Chapter 18 Ray Optics Chapter Goal: To understand and apply the ray model of light. Slide 18-1
Chapter 18 Preview Looking Ahead Text p. 565 Slide 18-2
Wavefronts and Rays When visible light or other electromagnetic waves interact with systems much larger than the wavelength (say 1 mm for visible light), it s a good approximation to Neglect the wave nature of light thus ignore diffraction. Assume that the light travels in straight lines, called rays, while in a uniform medium. Rays are perpendicular to the wavefronts and the rays direction is that of the wave propagation. The ray approximation is also known as geometrical optics. Slide 18-3
Reflection and Refraction Two things happen when a light ray crosses the boundary between the air and the glass: 1. Part of the light reflects from the boundary, 2. Part of the light refracts through the glass (and then refracts again as it goes back into into the air). Slide 18-4
Reflection When light reflects from a surface, the incident and reflected rays make the same angle with the normal to the surface. For smooth surfaces, parallel rays all reflect at the same angle. This is specular reflection. The incident angle=reflected angle. For rough surfaces,the light rays reflect in seemingly random directions. This is called diffuse reflection. The angles are equal (locally) even for rough surfaces. fuse reflection. Slide 18-5
The Plane Mirror Slide 18-6
QuickCheck 18.4 An object is placed in front of a mirror. The observer is positioned as shown. Which of the points shown best indicates where the observer would perceive the image to be located? Mirrror Mirror B Observer A Object C Slide 18-7
Question You would like to buy a full-length mirror which allows you to see yourself from head to toe. The minimum height of the mirror is A) half your height. B) two-thirds of your height. C) equal to your height. D) depends on distance you stand from mirror Slide 18-8
Refraction Refraction is the bending of light as it crosses the interface between two different transparent media. Refraction occurs because the wave speed differs in different media. For light, the index of refraction n describes the speed change. The speed of a wave in a medium is v = c/n. The angles of incidence and refraction are related by Snell s law: n 1 sin 1 = n 2 sin 2 Slide 18-10
Derivation of Snell s Law 1 x =sin 1 2 x =sin 2 1 sin 1 = i = 2 sin 2 tc/n i tc n 1 sin 1 = tc n 2 sin 2 n 1 sin 1 = n 2 sin 2 Slide 18-11
QuickCheck 18.5 A laser beam passing from medium 1 to medium 2 is refracted as shown. Which is true? A. n 1 < n 2 B. n 1 > n 2 C. There s not enough information to compare n 1 and n 2 Slide 18-12
From:http://www.physicsclassroom.com/class/refrn/Lesson-1/Refraction-and- Sight
QuickCheck 18.7 A light ray enters a glass prism as shown. Which is a possible path for the ray through the prism? A B C D Slide 18-14
Question 23.8 Gone Fishin I To shoot a fish with a gun, should you aim directly at the image, slightly above, or slightly below? 1) aim directly at the image 2) aim slightly above 3) aim slightly below Slide 18-15
Example: Sunlight strikes the surface of a lake. A diver sees the Sun at an angle of 42.0 with respect to the vertical. What angle do the Sun s rays in air make with the vertical? incident wave n 1 = 1.00; air n 2 = 1.33; water Transmitted wave 42 θ 1 Normal n 1 surface sinθ 1 = sinθ ( 1.00) sinθ = ( 1.333) 1 sinθ1 = 0.8920 θ = 63.1 1 n 2 2 sin 42
QuickCheck 18.8 A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, the distance to the cat appears to be A. Less than the actual distance. B. Equal to the actual distance. C. More than the actual distance. Slide 18-17
Clicker question The figure shows the path of a light ray through three different media. Rank the media in order of their refractive indices. A. B. C. D. n > n > n 1 2 3 n3 > n1 > n2 n > n > n 3 2 1 n > n > n 2 1 3 Slide 18-18
Mirages
Total Internal Reflection Total internal reflection (TIR) occurs when a light ray is unable to refract through a boundary. Instead, 100% of the light reflects from the boundary. Slide 18-20
Total internal reflection The angle of incidence for when the angle of refraction is 90 is called the critical angle. n n 1 1 sinθ = sinθ sinθ 1 c c = = n n 2 2 n n 2 1 sinθ 2 sin90 = n 2 If the incidence angle is greater than the critical angle, the beam can not refract but is completely reflected Slide 18-21
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Fiber Optics Fiber optics use total internal reflection for the transmission of light through optical fibers. Slide 18-23
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Thin Converging Lenses Focal length is distance from the lens parallel incoming rays converge. Diverging rays emitted from the focal point will be parallel after going through the converging lens. Slide 18-25
Real Images Slide 18-26
QuickCheck 18.13 Which of these ray diagrams is possibly correct? Slide 18-27
Thin Lens Equation Triangle PQO is similar to P Q O. Thus y s = y0 s 0 ) y0 y = s0 s m = y0 y = s0 s Also triangles OAF 2 and P Q F 2 are similar: y f = y 0 s 0 f ) y0 y = s0 f f Equating these two equations: 1 s + 1 s 0 = 1 f Slide 18-28
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QuickCheck 18.14 A lens creates an image as shown. In this situation, the object distance s is Larger than the focal length f Equal to the focal length f Smaller than focal length f Slide 18-30
QuickCheck 18.15 A lens creates an image as shown. In this situation, the image distance sʹ is Larger than the focal length f Equal to the focal length f Smaller than focal length f Slide 18-31
QuickCheck 18.11 A lens produces a sharply focused, inverted image on a screen. What will you see on the screen if a piece of dark paper is lowered to cover the top half of the lens? An inverted but blurry image An image that is dimmer but otherwise unchanged Only the top half of the image Only the bottom half of the image No image at all Slide 18-32
QuickCheck 18.12 A lens produces a sharply focused, inverted image on a screen. What will you see on the screen if the lens is covered by a dark mask having only a small hole in the center? An inverted but blurry image An image that is dimmer but otherwise unchanged Only the middle piece of the image A circular diffraction pattern No image at all Slide 18-33
Example 18.7 Finding the image of a flower A 4.0-cm-diameter flower is 200 cm from the 50-cm-focal-length lens of a camera. How far should the plane of the camera s light detector be placed behind the lens to record a well-focused image? What is the diameter of the image on the detector? Slide 18-34
Magnification The image can be larger or smaller than the object, depending on the location and focal length of the lens. The magnification m describes the orientation of the image relative to the object and its size. 1. The absolute value of m fives the ratio of image height to object height: hʹ /h = m. 2. A positive value of m indicates that the image is upright relative to the object. A negative value of m indicates that the image is inverted. Slide 18-35
Virtual Images When the object is inside the focal point, diverging rays from the object will not be focused by the lense. The rays are still diverging from each other on the right side of the mirror. Extrapolating these rays to the left side of the lens gives a virtual image to the left of the object. Slide 18-36
Virtual Images and Magnifying glasses. The magnification is positive since the virtual image is upright. Since m = sʹ /s, we define the image distance s to be negative for a virtual image. Slide 18-37