14.3 LensesImages formed by refractionImages formed by a thin lensImage formed by refraction• Light rays are deflected by refraction through media with different refractive indexes. • An image is formed by refraction across flat or curved interfaces and by passage through lenses.Image formed by refraction through a refracting surface.Real image formed by refraction.Light iscaused toConverge.Rotationof the ray at the interfaceImage formed by refraction through a refracting surface.Virtual Image formed by refraction.Light is causedto diverge ina different direction.Why is the pencil bent?ooImage of the tipConverging LensesFatter in the middle. Cause light to converge toward the optic axis2Diverging LensesThinner in the middleCause light to diverge away from the optic axisParallel light though a converging lens is focused at the focal point.A real image is formedRay tracing for lenses• A line parallel to the lens axis passes through the focal point• A line through the center of the lens passes through undeflected.fRay diagram for a converging lensesObjectImageImages formed by a converging lensconverging lightconverging lightdiverging lightRealInvertedreducedRealInvertedEnlargedVirtualUprightEnlargedAt the focal point the image changesfrom real to virtualQuestionHow will an object viewed through a converging lens appear as the lens is brought closer to the object?3Real ImageInvertedReal ImageInvertedMagnifiedVirtual ImageUprightMagnifiedVirtual imageUprightParallel light though a diverging lens appears to go through the focal point.A virtual image is formed.Image formed by a diverging lensVirtual UprightReduced4QuestionHow will the image of an object formed by a diverging lens change as the lens is brought closer to the object?Virtual ImageUprightReducedVirtual imageUprightReducedVirtual imageUprightReducedThin lens equation.111pqf+=p is positive for real objectsf is positive for converging lensesf is negative for diverging lensesq is positive for real imagesq is negative for virtual images.p and q are positive if light passes throughMagnificationh' qMhp=−=−for real image q is positive – image is invertedfor virtual imageq is negative – image is uprightM positive- uprightM negative- inverted5ExampleAn object is placed 30 cm in front of a converginglens with focal length 10 cm. Find the object distanceand magnification.ExampleAn object is placed 30 cm in front of a converginglens with focal length 10 cm. Find the object distanceand magnification.Ray diagram.111pqf+=111qfp=−fpqpf=−(10)(30)15cm30 10==−q15M0.5p30=− =− =−Real imageInvertedReducedExample•F30 cm10 cmAn object is placed 30 cm in front of a diverginglens with a focal length of -10 cm. Find the image distance and magnification111pqf+=111qfp=−fpqpf=−( 10)(30)7.5cm30 ( 10)−==−−−Virtual imageq7.5M 0.25p30−=− =− =Upright imagereducedProjector lensSuppose you want to project the image of a transparency 35 mm high on to a screen that is 1.5 m high using a lens with a focal length of 10 cm. Where would you position the film? How far from the lens would you place the screen? .35 mm1.5 mf=10cmProjector lensFind ratio of p and q..35 mm1.5 mf=10cmUse magnificationh' qMhp==−3h' 1.542.8h35x10−=− =−q 42.8p=Thin lens eq.111pqf+=111p42.8pf+=solve42.8 1 43.8p f 10 10.23cm42.8 42.8+⎛⎞⎛⎞===⎜⎟⎜⎟⎝⎠⎝⎠just outside of fq 42.8p 42.8(10.23) 438cm==