Chapter 17 - Light and Image Formation
Chapter 17 - Light and Image Formation
Chapter 17 - Light and Image Formation
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Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final Questions<strong>Chapter</strong> <strong>17</strong><strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong><strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsHomework Assignment 10Homework for <strong>Chapter</strong>s 16 <strong>and</strong> <strong>17</strong> (due at the beginning of class on Tuesday, November 23)<strong>Chapter</strong> 16: Q9, Q13, Q23, E2, E8<strong>Chapter</strong> <strong>17</strong>: Q2, Q10, Q14, Q18, E6, E10<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsOne-half ruleWhen unpolarized light reaches a polarizing sheet, the intensity I of the emerging polarized light isI = 1 2 I 0Malus’ lawWhen polarized light reaches a polarizing sheet, the intensity I of the emerging polarized light isI = I 0 cos 2 θwhere θ is the angle between the incoming light’s polarized direction <strong>and</strong> the polarizing direction of the sheet<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsReflectionWhen a light ray traveling in one medium encounters aboundary to another medium, part of the incident lightis reflectedReflection of light from a smooth surface is calledspecular reflectionReflection of light from a rough surface is called diffusereflectionA surface behaves as a smooth surface as long as thesurface variations are much smaller than the wavelengthof the incident lightWe will restrict our study to specular reflection <strong>and</strong> usethe term reflection to mean specular reflectionThe law of reflectionThe angle of reflection θ r is equal to the angle of incidence θ iθ i = θ r<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsRefractionsinθ 2sinθ 1= v 2v 1<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsRefractionsinθ 2sinθ 1= v 2v 1Index of refractionThe speed of light in any material is less than its speedin vacuumWe define the index of refraction n of a material asn = c v<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsRefractionsinθ 2sinθ 1= v 2v 1Index of refractionThe speed of light in any material is less than its speedin vacuumWe define the index of refraction n of a material asn = c vLaw of refraction (Snell’s law)n 1 sinθ 1 = n 2 sinθ 2<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsRefractionWhen light passes from one medium to another, its speed instantly changesAs light travels from one medium to another, its frequency does not change but its wavelength doesIndex of refractionThe index of refraction is a dimensionless numberThe index of refraction of a medium is inversely proportional to the wave speed in that mediumThe higher the index of refraction of a medium, the more it slows light down <strong>and</strong> the more itbends lightObservationsIf n 2 is equal to n 1 , then θ 2 is equal to θ 1 . In this case, refraction does not bend the light beam, whichcontinues in an undeflected directionIf n 2 is greater than n 1 , then θ 2 is less than θ 1 . In this case, refraction bends the light beam away fromthe undeflected direction <strong>and</strong> toward the normalIf n 2 is less than n 1 , then θ 2 is greater than θ 1 . In this case, refraction bends the light beam away fromthe undeflected direction <strong>and</strong> away from the normal<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsFlat mirrors<strong>Image</strong> formation by mirrors can be understood through the analysis of light rays under reflectionConsider a point source located at point O a distance p (the object distance) in front of a flat mirrorDiverging rays leave the point source <strong>and</strong> are reflected by the mirrorHowever, these diverging rays appear to originate at some point I (the image), a distance q (the imagedistance) behind the mirror<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final Questions<strong>Image</strong>s<strong>Image</strong>s are classified as real or virtualA real image is formed when light rays pass through <strong>and</strong> diverge from the image pointA virtual image is forced when the light rays do not pass through the image point but only appear todiverge from that pointReal images can be displayed on a screen; virtual images cannot be displayed on a screenThe image of an object seen in a flat mirror is always virtual<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsMagnificationWe define the lateral magnification M of an image asM =image heightobject height = h′hFor a flat mirror, M = +1 for any image<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsSpherical mirrorsNow, we will study images formed by curved mirrorsAlthough a variety of curvatures are possible, we will restrict our investigation to spherical mirrorsWe first consider reflection from the inner, concave surface of a spherical mirrorThis type of reflecting surface is called a concave mirror (also called a converging mirror)<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsMagnificationFor any mirror, we find that the magnification M of the image isM = h′h = −q pwhere p <strong>and</strong> q are the object <strong>and</strong> image distances, respectivelyIf M is negative, the image is inverted (upside down)<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsMagnificationFor any mirror, we find that the magnification M of the image isM = h′h = −q pwhere p <strong>and</strong> q are the object <strong>and</strong> image distances, respectivelyIf M is negative, the image is inverted (upside down)The mirror equationFor a mirror we find that1p + 1 q = 1 fwhere f is the focal lengthThe focal length of a mirror depends only on the shape of the mirror <strong>and</strong> not on the material from whichthe mirror is made<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsConvex mirrorsWe can also form an image using a convex mirror, that is, one that reflects light from its outer, convexsurfaceIt is sometimes referred to as a diverging mirror, because rays from any point on an object diverge afterreflectionWe will not derive any equations for convex spherical mirrors because our earlier equations can be used forboth concave <strong>and</strong> convex mirrorsSign conventions for mirrorsQuantity Positive When ... Negative When ...Object location (p) object is in front of mirror object is in back of mirror(real object) (virtual object)<strong>Image</strong> location (q) image is in front of mirror image is in back of mirror(real image) (virtual image)Magnification (M) image is upright image is invertedFocal length (f) mirror is concave mirror is convex<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsScenarioA mirror has a focal length of +10.0 cm.<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsScenarioA mirror has a focal length of +10.0 cm.QuestionLocate <strong>and</strong> describe the image for an object distance of 25.0 cm.<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsScenarioA mirror has a focal length of +10.0 cm.QuestionLocate <strong>and</strong> describe the image for an object distance of 25.0 cm.AnswerBecause the focal length of the mirror is positive, it is a concave mirrorWe expect the possibilities of both real real <strong>and</strong> virtual imagesBecause the object distance is larger than the focal length, we expect the image to be realTo find the image distance, we use 1/q = 1/f − 1/p <strong>and</strong> solve for qq = pfp − f(25.0 cm)(10.0 cm)= = 16.7 cm.25.0 cm − 10.0 cmThe magnification of the image isM = − q p= −16.7cm25.0 cm = −0.667.<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final Questions<strong>Image</strong>s formed by refractionWe will now examine how images are formed when light rays follow the wave under refraction model at theboundary between two transparent materialsIn particular, we will be studying the images formed by thin lensesA thin lens is one whose thickness is negligible compared to its focal length<strong>Image</strong>s formed by thin lenses satisfy the same equations as those for mirrors<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final Questions<strong>Image</strong>s formed by refractionWe will now examine how images are formed when light rays follow the wave under refraction model at theboundary between two transparent materialsIn particular, we will be studying the images formed by thin lensesA thin lens is one whose thickness is negligible compared to its focal length<strong>Image</strong>s formed by thin lenses satisfy the same equations as those for mirrorsSign conventions for mirrorsQuantity Positive When ... Negative When ...Object location (p) object is in front of mirror object is in back of mirror(real object) (virtual object)<strong>Image</strong> location (q) image is in front of mirror image is in back of mirror(real image) (virtual image)Magnification (M) image is upright image is invertedFocal length (f) mirror is concave mirror is convexSign conventions for thin lensesQuantity Positive When ... Negative When ...Object location (p) object is in front of lens object is in back of lens(real object) (virtual object)<strong>Image</strong> location (q) image is in back of lens image is in front of lens(real image) (virtual image)Magnification (M) image is upright image is invertedFocal length (f) a converging lens a diverging lens<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsQuestionA converging lens has a focal length of 10.0 cm. An object is placed 30.0 cm from the lens. Find the imagedistance <strong>and</strong> describe the image.<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsQuestionA converging lens has a focal length of 10.0 cm. An object is placed 30.0 cm from the lens. Find the imagedistance <strong>and</strong> describe the image.AnswerBecause the lens is a converging lens, its focal length is positiveWe expect the possibilities of both real <strong>and</strong> virtual imagesUsing the thin lens equation 1/p + 1/q = 1/f <strong>and</strong> solving for q we find thatq =( 1f= +15.0 cm−p) 1 −1 ()1=10.0 cm − 1 −130.0 cmBecause this image distance is positive, the image is realThe magnification of the image isM = − q p= −15.0cm30.0 cm = −0.500The magnification of the image tells us that the image is reduced in height by one half, <strong>and</strong> the negativesign for M tells us that the image is inverted<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsQuestionA diverging lens has a focal length of 10.0 cm. An object is placed 30.0 cm from the lens. Find the image distance<strong>and</strong> describe the image.<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsQuestionA diverging lens has a focal length of 10.0 cm. An object is placed 30.0 cm from the lens. Find the image distance<strong>and</strong> describe the image.AnswerBecause the lens is a diverging lens, its focal length is negativeBecause the lens is diverging, we expect it to form an upright, reduced, virtual image for any object positionUsing the thin lens equation 1/p + 1/q = 1/f <strong>and</strong> solving for q we find thatq =( 1q − 1 ) −1 ()1=p −10.0 cm − 1 −130.0 cm= −7.50 cmThe magnification of the image isM = − q p= −−7.50cm30.0 cm = +0.250This result confirms that the image is virtual, smaller than the object, <strong>and</strong> upright<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsHomework Assignment 10Homework for <strong>Chapter</strong>s 16 <strong>and</strong> <strong>17</strong> (due at the beginning of class on Tuesday, November 23)<strong>Chapter</strong> 16: Q9, Q13, Q23, E2, E8<strong>Chapter</strong> <strong>17</strong>: Q2, Q10, Q14, Q18, E6, E10<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
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