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1、-95-Chapter 4 Geometric Optics(Zhaos Chap I,4,5,6,7,9.;Hechts.5.1-5.4;6.1,6.2)(1)Why is it called geometric?The problems can be solved geometrically,at most by analytical geometry(such as the matrix method of ray tracing).(2)Why it can be solved geometrically?Treat the light as0.This means there is
2、no concern of wave characters of light,such as interference,diffraction,etc.We shall treat light as“rays”traveling in straight lines(the wavelength of light is negligible compared to the dimension of the problem in concern).The“ray”is a line representing the energy flow of the light wave and in isot
3、ropic media,it is same as direction of wave vector k.(3)Why Bother?G.O.though lacks of scientific(from theoretical point of view)significance,it is still having probably the broadest applications in technology development in optics (4)What can we do with G.O Modify the incident wavefront -96-Example
4、:Change the wave front and Image formation:In experiments in physics,there are numerous occasions that we need to detect the light emitting or scattered from an object,not even considering that our own eyes are a pair of complicated lens(or camera)system.The most common set up would be:We need to me
5、asure or record(a)Intensity such as molecular spectroscopy(b)Image For purpose(a),we need to collect as much light as possible.For purpose(b),we need a device to convert every point of the object into -97-every single point of the image,while keeping the proportion(one to one correspondence without
6、distortion).In sections below,we are going to study what are these devices,why they work and what are their limitations.4-1 Jargons in geometric optic Focal points,concave,convex,real,virtual images,objects,object-Image conjugates Take examples that we are already familiar with,a)convex lens light p
7、asses through a convex lens will be more converged Optical axis:The line passes the center of the lens and -98-perpendicular to it.of:a special point on the optical axis;lights from of after optical elements will be parallel beams.Or of has image at(infinity).if:another special point on optical axis
8、;beams parallel to the optical axis after pass optical element will focus toif.Or object at will have its image atif.b)concave lens:Light passes through a concave lens will be more diverged.The of is a virtual object on the optical axis whose image is at infinity;ifis the virtual image whose corresp
9、onding object is at infinity.-99-1)Real,Virtual Object:In case(C),(D),object AB are real objects for1L.Real means for the optical elements(such as1L),the incoming lights from the object are diverging.In case(C),if we insert a second lens into the system(2L),at the axial location as shown,A B(the rea
10、l image of AB through1L)now becomes a virtual object for2L,the incoming light to the 2L(another saying is that 2Lsees the incoming light)are converging.In case(D),A B(the virtual image of AB through1L)is a real object for2L.2)Real,Virtual Image A B is a real image in case(c)formed by1L,the lights th
11、at form the image are converging.The real image can be projected on a real screen.-100-A B in case(D)is a virtual image by1L,the lights“forming”it are diverging.If the objective points and image points have a 1 to 1 relation(i.e.each objective point corresponds to one and the only one image point),i
12、t means all the light from the object after passing through the optical element(lens here)will converge to its image point(for the real image case;or diverge to virtual image point).The corresponding objective and image points are called conjugate points.In(C),(D):1LAA 1LBB They are conjugate points
13、 by 1L 4-2 Fermats Principle The OPL for all light rays between conjugate points are equal.It is fairly easy to prove for real object and real image ease.The light rays connecting the object point O and image point I through the optical element have to be stationary.The OPL of these different rays c
14、annot be -101-maximum or minimum,have to be a constant.For virtual object or image,their OPL have to be counted as negative(try proving this as an exercise,hint:set up using extra lens to make virtual image as object again and forming real image with second lens),and the refractive index n in OPL=nl
15、,has to be taken corresponding to the objective or image side respectively.So if any interface can satisfy the constant OPL between pointsQ andQ,they will be a pair of Object Image conjugate with respect to that interface.Such interface can be derived by requiring:(Hechts 5.2.1,Zhaos 4.4)QMMQconstQM
16、MQconst+=contour of point M for real or virtual image(object)The contour of M,satisfying the above equation is generally an aspheric curve(or surface,such as elliptical,hyperbolic).These aspheric optical elements are more difficult to manufacture in great precision than spherical ones.Since spherica
17、l optical elements are widely used and easier to make,we are going to study in detail of such optical lenses or mirrors,to see -102-under what condition that the spherical ones can be treated as ideal.Ideal optical Element:(1)All light rays from one point source can be transformed into another point
18、.11 relation between image and object,so to the lines and planes.(2)The optical element can cover wide range of objective distance,not only for a specific pair of object image points.(3)No distortion,i.e.keep the original shape of the object,a constant magnification value in image plane.4-3 Refracti
19、on at a Single Spherical Surface (Fig 5.6 Hechts or Fig 5.1 Zhaos)We need to find relation betweenoS,iS.If it is independent of (orh,1i,2i)then all rays coming from Q can be converged intoQ,i.e.Q,Q -103-are conjugate.The position of the image is most generally in a functional form:0012(,.)iiSF S n n
20、 ri i=.Of course,not all the variables are independent,they are related by Snells law,triangular relations etc.Neglect the detail derivation(refer to Hechts 5.2.2),we get:22202222220000114 sin()()2()()iiiiiSSrnSrnSrnSrnSr=+(4-1)Generally,0S and iS are dependent on(that is where the ray hit the surfa
21、ce),so rays from 0S of different angle generally interact the axis at different distance,except for a special pair0S,iS where 22022220022000()()110()()iiiiiSSnSrnSrnSrnSr=+=+Such0S,iS defines the location of Stigmatic Points for the spherical surface(these special pair can achieve so called broad be
22、am image formation.i.e.all light from the object at O are converged to the image point I).So(4-1)shows that the spherical surface is not an ideal optical element except for a special pair of points.4-4 Paraxial Approximation.Under paraxial condition for points on the axis,i.e.-104-220hS,2iS,2r or 2,
23、2,21 Then 22sin022 in(4-1)2202222000()()iiiSSnSrSrn=+(4-2)000()()iiiSSn SrSrn=+000iiin rnrnnSS+=000iiinnnnSSr+=(4-3)For image focal point,if:0S ,iiSf 0iiinnnfr=0iiinrfnn=(4-4)Similarly for object focal point:iS ,00Sf 000in rfnn=(4-5)(4-3)object-image relation can be rewritten in the more familiar fo
24、rmula(Gausss formula):-105-001iiffSS+=(4-6)Under paraxial condition,that each point on the axis0S,will form an image at iS on the axis based on(4-6)*Sign convention for spherical refraction and thin lenses(Light entering the lens from Left side)0S,0f +left of vertex A iS,if +right of A R +c is right
25、 of A 0y,iy +above the optical axis ox +left of 0f(x is used in Newton formular)ix +right of if +optical axis to ray is counter clock wise.Choose the convention and stick to it,then from the computation to determine the positions of the image relative to the optical elements.This is the basics of ge
26、ometric optics.Example:-106-The calculation based on the above sign convention is straightforward in this simple case.What I want to demonstrate is that they agree with the geometric method(draw directly from Snells law).Also noticed that the converging or diverging of the convex and concave surface
27、 depend on relative value betweenin,0n ifionn0 according to our convention)110()oinnnnSSr+=11ionSSn=1i is the image formed by O through surface 1.-108-10iS,means 1i is at the same side asO,i.e.it is a virtual image,1onSn to the left of 1.For surface 2,1i becomes a real object(The incident light towa
28、rds surface 2 are diverging)The distance between surface 2 and 1i is 211ooioonSdSdSn=+(Notice this is n the incident light to 2 is from n side)220oinnSS+=2210ioonnSSSdnn=2i is also a virtual image(2iS0),to the left of 2,with distance 10onSdn+21()ionnSSSddn=+=Summary on Image Formation by a Spherical
29、 Refracting Surface Generally a spherical surface is not an ideal optical element for image formation.However,under Paraxial Approximation,it is close to ideal.Relation between image-object conjugate is given by:-109-ioioionnnnSSr+=Or 1oioiffSS+=Where ooion rfnn=iiionrfnn=Important Note:a)The defini
30、tion of+,-sign for os,is,if,of,r(the sign convention)b)Their correspondence to the real,virtual object(or Image,focal point);c)And their location with respect to the spherical surface.The above conclusion is easily verified by Snells law(on diverging or converging behavior of light)and the formula o
31、fof,if given earlier.The real object(image)+(,)oiS S Virtual object(image)-(,)oiS S -110-Their location with respect to the surface is given sign convention and summarized in the notes above and also in table 5.1 Hechts:4-5 Finite Imagery and Transverse Magnification.This is the question of image fo
32、rmation for points off the optical axis:Due to the spherical symmetry,PPSS are also object-Image conjugate.If the is very small(paraxial condition),then Curve?PQS and its conjugate image?PQS will be close to a plane perpendicular()to the optical axis.Equivalently this requires:-111-22222,oioiyySSr m
33、agnified Image 1v Erected Image 0v)or diverging(0f)Object Image Location Type Location Orientation T.M 2oSf Real 2ifSf Real 2iSf Inverted Magnified oSf=oSf Erect Magnified 000111iiissfysVys+=Concave(,0ioff)Object Image Location Type Location Orientation T.M Any Virtual iSf,so will be theooxSf=,and t
34、he area of -144-image will be approximately proportional to2f.The energy flow through the system as we had seen previously will be proportional to the square of the diameter of the lens or the aperture stop:2TED.So the energy flow density or energy per unit time per unit area on the image will be pr
35、oportional to 22Df The f number of the lens system is defined as:#ffD (4-37)The reason defined as 4-37 as reciprocal of the energy flow density is in photography,this f number is used to determine the exposure time.The larger the f number(the smaller the energy flow rate)means longer the exposure ti
36、me to get a bright picture.(4)Field Stop Previously we focus on the axial point sources and concern how much energy will pass the optical system.Here we shall consider off-axis point and estimate how much energy will pass through system for these off axis point,this relates to the problem of field o
37、f view for the imaging system.For a single lens system,the lens itself will be the aperture stop,the entrance and exit pupil as well.The light from off axis points can be -145-collected by the lens and the its field of view is only limited by the paraxial requirement and slow decrease of steric angl
38、e spanned by the lens as the point moves further away from axis thus the decrease of energy collected.For the multiple lens system,the limitation of field of view can arise from an extra factor,so called field stop.This is better illustrated by an example:In the example of two-lens setup,the second
39、lens AB is the aperture stop and its entrance pupil is the image by lens 1 AB as shown in the figure.For the off axis point O and O”,we shall focus on their chief ray,the chief ray is defined as lines connect the off axis points and the center of the entrance pupil(the colored lines in the figure).T
40、his chief ray is approximately the direction of energy of input light to the system,it passes through the center of entrance pupil and it may pass through the exit pupil to have an output.If it passes the exit pupil,then ABABOOO-146-we will say the point can be“seen”by the system.That is to say we s
41、elect the chief ray as criteria for off-axis point to determine whether it can be“seen”.Of course even when chief ray is passing,the off-axis point still suffers some energy loss(in the figure the light from o”-A may not pass).In the figure above,the chief rays of both O and O”will clear the system,
42、but the O”is the limiting case.For any point further away from the O”,the chief ray of such points will be unable to pass the lens 1 and thus will not pass the exit pupil to have an output.We call the lens 1 in this example the field stop.The field stop will decide the field of view and in our examp
43、le,the angle spanned by the center of entrance pupil(axial point of AB)and the field stop(lens1)is the field of view in this example.Of course here we determine the field of view solely by the chief ray.For the off-axial point further away than O”,there will still be some light from them can pass th
44、rough the system,for instance,the ray such as O”B can clear the system,but you can see those point will suffer energy loss since most of its energy cannot pass,so the field of view is not a clear cut(the jargon for such phenomena is vignetting)which agrees with our experience(you do not experience s
45、udden complete darkness from side of your eyes,do you?).-147-(5)Entrance Window and Exit Window In the above simple example of two lenses,determine the field stop is relative easy.For complicated lens system,as in the aperture stop case,we need a systematic way to find field stop.This part is also v
46、ery similar to the relation between aperture stop to those of entrance and exit pupil.We use similar image formation to find out entrance and exit window from field stop.Also first illustrate with example:In the example,the iris is the aperture stop,and entrance pupil is the image formed by it with
47、the optical elements to the left of it which is only L1 in this case.We shall also make image of L2 with respect to the optical elements to the left of it and the image is L2.The angle spanned by center of the entrance pupil to those of L1 and L2 are also explicitly displayed and in this case,the L2
48、 forms smaller angle and thus limits the field of view.The L2 will be the field stop and its image L2 is called entrance window.1LASEntrancePupil2L2LEntarnceWindow-148-Entrance window:The image of the field stops with respect to the optical elements preceding(to the left in our conventional setup)it
49、.Exit window:The image of the field stops with respect to the optical elements after(to the right in out convention)it.The field of view will be determined by the angle spanned by the center of the entrance pupil to the entrance window.Its image conjugate would be angle spanned by the center of the
50、exit pupil to exit window.In Summary of the image system by geometrical lenses,we see that the multiple lens system is equivalent to one single thick lens(though in practice it is straightforward to work out one lens at a time).The problem regarding to energy flow through the system is dealt with by