March 26, 2005 Physics for Scientists&Engineers 2 1Physics for Scientists &Physics for Scientists &EngineersEngineers 22Spring Semester 2005Lecture 36March 26, 2005 Physics for Scientists&Engineers 2 2Geometric OpticsGeometric Optics! The study of light divides itself into three fields• geometric optics• wave optics• quantum optics! In the previous chapter, we learned that light is anelectromagnetic wave! In the next chapter, we will deal with the wave propertiesof light! In this chapter we will deal with geometric optics in whichwe will treat light that travels in straight lines called lightrays! Quantum optics makes use of the fact that light isquantizedMarch 26, 2005 Physics for Scientists&Engineers 2 3Spherical WavesSpherical Waves! Electromagnetic waves spread spherically from a source! The concentric yellow spheres shown below represent the spreadingspherical wave fronts of the light emitted from the light bulb! The black arrows are the light rays, which are perpendicular to thewave fronts at every point in space.March 26, 2005 Physics for Scientists&Engineers 2 4Plane WavesPlane Waves! We can treat light waves from far away sources as plane waves whosewave front is traveling in a straight line! We can further abstract these traveling planes by vectors or arrowsperpendicular to the surface of these planes! These planes can then be represented by a series of parallel rays orjust one ray! In this chapter we will treat light as aray traveling in a straight line! We can solve many problems geometricallyand by various constructions! Thus for the remainder of this chapterwe will ignore our knowledge of thestructure of light and attack a broadrange of practical problemsMarch 26, 2005 Physics for Scientists&Engineers 2 5ReflectionReflection and Plane Mirrorsand Plane Mirrors! A mirror is a surface that reflects light! A plane mirror is a flat mirror! For reflection from plane mirrors we have a simple rule for light raysincident on the surface of the mirror! This rule states that the angle of incidence, !i, is equal to the angle ofreflection, !r! These angles are always measured from the normal, which is defined tobe a line perpendicular to the surface of the planePlane mirrorIncident rayReflected rayNormal to surface of mirrorMarch 26, 2005 Physics for Scientists&Engineers 2 6Reflection and Plane Mirrors (2)Reflection and Plane Mirrors (2)! Parallel rays incident on a plane mirror will bereflected such that the reflected rays arealso parallel, because every normal to thesurface is also parallel! The law of reflection is given by! Images can be formed by light reflected fromplane mirrors! For example, when you stands in front of a mirror,you see your image in the mirror! This image appears to be behind the mirror! This type of image is referred to as a virtual image because itcannot be projected on a screen!r=!iMarch 26, 2005 Physics for Scientists&Engineers 2 7Mirror ImageMirror Image! Images formed by plane mirrors appear to be reversed because thelight rays incident on the surface of the mirror are reflected back onthe other side of the normal! A mirror image looks correct vertically! The image seen by the person can beconstructed with two light rays asshown• of course light rays are coming fromevery visible point of the person! The image is• upright (meaning not “upside-down”)• virtual (implying that the image is formed behind the surface of the mirror! The distance the person is standing from the mirror is called theobject distance, do, and the distance the image appears to be behindthe mirror is called the image distance, di, and for a plane mirror, do = didodiMarch 26, 2005 Physics for Scientists&Engineers 2 8Mirror Image (2)Mirror Image (2)! Now let’s discuss the left-right question for the mirror image! Again we construct the virtual image with two rays• all rays behave the same way! One can see that the real live person has his watch on his left hand andhe sees that his virtual self has his watch on his right hand! Thus when one looks in a mirror one sees an image that is upright butflipped left and right forming a “mirror image”March 26, 2005 Physics for Scientists&Engineers 2 9Example: Full-lengthExample: Full-length MirrorMirror!Question:! A 184 cm (6 ft 1/2 inch) tall person wants to buy a mirror so that hecan see himself full length. His eyes are 8 cm from the top of his head! What is the minimum height of the mirror?! Let us abstract the person as a 184 cm tall pole with eyes 8 cm fromthe top of the poleMarch 26, 2005 Physics for Scientists&Engineers 2 10Example: Full-lengthExample: Full-length MirrorMirror! The distance from the floor to the bottom of the mirror is• (184 cm – 8cm)/2 = 88 cm• the angle of incidence equals the angle of reflection so the point on themirror where the person can see the bottom of his feet will be half waybetween his eyes and the bottom of his feet! Similarly the difference between the top ofthe mirror and 184 cm is• (8 cm)/2 = 4 cm! So the minimum length of the mirror is• 184 cm – 88 cm – 4 cm = 92 cm! The required length of the “Full-lengthmirror” is just half the height of the personwanting to see himself full-lengthMarch 26, 2005 Physics for Scientists&Engineers 2 11Curved MirrorsCurved Mirrors! When light is reflected from the surface of a curved mirror, the lightrays follow the law of reflection at each point on the surface! However, unlike the plane mirror, the surface of a curved mirror is notflat! Thus light rays that are parallel before they strike the mirror arereflected in different directions depending on the part of the mirrorthat they strike! Depending of the shape of the mirror, the light rays can be focused ormade to diverge! Suppose we have a spherical mirror where the reflecting surface is onthe inside of the sphere! Thus we have a concave reflecting surface and the reflected rays willconvergeMarch 26, 2005 Physics for Scientists&Engineers 2 12Concave Spherical MirrorsConcave Spherical Mirrors! We can abstract this sphere to a two dimensionalsemicircle! The optical axis of the mirror is a line through thecenter of the sphere, represented in this drawingby a horizontal dashed line! Imagine that a horizontal light ray above the opticalaxis is incident on the surface of the mirror! At the point the light ray strikes the mirror, the law of reflectionapplies•!i = !r! The normal to the surface is a radius line that points to the center ofthe sphere marked as CMarch 26,
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