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Lecture 15 Reflection & Refraction

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Lecture 15 Reflection & Refraction LECTURE 15 REFLECTION & REFRACTION 18.1 The ray model of light Endoscopes use fiber optics. Seeing objects Shadows 18.2 Reflection The plane mirror 18.3 Refraction Total internal reflection Learning objectives 2 Use the ray model of light—how light is emitted from objects, how we see objects, and how shadows are formed. Use the law of reflection to locate images formed in plane mirrors. Apply the laws of reflection and refraction to relate indices of refraction, and angles of incidence, reflection, and refraction. Describe how total internal reflection occurs and calculate or use the critical angle. 18.1 Seeing objects / Demo Demo: Laser beam Why can you not see the laser beam? In order for our eye to see an object, rays from that object must enter the eye. https://tvtropes.org/pmwiki/pmwiki.php/Main/LaserHallway Group exercise: 18.1-1 (Evolution of eyes) 4 a) The eye can sense light and dark. b) How does this eye allow directional sensitivity? Group exercise: 18.1-1 (Evolution of eyes) answer 5 a) The eye can sense light and dark. b) How does this eye allow directional sensitivity? Evolution of eyes 6 c) Pinhole eyes have tradeoff of light collection and sharp imaging. d) Humor allows refraction to converge rays. f) Our eyes see by focusing a bundle of rays using lens. 18.1 Shadows An opaque object can intercept rays, creating a shadow behind it. A point source creates a completely dark shadow with a sharp edge. An extended source often creates a true shadow that no light reaches, surrounded by a fuzzy region of increasing brightness. 18.2 Reflection 8 The incident and reflected rays and the normal lie on the same plane, and the law of reflection states: 휃r = 휃i Specular reflection is the reflection from a smooth surface. Reflection from a rough surface is diffuse reflection. 18.2 The plane mirrors 9 A plane mirror is a flat mirror. The virtual image of P is at Point P. ′ The image distance 푠 is equal to the object distance 푠: 푠′ = 푠 Quiz: 18.2-1 10 An eye at point E is facing a mirror and observes an object O. mirror Where does the observer E perceive the mirror image of the object to be located? O Quiz: 18.2-1 answer / Demo 11 Trace the light rays from the object to the mirror and then to the eye, following the law of reflection. mirror Since the brain assumes that light travels in a E straight line, extend the rays back behind the mirror. The image is located the same distance from the mirror as the object. Demo: Location of image O Quiz: 18.2-2 12 At a hair salon, you are trying to see the back of your head. You hold a 1.0 m 0.5 m small mirror 0.5 m in front of you and look at your reflection in a big mirror 1.0 m behind you. How far behind the small mirror do you see the closest image of the back of your head, in meters? Quiz: 18.2-2 answer 13 The image of the head reflected in the big mirror appears 1.0 m behind the full big mirror. This image (which is the object for the small mirror) is 2.5 m away from the small mirror. The final image is 2.5 m behind the small mirror. 1.0 m 1.0 m 0.5 m big mirror small mirror Image of head in Head Image of head in big mirror small mirror 2.5 m 2.5 m 18.3 Refraction 14 Refraction is a change in direction of light as it transmits from one medium to another. Angle of refraction is the angle of transmitted ray, 휃2. The angles of incidence and refraction, and indices of refraction of two media are related by Snell’s Law: 푛1 sin 휃1 = 푛2 sin 휃2 Refraction occurs because light travels from one point to another along the path that takes least time. Lifeguard analog Quiz: 18.3-1 You are trying to catch fish by using a spear. You observe a large fish a couple of meters in front of you and a meter below the water surface. Assuming that the fish is stationary, in order to hit the fish with your spear you must aim A. directly at the point where you see the fish. B. slightly above the point where you see the fish. C. slightly below the point where you see the fish. Hint: draw a ray diagram with the fish, your eye, and the water-air boundary. Quiz: 18.3-1 answer For you to see the fish, the light must travel from the fish to your eyes. The ray is refracted at the water-air boundary. The angle of incidence is smaller than the angle of refraction since light travels faster in air than in water: 푛water sin 휃i = 푛air sin 휃r You must aim slightly below the point where you see the fish. Follow-up: To shoot a fish with a laser gun instead, should you aim directly at the image, slightly above, or slightly below? 휃r 휃i 18.3 Total internal reflection / Demo 17 If 푛2 < 푛1, total internal reflection (TIR) occurs when the incident angle is greater than the critical angle, 휃c, and no refracted ray exists. 푛2 sin 휃c = 푛1 Demo: TIR Quiz: 18.3-2 18 In an optical fiber, what can be said about the index of refraction of the core, 푛core, versus the index of refraction of the cladding, 푛cladding? Quiz: 18.3-2 answer / Demo 19 In order for there to be total internal reflection light must travel in a higher index of refraction and incident on a boundary with medium of a lower index of refraction. 푛cladding < 푛core Total internal reflection in fiber optic light pipes makes it possible to transport light and light- encoded signals over long distances without significant loss. Endoscopes made from optical fibers are used for anthroscopic surgery. Demo: Fiber optics Quiz: 18.3-3 A laser beam undergoes two refractions at the interface between medium 1 and 2, and total internal reflection at the interface between medium 2 and medium 3 as shown. Rank the indices of refraction of media 1, 2, and 3, smallest first. Quiz: 18.3-3 Since there is TIR at the the interface between medium 2 and medium 3, 푛3 < 푛2. Since the angle of the ray in medium 1 is smaller than the angle in medium 2, 푛2 < 푛1 from Snell’s law: 푛1 sin 휃1 = 푛2 sin 휃2. 푛3 < 푛2 < 푛1.
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