. Image Formation by a Plane Mirror • For an extended object, each point on the object gives rise to a corresponding image point. • As shown, the image of the top of the object is above the normal. The same is true for the images of all other points on the object. Hence, the image is upright, i.e., the same orientation as the object. • The geometry also shows that the image is the same height as the object. This is true for the width and thickness as well, so the image is the same size as the object. 11 10 Images Due to Two Perpendicular Mirrors Images P1’ and P2’ are each due to reflections from one mirror. P3’ is due to reflections from both mirrors. 12 An observer O, facing a mirror, observes a light source S. Where does O perceive the mirror image of S to be located? 1. 1 2. 2 3. 3 4. 4 5. Some other location. 6. The image of S cannot be seen by O when O and S are located as shown. 13 Workbook: Chapter 18, Question 7. 14 Refraction • When light waves arrive at the interface between two transparent media, the waves are partially reflected and partially transmitted. In the context of optics, the transmitted ray is called the refracted ray. • The direction of the refracted ray for a given direction of the incident ray depends on the index of refraction of each medium. • The index of refraction of a medium is defined by n= speed of light in vacuum =c. speed of light in medium v 15 Refraction • The speed of light is maximum in vacuum, so n > 1 for every material medium. 16 Law of Refraction (Snell’s Law) • Consider a ray incident on the interface between two transparent media, 1 and 2. The law of refraction, or, Snell’s law, is given by nn112sinθ = sinθ2 , where is the angle with the θ1 normal in medium 1 and is θ2 the angle in medium 2. n1 and n2 are the corresponding indices of refraction. 17 Law of Refraction (Snell’s Law) nnsin sin . 1122θ = θ • If the medium of incidence (medium 1) has a smaller index of θ1 refraction than the second 1 medium, i.e., n1 < n2, then 2 according to the law of refraction, θ2 θ1 > θ2, i.e., the refracted ray is bent toward the normal. • If the medium of incidence has a greater index of refraction than θ1 the second medium, i.e., n1 > n2, 1 then according to the law of 2 θ2 refraction, θ1 < θ2, i.e., the refracted ray is bent away from the normal. 18 Some Phenomena Due to Refraction • “Bent” pencil when it is placed in water. plus.maths.org/issue15/news/refract/index.html • “Flattened” sun when it is near the horizon. Flattened Sun 19 Workbook: Chapter 18, Question 10 Textbook: Chapter 18, Problem 15 20 Refraction by a Prism b a Which medium has the greater index of refraction? 1. a 2. b 21 3. Same. Total Internal Reflection • As seen before, if the medium of incidence has a greater index of refraction than the second medium, light is bent away from the normal. If the angle of incidence is increased continuously, the angle of refraction increases until it becomes 90°. At this point, the refracted ray travels along the interface between the two media. If the angle of incidence is increased further, the refracted ray disappears and the incident ray is totally reflected. This phenomenon is called total internal reflection. 22 Total Internal Reflection • From Snell’s law, n 2 sinθθ= n sin . 121 The critical angle is the angle of incidence for which the angle of refraction is 90° . Thus, nn 22 sinθcrit =°=nn sin90 . 11 n 2 sinθcrit =<n . (nn ) 1 21 For a water-air interface, sinθ = 1 =0.750. crit 1.333 θcrit =°48.6 . 23 Total Internal Reflection • A fish or person looking up at the water’s surface at an angle greater than 48.6° will see only reflections of objects below the surface! 24 Applications of Total Internal Reflection Binoculars Optical Fiber 25 Workbook: Chapter 18, Questions 11, 13 Textbook: Chapter 18, Problem 18 P18.18. Prepare: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection. The critical angle of incidence is given by Equation 18.3. Solve: −1 ⎛ ncl adding ⎞ −1 ⎛1.48⎞ θc = sin ⎜ ⎟ = sin ⎜ ⎟ = 67.7° ⎝ ncore ⎠ ⎝1.60⎠ Thus, the maximum angle a light ray can make with the wall of the core to remain inside the fiber is 90° − 67.7° = 22.3°. 26 Assess: We can have total internal reflection because ncore > ncladding. .
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