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UCF PHY 2054C - Geometrical Optics

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Geometrical Optics – Part IIGoing BackwardsStuffSlide 4Concave Mirror/Paraxial ApproximationImage FormationThe geometry……Let’s try an exampleA concave spherical mirror has a radius of 10 cm. Calculate the location and size of an 8mm object a distance 15 cm from the mirror.A concave spherical mirror has a radius of 10 cm. Calculate the location and size of an 8mm object a distance 10 cm from the mirror.A concave spherical mirror has a radius of 10 cm. Calculate the location and size of an 8mm object a distance 2.5 cm from the mirror.The Concave MirrorMore Convex MirrorGraphical Methods are very useful to check your work.Moving on to refractive surfacesSpherical Refractive SurfacesA closer look at the Math ….No for the height of the imageCheck this out – how big is R?From the math:Slide 21The Thin LensFrom whence it cameThe thin lens - geometryMore GeometrySlide 26Slide 27Thin Lens (con’t)Slide 29FINALLY – with some algebra and obvious substitutions, we get:Two Ways to do this STUFFGraphical Methods:Slide 33Most important case: converging lens Object to left of F1Most important case: converging lensSlide 36Slide 37Slide 38Slide 39Geometrical Optics – Part IIChapter 241Going Backwards2'0'1)2(1/s 22'11ssRRsRssStufWe continue with mirrors and lenses and even refractive surfaces.Quiz on FridayFor a while, office hours will be in, of all places, my office. We really don’t need MAP-318 except before exams. And the hours are too confusing.Next Exam is on Wednesday, December 2nd.I give up on the remaining evil clickers. Clicker grade=0.Let’s move on.34When the Center of Curvatureis on the same side of theoutgoing ray, RR is positive.Otherwise, if the center of curvature is not on the same sideas the outgoing ray, RR is negative.Concave Mirror/Paraxial ApproximationConsequentlyRssRhshsh2'112'MIRROREQUATION5Image Formation60'00sRs‘‘y’<0(from the diagram) so image is inverted.The geometry……7ss'-mand ''syso Triangles,Similar diagram)in image inverted fromsign (- 'syyymLet’s try an example8A concave spherical mirror has a radius of 10 cm. Calculate the location and size of an 8mm object a distance 15 cm from the mirror.910 cm 5 cmNormal to mirrorand bounces backalong incomingpath.mmyssmsfRss 45.'5.7'12'11A concave spherical mirror has a radius of 10 cm. Calculate the location and size of an 8mm object a distance 10 cm from the mirror.1010 cm 5 cmmmyssmcmsfRss 80.1'10'12'11A concave spherical mirror has a radius of 10 cm. Calculate the location and size of an 8mm object a distance 2.5 cm from the mirror.1110 cm 5 cmeyemmyssmcmsfRss 80.2'5'12'11virtualimageThe Concave Mirror12More Convex Mirror13Graphical Methods are very useful to check your work.14Moving on to refractive surfaces15Spherical Refractive Surfaces16air glassA closer look atthe Math ….17bbaabann)( ' sh babaabnnnnRhshIgnoringRnnsnsnRhnnshnshnbnnnnnnnnnabbaabbabaabaabbba)(')(')()(No for the height of the image18snsnyymsynnsynnsysysabbbaaabaa''''''tanCheck this out – how big is R?19From the math:200')('snsnRnnsnsnbaabbasnsnyymsa''11'''0'msnsnsnsnsnsnsnsnabbababa21The Thin LensWe ignore the thickness of the lens.We will use mostly geometrical methods.Any ray that bends is assumed to bend only once at the center of the lens.22From whence it came23Surface 1Surface 2n=1 n=1.5 n=1Surface 2n>1The thin lens - geometry24parallelMore GeometryLens is thinActual thickness of the lens is ignored.Image from first surface provides the object for the second surface.Paraxial Ray Approximationsin(x)=tan(x)=xcos(x)=1x is in RADIANS25More Geometry26Triangle PQO andtriangle P’Q’O aresimilar.ssyy ''We will show that fora very thin lens:F1=F2=ffssssmffsyyorfsyfy1'11'''''AOP'at LookingThe Thin Lens Equation27This, of coursedepends on where theobject is placed with respect to f.Thin Lens (con’t)28Image thatwould form if material “a”was all on thisside of the lens.Object for secondsurface.29Procedure for equation•Solve for image position for first surface•Use image as object for the second surface.•Use the refraction equation in both cases.222b111a'sn:surface secondFor 'sn:#1 SurfaceConsider RnnsnRnnsnbccabbFor a lens. na=nc=1So we can call the middle one just n12' : ssNote Mess with the algebra and you will get:2211111'1s'n-1's1RnsRnsnFINALLY – with some algebra and obvious substitutions, we get:30fRRnss111)1('1121The Lensmaker’s EquationTwo Ways to do this STUFFAlgebraically using the lens equation (with the 1/f if you know it)Using graphical Methods31Graphical Methods:32Graphical Methods:33Most important case: converging lensObject to left of F134Most important case: converging lens35Most important case: converging lens36Most important case: converging lens37Most important case: converging lens38Most important case: converging


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UCF PHY 2054C - Geometrical Optics

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