Ch 23 1 Chapter 23 Light: Geometric Optics © 2006, B.J. Lieb Some figures electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey Giancoli, PHYSICS,6/E © 2004.

Ch 23 2 Wave fronts and Rays Light is a wave phenomena which can be represented by “wave fronts” as shown in blue. In most cases in is better to use “rays” shown in red, which are always perpendicular to the wave fronts.

Ch 23 3 Reflection For light reflected from a smooth surface: Angle of incidence equals angle of reflection. Note that angles are measured relative to the normal to the surface.

Ch 23 4 Virtual Image In the figure below, rays from the bottle reflect off the mirror and enter the eyes of the person. The reflected rays appear to come from a point inside of the mirror where there is no light as indicated by the dotted lines. This is a virtual image.

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6 Real Image A real image is formed when rays actually converge at a point. If you placed a screen there, you would see an image reflected on the screen. In this figure, the incoming rays are parallel which often means that the object emitting them is far away.

Ch 23 7 Index of Refraction All electromagnetic radiation travels at the same speed in vacuum, often called the speed of light: c = 3.00 x 10 8 m/s. In materials, light slows down and this is indicated by the index of refraction n where: Indices of Refraction Vacuum Air Water1.33 Crown Glass1.52 Plexiglas1.51 Diamond2.42

Ch 23 8 Refraction As light enters a medium where n 2 > n 1 the velocity is less, but the frequency remains the same, thus 2 = 1 / n 2 if n 1 = 1. The wave fronts bend and the ray is bent toward the normal.

Ch 23 9 Snell’s Law of Refraction Given the angles defined in the picture below:

Ch Example 23-1: A layer of ice (n ice = 1.309) floats on water. The sides of the ice are parallel. If light is incident on the upper surface of the ice at an angle of 30.0º, what is the angle of refraction in the water?

Ch Total Internal Reflection In the figure below, light is refracted at a boundary where n 2 < n 1. As the angle of incidence increases, eventually a critical angle θ C is reached where the angle of refraction is 90 O. Beyond θ C no light can exit the material. This is total internal refraction. And since sin 90 o = 1,

Ch Applications of Total Internal Reflection Total internal reflection makes it possible for light to pass down a glass fiber with minimal losses. This facilitates applications in Communications – because of its high frequency, a light beam can carry much more information then current in a wire. Medicine – because a bundle of these fibers can transmit a clear view of internal parts of the body such as the lungs (bronchoscope).

Ch Example 23-2: Calculate the critical angle for total internal reflection for light that is inside of a diamond and incident on the interface between the diamond and air. The small critical angle of the diamond allows light to become trapped inside the diamond and will eventually emerge essentially perpendicular to one of the many cut faces. The small critical angle of the diamond is thus responsible for its sparkle.

Ch Lenses Converging Lenses are convex (thicker in the center and have positive focal length. Diverging Lenses are concave (thinner in the center) and have negative focal length.

Ch Focal Length ( f ) Almost everyone has used a lens to focus light from the sun to burn a piece of paper. The sun is far away, so the rays from the sun are parallel. The distance from the lens to the paper is f the focal length. The focal length of a diverging lens is less obvious-it is the distance from the lens to the point where parallel rays appear to diverge from. Focal length is a distance so the units are m.

Ch Power of a Lens People who use lenses to help people see (optometrists, etc.) use the “power” (P) of the lens, which is the inverse of the focal length. The unit of “power” is the diopter (D) which is the inverse of a meter. Example: If the focal length ( f ) is 2.0 m then

Ch Ray Tracing Sketch the lens and place the bottom of the object on the axis of the lens. Mark the focal distance on each side of the lens. Ray 1 leaves the tip of the object and is parallel to the lens axis, so it refracts through focal point. Ray 2 passes through F’, so it emerges parallel to the lens axis Ray 3 passes through the center of the lens, so is not bent. The intersection of the three rays gives the image of the tip of the object. The bottom of the image is on the axis of the lens. This is a real image because the rays actually converge to a point.

Ch Object inside Focal Point

Ch Derivation of the Lens Equation h o = height of object h i = height of image d o = object distance (from object to center of lens) d i = image distance (from image to center of lens) This equation follows from the two similar triangles highlighted in yellow. This equation follows from the two similar triangles OAO’ and IAI’ Setting the right hand side of these equations equal gives the lens equation.

Ch The Lens Equation f = focal length (positive for converging lens, negative for diverging lens) h o = height of object (always positive) h i = height of image (positive for upright image) d o = object distance (positive if object is on side of lens from which light is coming. True except in some lens combinations) d i = image distance (positive for real image and thus it is on the side of the lens to which the light is going. Lateral Magnification m is positive for an upright image; negative for an inverted image.

Ch Hints for Solving Lens Problems 1.Read problem and identify “givens” 2.Draw rough ray diagram and estimate what the answer is 3.Re-read problem and check your “givens” and the ray diagram. 4.Make sure you have applied the sign conventions to the “givens” properly. 5.Plug in numbers and don’t forget to take reciprocal. 6.Does your answer agree with your estimate from #2 above?

Ch Example 23-3: Determine the image distance and image height of a 4.0 cm high object formed by a converging lens of focal length 10.0 cm if the object is placed (a) 30 cm, (b) 10.0 cm and (c) 5.0 cm from the lens. a) b) Object is at infinity X xx

Ch Example 23-3c: Determine the image distance and image height of a 4.0 cm high object formed by a converging lens of focal length 10.0 cm if the object is placed (a) 30 cm, (b) 10.0 cm and (c) 5.0 cm from the lens. c) xx Negative d i indicates it is on the side from which the light is coming

Ch Combinations of Lenses For combinations of lenses: 1.Ignore the second lens and solve for the image distance ( d i ) and image height ( h i ) of the first lens that the light enters. 2.The image of the first lens becomes the object for the second lens. 3.Use the image distance from the first lens to calculate the object distance for the second lens. 4.It is important to note that for combinations of lenses it is possible to have a negative object distance ( d o2 ). 5.The total magnification of the combination is the product of the magnifications of the two lenses.

Ch Example 23-4: Two converging lenses are placed 20 cm apart. If the first lens has a focal length of 10 cm and the second lens has a focal length of 20 cm, determine the location of the image of an object placed 30 cm in front of the first lens. Ignore the second lens and find the image of the first. Image from first lens is inverted and half the size of the object.

Ch Example 23-4 continued: Two converging lenses are placed 20 cm apart. If the first lens has a focal length of 10 cm and the second lens has a focal length of 20 cm, determine the location of the image of an object placed 30 cm in front of the first lens. The image from the first lens is the object for the second lens. Virtual image

Ch Spherical Mirrors In this course we concentrate on lenses. Much of what you learn about lenses also applies to spherical mirrors The equation for determining the location of the image is the same as for lenses The rules for the signs of the image and object distance are similar.

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Ch Spherical Mirrors A real image can only be formed on the reflecting side of the mirror (since no light passes through the mirror) The focal length of a mirror is ½ of the radius of curvature of the mirror