2Objectives: After completing this module, you should be able to: Define and illustrate the following terms: real and virtual images, converging and diverging mirrors, focal length, and magnification.Understand and apply the sign conventions that apply to focal lengths, image distances, image heights, and magnification.Predict mathematically the nature, size, and location of images formed by spherical mirrors.Determine mathematically the magnification and/or the focal length of spherical mirrors.
3Analytical OpticsIn this unit, we will discuss analytical relationships to describe mirror images more accurately. But first we will review some graphical principles covered in Module 34a on light reflection.
4The Plane Mirror = Object distance Image distance p = q p q Image is virtualObject distance: The straight-line distance p from the surface of a mirror to the object.Image distance: The straight-line distance q from the surface of a mirror to the image.
5Spherical MirrorsA spherical mirror is formed by the inside (concave) or outside (convex) surfaces of a sphere.Concave MirrorRadius of Curvature RVertex VCenter of Curvature CLinear apertureVCRAxisA concave spherical mirror is shown here with parts identified.The axis and linear aperture are shown.
6The Focal Length f of a Mirror Since qi = qr, we find that F is mid- way between V and C; we find:Incident parallel rayqrqiRFCVThe focal length f is:axisfFocal pointThe focal length, fThe focal length f is equal to half the radius R
7Converging and Diverging Mirrors Concave mirrors and converging parallel rays will be called converging mirrors.Convex mirrors and diverging parallel rays will be called diverging mirrors.CFConverging MirrorConcaveCFDiverging MirrorConvex
8DefinitionsFocal length: The straight-line distance f from the surface of a mirror to focus of the mirror.Magnification: The ratio of the size of the image to the size of the object.Real image: An image formed by real light rays that can be projected on a screen.Virtual image: An image that appears to be at a location where no light rays reach.Converging and diverging mirrors: Refer to the reflection of parallel rays from surface of mirror.
9Image Construction Summary: Ray 1: A ray parallel to mirror axis passes through the focal point of a concave mirror or appears to come from the focal point of a convex mirror.Ray 2: A ray passing through the focus of a concave mirror or proceeding toward the focus of a convex mirror is reflected parallel to the mirror axis.Ray 3: A ray that proceeds along a radius is always reflected back along its original path.
10Examples of Image Construction The three principal rays for both converging (concave) and diverging (convex) mirrors.CFConverging mirrorCDiverging mirrorFRay 1Ray 3Ray 1Ray 3Ray 2Ray 2Image
11Review of Imaging Facts For plane mirrors, the object distance equals the image distance and all images are erect and virtual.For converging mirrors and diverging mirrors, the focal length is equal to one-half the radius.All images formed from convex mirrors are erect, virtual, and diminished in size.Except for objects located inside the focus (which are erect and virtual), all images formed by converging mirrors are real and inverted.
12Questions About Images 1. Is the image erect or inverted?2. Is the image real or virtual?3. Is it enlarged, diminished, or the same size?4. What are object and image distances p and q?5. What is the height y’ or size of image?6. What is the magnification M = y’/y of image?
13Definition of SymbolsBy applying algebra and geometry to ray-tracing diagrams, such as the one below, one can derive a relationship for predicting the location of images.yY’RqpfObject dist. pImage dist. qFocal length fRadius RObject size yImage size y’
14Mirror EquationThe following equations are given without derivation. They apply equally well for both converging and diverging mirrors.yY’Rqpf
15Sign Convention1. Object distance p is positive for real objects and negative for virtual objects.2. Image distance q is positive for real images and negative for virtual images.3. The focal length f and the radius of curvature R is positive for converging mirrors and negative for diverging mirrors.
16Example 1. A 6 cm pencil is placed 50 cm from the vertex of a 40-cm diameter mirror. What are the location and nature of the image?CFpqfSketch the rough image.p = 50 cm; R = 40 cm
17Example 1 (Cont. ). What are the location and nature of the image Example 1 (Cont.). What are the location and nature of the image? (p = 50 cm; f = 20 cm)CFpqfq = cmThe image is real (+q), inverted, diminished, and located 33.3 cm from mirror (between F and C).
18Working With Reciprocals: The mirror equation can easily be solved by using the reciprocal button (1/x) on most calculators:Possible sequence for finding f on linear calculators:P q1/x+=Finding f:Same with reverse notation calculators might be:Finding f:P q1/x+Enter
19Alternative Solutions It might be useful to solve the mirror equation algebraically for each of the parameters:Be careful with substitution of signed numbers!
20Example 2: An arrow is placed 30 cm from the surface of a polished sphere of radius 80 cm. What is the location and nature of image?Draw image sketch:p = 30 cm; R = -80 cmSolve the mirror equation for q, then watch signs carefully on substitution:
21Example 2 (Cont.) Find location and nature of image when p = 30 cm and q = -40 cm. The image is virtual (-q), erect, and diminished. It appears to be located at a distance of 17.1 cm behind the mirror.
22Magnification of Images The magnification M of an image is the ratio of the image size y’ to the object size y.Magnification:Obj.Img.M = +2M = -1/2y y’y and y’ are positive when erect; negative inverted.q is positive when real; negative when virtual.M is positive when image erect; negative inverted.
23Example 3. An 8-cm wrench is placed 10 cm from a diverging mirror of f = -20 cm. What is the location and size of the image?Y’YpqVirtual imageConverging mirrorFCWrenchq = cmVirtual !Magnification:M =Since M = y’/y y’ = My or:y’ = cm
24Example 4. How close must a girl’s face be to a converging mirror of focal length 25 cm, in order that she sees an erect image that is twice as large? (M = +2)Also,Thus,f = -2(p - f) = -2p + 2fp = 12.5 cmf = -2p + 2f
25SummaryThe following equations apply equally well for both converging and diverging mirrors.yY’Rqpf
26Summary: Sign Convention 1. Object distance p is positive for real objects and negative for virtual objects.2. Image distance q is positive for real images and negative for virtual images.3. The focal length f and the radius of curvature R is positive for converging mirrors and negative for diverging mirrors.4. The image size y’ and the magnification M of images is positive for erect images and negative for inverted images.
27Summary: Magnification The magnification M of an image is the ratio of the image size y’ to the object size y.Magnification:Obj.Img.M = +2M = -1/2y y’y and y’ are positive when erect; negative inverted.q is positive when real; negative when virtual.M is positive when image erect; negative inverted.
28CONCLUSION: Chapter 34B Reflection and Mirrors II (Analytical)