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1.1Chapter 1.1

Geometrical optics

Menn

1.1.1. Ray optics conventions andpractical rules. Real and virtualobjects and images

Electro-optical systems are intended for the transfer andtransformation of radiant energy. They consist of activeand passive elements and sub-systems. In active ele-ments, like radiation sources and radiation sensors, con-version of energy takes place (radiant energy is convertedinto electrical energy and vice versa, chemical energy isconverted in radiation and vice versa, etc.). Passive ele-ments (like mirrors, lenses, prisms, etc.) do not convertenergy, but affect the spatial distribution of radiation.Passive elements of electro-optical systems are fre-quently termed optical systems.

Following this terminology, an optical system itselfdoes not perform any transformation of radiation intoother kinds of energy, but is aimed primarily at changingthe spatial distribution of radiant energy propagated inspace. Sometimes only concentration of radiationsomewhere in space is required (like in the systems formedical treatment of tissues or systems for materialprocessing of fabricated parts). In other cases the abilityof optics to create light distribution similar in some wayto the light intensity profile of an ‘‘object’’ is exploited.Such a procedure is called imaging and the correspondingoptical system is addressed as an imaging optical system.

Of all the passive optical elements (prisms, mirrors,filters, lenses, etc.) lenses are usually our main concern. Itis lenses that allow one to concentrate optical energy orto get a specific distribution of light energy at differentpoints in space (in other words, to create an ‘‘image’’).

In most cases experienced in practice, imaging systemsare based on lenses (exceptions are the imaging systemswith curved mirrors).

The functioning of any optical element, as well as thewhole system, can be described either in terms of rayoptics or in terms of wave optics. The first case is usuallycalled the geometrical optics approach while the secondis called physical optics. In reality there are many situa-tions when we need both (for example, in image qualityevaluation, see Chapter 2). But, since each approach hasadvantages and disadvantages in practical use, it is im-portant to know where and how to exploit each one inorder to minimize the complexity of consideration and toavoid wasting time and effort.

This chapter is related to geometrical optics, or, morespecifically, to ray optics. Actually an optical ray isa mathematical simplification: it is a line with no thick-ness. In reality optical beams which consist of an endlessquantity of optical rays are created and transferred byelectro-optical systems. Naturally, there exist three kindsof optical beams: parallel, divergent, and convergent (seeFig. 1.1.1). If a beam, either divergent or convergent, hasa single point of intersection of all optical rays it is calleda homocentric beam (Fig. 1.1.1b,c). An example ofa non-homocentric beam is shown in Fig. 1.1.1 d. Such

Fig. 1.1.1 Optical beams: (a) parallel, (b,c) homocentric and(d) non-homocentric.

Practical Optics; ISBN: 9780124909519

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a convergent beam could be the result of different phe-nomena occurring in optical systems (see Chapter 2 formore details).

Ray optics is primarily based on two simple physicallaws: the law of reflection and the law of refraction. Bothare applicable when a light beam is incident on a surfaceseparating two optical media, with two different indexesof refraction, n1 and n2 (see Fig. 1.1.2). The first law isjust a statement that the incident angle, i, is equal to thereflection angle, i’. The second law defines the relationbetween the incident angle and the angle of refraction, r:

sinðiÞ=sinðrÞ ¼ n2=n1: (1.1.1)

It is important to mention that all angles are measuredfrom the vertical line perpendicular to the surface at thepoint of incidence (so that the normal incidence of lightmeans that i ¼ i0 ¼ r ¼ 0).

In the geometrical optics approach the following as-sumptions are conventionally accepted:

(a) radiation is propagated along a straight line trajec-tory (this means that diffraction effects are nottaken into account);

(b) if two beams intersect each other in space there isno interaction between them and each one is prop-agated as if the second one does not appear (thismeans that interference effects are not taken intoaccount);

(c) ray tracing is invertable; in other words, if the raytrajectory is found while the ray is propagatedthrough the system from input to output (say, fromthe left to the right) and then a new ray comes to thesame system along the outgoing line of the first ray,but propagates in the reverse direction (from theright to the left), the trajectory of the second rayinside and outside of the system is identical to thatof the first ray and it goes out of the system along theincident line of the first ray.

Normally an optical system is assumed to be axisym-metrical, with the optical axis going along OX in thehorizontal direction. Objects and images are usually lo-cated in the planes perpendicular to the optical axes,meaning that they are along the OY (vertical) axis. Raytracing is a procedure of calculating the trajectory of

optical rays propagating through the system. Radiationpropagates from the left to the right and, consequently,the object space (part of space where the light sources orthe objects are located) is to the left of the system. Theimage space (part of space where the light detectors orimages are located) is to the right of the system.

All relevant values describing optical systems can bepositive or negative and obey the following sign conven-tions and rules:

ray angles are calculated relative to the optical axis; theangle of a ray is positive if the ray should be rotatedcounterclockwise in order to coincide with OX,otherwise the angle is negative;

vertical segments are positive above OX and negativebelow OX;

horizontal segments should start from the optical systemand end at the relevant point according to the seg-ment definition. If going from the starting point tothe end we move left (against propagated radiation),the segment is negative; if we should move right(in the direction of propagated radiation), thecorresponding segment is positive.

Examples are demonstrated in Fig. 1.1.3. The angle u isnegative (clockwise rotation of the ray to OX) whereas u0

is positive. The object Y is positive and its image Y0 isnegative. The segment S defines the object distance. Itstarts from the point O (from the system) and ends atthe object (at Y). Since we move from O to Yagainst thelight, this segment is negative (S < 0). Accordingly, thesegment S0 (distance to the image) starts from the system(point O’) and ends at the image Y0. Since in this case wemove in the direction of propagated light (from left toright) this segment is positive (S0 > 0).

The procedure of imaging is based on the basic as-sumption that any object is considered as a collection ofseparate points, each one being the center of a homo-centric divergent beam coming to the optical system. Theoptical system transfers all these beams, converting eachone to a convergent beam concentrated in a small spot(ideally a point) which is considered as an image of thecorresponding point of the object. The collection of such‘‘point images’’ creates an image of the whole object (seeFig. 1.1.4).

An ideal imaging is a procedure when all homocentricoptical beams remain homocentric after traveling

Fig. 1.1.2 Reflection and refraction of radiation.

Fig. 1.1.3 Sign conventions.

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S E C T I O N O N E Optical Theory

through the optical system, up to the image plane (thiscase is demonstrated in Fig. 1.1.4). Unfortunately, in realimaging the outgoing beams become non-homocentricwhich, of course, ‘‘spoils’’ the images and makes it im-possible to reproduce the finest details of the object (thisis like a situation when we try to draw a picture usinga pencil which is not sharp enough and makes only thicklines – obviously we fail to draw the small and fine detailson the picture). The reasons for such degradation inimage quality lie partially in geometrical optics (thenthey are termed optical aberrations) and partially are dueto the principal limitations of wave optics (diffractionlimit). We consider this situation in detail in Chapter 1.2.Here we restrict ourselves to the simple case of idealimaging.

In performing ray tracing one should be aware thatdoing it rigorously means going step by step from oneoptical surface to another and calculating at each step theincident and refraction angles using Eq. (1.1.1). Sincemany rays should be calculated, it is a time-consumingprocedure which today is obviously done with the aid ofcomputers and special programs for optical design.However, analytical consideration remains very difficult(if possible at all). The complexity of the procedure iscaused mainly by the nonlinearity of the trigonometricalfunctions included in Eq. (1.1.1). The situation can besimplified drastically if we restrict ourselves to consid-ering small angles of incidence and refraction. Thensin(i)z i; sin(r) z r; r ¼ i/n and all relations becomelinear. Geometrically this approximation is valid only ifthe rays are propagated close to the optical axis of thesystem, and this is the reason why such an approximationis called paraxial. A paraxial consideration enables one totreat optical systems analytically. Because of this, it isvery fruitful and usually is exploited as the first approx-imation at the early stage of design of an optical system.

Even in the paraxial approach we can further simplifythe problem by neglecting the thickness of optical lenses.Each lens consisting of two refractive surfaces (sphericalin most cases, but sometimes they could be aspherical)separated by glass (or other material) of thickness t isconsidered as a single ‘‘plane element’’ having no thick-ness, but still characterized by its ability to concentratean incident parallel beam in a single point (called the

focal point or just focus). In such a case the only pa-rameter of the lens is its focal length, f 0, measured as thedistance between the lens plane and the focus, F0. Eachlens has two focuses: the back (F0) and the front (F), thefirst being the point where the rays belonging to a parallelbeam incident on the lens from the left are concentratedand the second being the center of the concentrated rayswhen a parallel beam comes to the lens from the right.Obviously, if the mediums at both sides of the lens areidentical (for example, air on both sides or the lens beingin water) then f 0¼ � f. In the case when the mediums aredifferent (having refractive index n and n0 correspond-ingly) the relation should be

nf 0 ¼ �n0f : (1.1.2)

The optical power of a lens, defined as

f ¼ 1=f 0; (1.1.3)

is used sometimes in system analysis, as we shall see later.Imaging with a simple thin lens obeys the two fol-

lowing equations:

1

S0� 1

S¼ 1

f 0; (1.1.4)

V ¼ S0=S ¼ y0=y; (1.1.5)

where V is defined as the optical magnification. Thesetwo formulas enable one to calculate the positions andsizes of images created by any thin lens, either positive ornegative, if all values are defined according to the signconventions and rules described earlier in this section. Anumber of thin lenses which form a single system can alsobe treated using expressions (1.1.4) and (1.1.5) step-by-step for each component separately, the image of ele-ment i being considered as a virtual object for element(i þ 1). An example of such a consideration with detailsfor a two-lens system is presented in Problem P.1.1.7.

The next step in approaching the real configuration ofan optical system is to take into account the thickness ofits optical elements. Still remaining in the paraxial rangeone can describe the behavior of a single spherical surface(see Fig. 1.1.5) by the Abbe invariant (r is the radius ofthe surface):

n

�1

r� 1

S

�¼ n0

�1

r� 1

S0

�: (1.1.6)

Then, the ray tracing for an arbitrary number of surfacescan be performed with the aid of the following twosimple relations (see also Fig. 1.1.6):

ukþ1 ¼nk

nkþ1uk þ

hk

rknkþ1ðnkþ1 � nkÞ: (1.1.7)

Fig. 1.1.4 Concept of image formation.

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Geometrical optics C H A P T E R 1 . 1

hkþ1 ¼ hk � ukþ1dk ðk ¼ 1; 2.;NÞ: (1.1.8)

Given the radii of the spherical surfaces, the refractionindexes on both sides, and the distances between them,all angles, uk, and heights, hk, can be easily found, startingfrom initial values u1, h1.

To apply Eqs. (1.1.7) and (1.1.8) to a single lens de-fined by two spherical surfaces of radii r1 and r2 sepa-rated by the segment d, we first have to remind ourselvesof the definition of the principal planes, H, H0 and thecardinal points. As is seen from Fig. 1.1.7, the real raytrajectory ABCD can be replaced by ABMM’CD in sucha way that they are identical outside the lens, but insidethe lens the rays intersect two virtual planes H and H0 atthe same height (OM ¼ O0M0). Actually these principalplanes, H, H0, can represent the lens as far as ray tracing isconsidered. Furthermore, the focal distances, f, f 0, aremeasured from the cardinal points O, O0 to the front andback focuses, F and F0. The terms ‘‘back focal length’’(BFL) and ‘‘front focal length’’ (FFL) are related to thesegments SF, SF from the back and front real surfaces toF’ and F, respectively (see Fig. 1.1.7). Calculation of BFLand FFL enables one to determine the location of bothprincipal planes with regard to the lens surfaces. Leavingthe details of calculation to Problem P.1.1.5 we justindicate here the final results:

SF0 ¼ f 0�1� d

r1nðn� 1Þ

�; (1.1.9)

SF ¼ �f 0�1þ d

r2nðn� 1Þ

�; (1.1.10)

and for the focal distance

1

f 0¼ ðn� 1Þ

�1

r1� 1

r2

�þ dðn� 1Þ2

r1r2n: (1.1.11)

In many cases the second term of the last formula can beneglected since it is much smaller than the first one.

Problems

P.1.1.1. Find the image of the object OA in Fig. 1.1.8using the graphical method.

P.1.1.2. Find the image of the point source A anddirection of the ray AB after the positive lens L1

(Fig. 1.1.9a) and the negative lens L2 (Fig. 1.1.9b).P.1.1.3. Ray tracing in a system of thin lenses. Find the

final image of a point source A after an optical systemconsisting of thin lenses L1, L2, and L3 (f

0

1¼ f0

2 ¼ 15 mm;f0

3 ¼ 20 mm) if A is located on the optical axis 30 mm leftof the lens L1 and the distances between the lenses ared12 ¼ 40 mm, d23 ¼ 60 mm. [Note: Do this by raytracing based on Eq. (1.1.4).]

P.1.1.4. Method of measurement of focal length ofa positive lens. An image of an object AB created by a lensis displayed on a screen P distant from AB at L¼ 135 mm(Fig. 1.1.10). Then the lens is moved from the initialposition, 1, where the sharp image is observed at

Fig. 1.1.5 Refraction of rays at a single spherical surface.

Fig. 1.1.6 Ray tracing between two spherical surfaces.

Fig. 1.1.7 Principal planes of a thick lens.

Fig. 1.1.8 Problem P.1.1.1 – Imaging by the graphical method.

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magnification V1, to the position 2 where again the sharpimage is observed on the same screen, but at magnifica-tion V2¼ 1/V1. The distance between positions 1 and 2 isa¼ 45 mm. Find the focal length of the lens and estimatethe uncertainty of the measured value if the lens thick-ness, t, is about 5–6 mm.

P.1.1.5. Location of the principal planes of a thick lens.Find the positions of two principal planes H and H0, BFL,and FFL of a lens made of glass BK-7(n ¼ 1.5163) havingtwo spherical surfaces of radii R1 ¼50 mm and R2 ¼�75 mm and thickness t ¼ 6 mm.

P.1.1.6. Violation of homocentricity of a beam passedthrough a flat slab. A flat slab of glass is illuminated bya homocentric beam which fills the solid angle u ¼1.5 srwith the center at point A, 30 mm behind the slab(Fig. 1.1.11). The thickness of the slab t ¼ 5 mm andrefractive index n¼ 1.5. Find the location of the point A0

after the slab as a function of incident angle, i, and esti-mate the deviation of the outgoing beam fromhomocentricity.

P.1.1.7. A two-lens system in the paraxial range. A lensL1 of 100 mm focal length is followed by a lens L2 of75 mm focal length located 30 mm behind it. Consid-ering both lenses as a unified system find the equivalentoptical power and position of the focal plane.

P.1.1.8. A ball lens. Find the location of the principalplanes of a ball lens (a full sphere) of radius r¼ 3 mm andits BFL.

1.1.2. Thin lenses layout.Microscope and telescope opticalconfigurations

We will consider here the following basic configurations:(i) magnifier; (ii) microscope; and (iii) telescope. Allthree can be ended either by a human eye or by anelectro-optical sensor (like a CCD or other area sensor).

1.1.2.1. The human eye

Although the details of physiological optics are beyondthe scope of this book, we have to consider some im-portant features of the human eye (for further details,see Hopkins, 1962) as well as eye-related characteristicsof optical devices. Usually the ‘‘standard eye’’ (normaleye of an adult person) is described in terms of a simpli-fied model (so-called ‘‘reduced eye’’) as a single lenssurrounded by the air from the outside, and by the op-tically transparent medium (vitreous humor) of re-fractive index 1.336 from the inside. As a result, thefront focal length of the eye, f, differs from the back focallength, f 0 (see Eq. (1.1.2)). The front focal length isusually estimated as 17.1 mm where as f 0 is equal to22.9 mm. The pupil of the eye varies from 2 mm (min-imum size) to 8 mm (maximum size) according to thescene illumination level (adaptation). The lens createsimages on the retina which consists of huge numbers ofphotosensitive cells. The average size of the retina cellsdictates the angular resolution of the eye (ability ofseeing two small details of the object separately). Thelimiting situation is that the images of two points arecreated at two adjacent cells of the retina. This rendersthe angular resolution of a normal eye to be 1 arcminute(3 � 10�4 rad). The lens curvature is controlled by theeye muscles in such a way that the best (sharp) image is

Fig. 1.1.9 Problem P.1.1.2 – Imaging by the graphical method:(a) with a positive lens; (b) with a negative lens.

Fig. 1.1.10 Problem P.1.1.4 – Method of focal lengthmeasurement.

Fig. 1.1.11 Problem P.1.1.6 – Consideration of a parallel plate.

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always created on the retina, whether the object is far orclose to the eye (accommodation process). The distanceof best vision is estimated as 250 mm, which means thatthe eye focused on objects at distances of 250 mm is notfatigued during long visual operation and can still dif-ferentiate small details.

Three kinds of abnormality of eye optics are usuallyconsidered: myopia, hyperopia, and astigmatism. Thefirst one (also called near-sightedness) occurs whena distant object image is not created on the retina but infront of it. A corrective negative lens is required in sucha situation. In the second case (called far-sightedness) theopposite situation takes place: the images are formedbehind the retina and, obviously, the corrective lensshould be positive. Astigmatism means that the lenscurvature is not the same in different directions whichresults in differences in focal lengths, say in the hori-zontal and vertical planes. Correction is done by specta-cles with appropriately oriented cylindrical lenses.

The other properties of the eye related to visual per-ception are considered in Chapter 10.

1.1.2.2. Magnifications in opticalsystems

Generally, four adjacent magnifications can be defined forany optical system: (i) linear magnification, V, for objectsand images perpendicular to the optical axis; (ii) angularmagnification, W; (iii) longitudinal magnification, Q(magnification in the direction of the optical axis); and(iv) visible magnification, G (used only for systemsworking with the human eye).

Linear magnification, defined earlier for a single lensby Eq. (1.1.5), is still applicable for any complete opticalsystem. Angular magnification can be defined for anyseparate ray or for a whole beam incident on a system.For example, for the tilted ray shown in Fig. 1.1.12 W iscalculated as follows:

W ¼ tanðu0Þ=tanðuÞ: (1.1.12)

As can be shown, the product VW is a system invariant: itdoes not depend on the position of the object and image,but is determined by the refractive indexes on both sidesof the optical system (n and n0). If n ¼ n0 then VW ¼ 1.

Considering the segment l along the optical axisand two pairs of conjugate points, A and A0, C andC0 (Fig. 1.1.12), we can find the longitudinal magnifica-tion, Q:

Q ¼ l0=l: (1.1.13)

It can be shown that for small segments l, i0 one can usethe formula Q ¼ V2.

Finally, visible magnification is related to the size ofimages on the retina of an eye. It is defined as the ratio ofthe image created by the optical system to the imageof the same object observed by the naked eye directly.Since the image size is proportional to the observationangle (see Fig. 1.1.13), G is determined as follows:

G ¼ tanðg0Þ=tanðgÞ: (1.1.14)

1.1.2.3. A simple magnifier

This is usually operated with the eye. While observingthrough a magnifying glass an object is positioned be-tween the front focus of the lens and the lens itself(Fig. 1.1.14). The image is virtual and its position cor-responds to the distance of best vision of the eye (250mm). The closer the object to F, the higher the magni-fication. Therefore, approximately, we can define that

Fig. 1.1.12 Imaging of vertical and horizontal segments.

Fig. 1.1.13 Explanation of visible magnification.

Fig. 1.1.14 A simple magnifier.

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s z f (of course s< 0; f< 0), and for visible magnificationof the magnifier we have

G ¼ 250

f 0: (1.1.15)

Since the distance of best vision is much greater than thefocal distance of the eye, the rays coming to the eye pupilare almost parallel. In most cases they can be treated justas a parallel beam (or beams).

1.1.2.4. The microscope

Figure 1.1.15a demonstrates the basic layout of a micro-scope working with the eye and Fig. 1.1.15b showsa microscope working with an electro-optical detector(like a CCD or other video sensor). In both cases the firstlens L1 (called the objective) is a short-focus well-corrected lens creating the first real magnified image ofthe object (AB) in the plane P. The second lens is theeyepiece (Fig. 1.1.15a) or the relay lens (Fig. 1.1.15b).The eyepiece L2 functions like a simple magnifier and itsvisible magnification obeys Eq. (1.1.15). Magnification ofthe objective, V1, can be found from Eq. (1.1.5). Usuallythe object distance S1 is very close to f1 and the distanceS01 ¼ T from the lens L1 to the plane P is chosen as one ofseveral standardized values accepted by all manufac-turers of microscopes (160 mm or 180 mm or 210 mm).Therefore, for the total magnification of the microscopeworking with eye we get:

VM ¼ V1V2 ¼T

f 01� 250

f 02: (1.1.16)

If instead of an eyepiece a relay lens is exploited its actuallinear magnification, V2, should be taken into account,and then

VM ¼T

f 01V2: (1.1.17)

If the microscope is intended for measurements and notonly for observation then a glass slab with a special scale(a ruler, a crosshair, etc.) called a reticle is introduced inthe plane P. In such a case the eye observes the imageoverlapped with the scale.

In Fig. 1.1.15 the rays originating from two points ofthe object are drawn – from the center of the object(point O) and from the side (point A). As can be seenfrom Fig. 1.1.15a, each point gives a parallel beam afterthe eyepiece: one is parallel to the optical axis and theother is tilted to OX. Intersection of the beams occurs inthe plane M (exit pupil of the microscope) where theoperator’s eye should be positioned.

For the convenience of the operator the optical layoutin most cases is split after the plane P in two branches,each one having a separate eyepiece. Such an output as-sembly is called binocular and observation is done by twoeyes. It should be understood, however, that binocularitself does not render stereoscopic vision, since both eyesare observing the same image created by a single objectiveL1. To achieve a real stereoscopic effect two objectivesare required in order to observe the object from twodifferent directions. Each image is transferred througha separate branch (a pair of lenses L1 and L2).

The architecture shown in Fig. 1.1.16 is actually thecombination of the two layouts presented in Fig. 1.1.15and its output assembly is called trinocular – it createsimages on the area sensor as well as in the image plane ofboth eyepieces. The beam splitter, BS, turns the opticalaxis in the direction of the relay lens L3.

In the last few years microscopes have been designedas infinity color-corrected systems (ICS) which meansthat the object is located in the front focal plane of theobjective, its image is projected to infinity, and an addi-tional lens L4 (the tube lens) is required in order to createan intermediate image in the plane P. Such a layout is

Fig. 1.1.15 Layout of a microscope working with (a) the eye and(b) an area detector. Fig. 1.1.16 Microscope with a trinocular assembly.

9


demonstrated in Fig. 1.1.17. One of the important ad-vantages of ICS optics is that the light beams are parallelbetween L1 and L4 enabling one to introduce here opticalfilters with no degradation of the optical quality andwithout relocation of the image plane P.

1.1.2.5. The telescope

Telescopic systems are intended for observation ofremote objects. If the distance between the object andthe first lens of the system is much greater than the focallength of L1 all light beams at the entrance of the systemcan be considered as parallel, whether they are comingfrom the central point of the object or from the side.

Again, as in the above consideration of a microscope,the central point beam is parallel to the optical axiswhereas the side point generates an oblique parallelbeam. All incident beams are concentrated by the ob-jective in its back focal plane (passing through the backfocus F01). The second lens L2 is positioned in such a waythat its front focus F2 coincides with F01. Obviously allbeams after lens L2 become parallel again, but the exitangles of the oblique rays are different from those of thecorresponding beams at the entrance (see Fig. 1.1.18)and this causes the angular magnification of the telescopeto be (see Eq. (1.1.12))

W ¼ tanðb0Þ=tanðbÞ ¼ f10=f20: (1.1.18)

As follows from Eq. (1.1.18), the longer the focal lengthof the objective, the greater the magnification. However,along with this the necessary size of the lens L2 also in-creases, which might cause a limitation of the visible fieldof view (the part of the object space visible through thesystem). To solve this problem an additional lens L3 (the

field lens) is introduced in the system (see Fig. 1.1.19).This lens allows one to vary the vertical location of theoblique beam incident on L2.

The configurations shown in Figs. 1.1.18 and 1.1.19are built of positive lenses. In the Galilean architecturethe eyepiece L2 is negative (Fig. 1.1.20). As a result, thetotal length of the system is shortened. However, theintermediate image is virtual (both focal points F01 and F2

are behind the eyepiece) and there is no way to introducea measurement scale, if necessary. However, as it turnsout, this shortcoming becomes very useful if the Galileanconfiguration is exploited with high-power lasers (forbeam expanding).

Problems

P.1.1.9. If the angular resolution of the eye is 3 � 10�4

rad, what is the average size of the retina cells?P.1.1.10. A microscope is intended for imaging an

object located in the plane P simultaneously in twobranches: one for observation by eye and the other forimaging onto a plane area sensor (CCD). The objective ofthe microscope serving the two branches is of 20 mmfocal length and provides linear magnification V¼�10 tothe image plane of the eyepiece where a reticle M of19 mm diameter is positioned (Fig. 1.1.21). The CCDsensor is 4.8 mm (vertical)� 5.6 mm (horizontal) in size.In front of the CCD at a distance of 20 mm an additionalrelay lens L3 is introduced in order to reach the bestcompatibility of the field of view in both branches. As-suming the eyepiece L2 to be of 25 mm focal length andneglecting the thickness of the lenses, find:

(a) the working distance (location of the object plane Pwith regard to the objective);

Fig. 1.1.17 Layout of a microscope with ICS optics.

Fig. 1.1.18 Basic layout of a telescope.

Fig. 1.1.19 Telescope with a field lens.

Fig. 1.1.20 Galilean telescope.

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(b) the total magnification in the branch to the eye;

(c) the optical power of lens L3. [Note: Find two solu-tions and choose the one which provides theshortest distance between P and the CCD.]

P.1.1.11. Dual magnification system with negative relaylens. Such a system is widely used in the microelectronicsindustry where automatic processing of wafers is a mainconcern. The object (usually a wafer) is located in theplane P and imaged onto a CCD either at low magnifi-cation V1 ¼ �3 (through the right branch of the ar-rangement, exploited for initial alignment) or at highmagnification V2 ¼ 2 � V1 (fine alignment through theleft branch where lens L2 and retroreflector R are in-troduced). While switching the system between twoalignment procedures no optical element should bemoved, except the aperture D (Fig. 1.1.22). The retro-reflector R allows one to vary the high magnification ofthe system with minimum effort–just replacement of Rand L2, with no other changes in the arrangement. Thus,lens L2 serves as a negative relay lens of the system.Neglecting the thickness of the lenses and taking allnecessary distances from Fig. 1.1.22, find:

(a) the focal length of L2 and its position with regard tothe CCD and the other elements of the arrangement;

(b) the relocation of the retroreflector and the relay lensL2 from their initial positions required to increasethe magnification in the left branch by 10%.

1.1.3. Diaphragms in opticalsystems. Calculation of apertureangle and field of view. Vignetting

The size of each optical element of a system should beconsidered properly, since it influences: (i) the quantityof radiant energy passing through the system; (ii) thequality of images; and (iii) the cost of the system. Amongall the geometrical parameters the working diameter is ofprimary importance (remember that we assume that thesystem is rotationally symmetric) – it acts as the trans-parent part of the element.

Sometimes an additional diaphragm (a physical ele-ment called a stop which has a final size aperture andnegligible thickness) is introduced in the system. Anaperture stop is a diaphragm which actually limits thesize of light bundles passing through the system andconsequently it is responsible for the amount of energycollected at each point of the image. The aperture stop isillustrated in Fig. 1.1.23. Assume that the system con-sists of a number of elements (of which the first and thelast curved surfaces are shown in the figure) and alsoincludes the stop cd. The boundaries of each opticalsurface are also considered as diaphragms. First we‘‘transfer’’ all the diaphragms into the object space (e.g.,we find the size and location of the image of each di-aphragm through the rest of the optical elements to theleft of it, as if the light beams are propagated from rightto left). Such an image of the stop cd is c’d’; the image ofthe first diaphragm ab is ab itself, since there is no ele-ment left of it; the third diaphragm shown in the figure isthe image of some other optical surface, etc. Then weconnect the ray from the central point O of the object tothe side of each image and find the angle of each ray withthe optical axis. The smallest angle (in our example it isthe angle of the ray Oc’) is called the aperture angle, aap,

Fig. 1.1.21 Problem P.1.1.10 – Two-branch microscope.

Fig. 1.1.22 A dual magnification system. Fig. 1.1.23 Aperture stop and entrance and exit pupils.

11


and the corresponding physical diaphragm is called theaperture stop (cd in the case of Fig. 1.1.23). Its image inthe object space is called the entrance pupil and its imageto the image space is called the exit pupil (c0d0 and c00d00,respectively). Obviously, the aperture angle defines themaximum cone of light rays emerging from point O andpassing through the system with no obstacle up to pointO0 in the image plane. The corresponding angle a0ap is theaperture angle in the image space. Drawing the rays thatconnect any other point of the object with the entrancepupil we find the corresponding cone of light partici-pating in imaging of that point. The ray connecting theoblique point A with the center C of the entrance pupil iscalled the chief ray (shown by the dotted line inFig. 1.1.23). Its position in the image space is C0A0.

Now we consider the entrance pupil, ab, together withany other diaphragm (or its image, gh) in the object space(Fig. 1.1.24). It is understood that the conical bundleoriginating from point O of the object is not affected bygh at all. The same is true for any other point of theobject plane between O and A where the last one is foundwith the ray passing through the sides a and g of bothdiaphragms. For remote points above A (point B, forexample), the light cone filling the entrance pupil is cutpartially by the diaphragm gh (the dotted line originatingin B cannot be transferred). This means that the activecone of light passing through the system is reducedgradually until we achieve finally the point C from whichno ray can pass the system. The rays emerging from anypoint above C cannot achieve the image plane at all.Therefore, image formation can be performed only fora part of the object plane (the circle of radius OC). Thispart of the object plane is called the field of view and thediaphragm gh is called the field aperture. If gh is theimage of a real physical diaphragm GH located some-where in the system then GH is called the field stop.

Reduction of the light cones while moving out fromthe optical axis causes a decrease of the image brightnessin the corresponding parts of the image plane. Even if theobject plane is equally illuminated we get a reduction ofthe brightness in the image plane, as is illustrated by thegraph of intensity, I(r), in Fig. 1.1.24.

This phenomenon is known as vignetting, and it shouldbe carefully investigated if a new optical system is designed.

To find the field aperture it is necessary to image allphysical diaphragms of the system into the object space,to calculate the sizes and location of each image, and thento draw a ray connecting the center of the entrance pupilwith the side point of each image and to calculate thecorresponding angle with the optical axis. The minimumangle defines the field aperture (and consequently thefield stop). The procedure described is illustrated byFig. 1.1.25 where g2h2 serves as the field aperture. It isuseful to take into account the fact that to avoid vi-gnetting it is necessary to position the field stop at theplane of the intermediate image of the system.

Problems

P.1.1.12. The system of two thin lenses L1 (focal length100 mm, diameter 20 mm) and L2 (focal length 50 mm,diameter 20 mm) shown in Fig. 1.1.26 forms an image ofthe object plane P on a screen M at magnification V ¼ 3.The distance between P and L1 is 200 mm.

(a) How can the field stop ab of the system be posi-tioned in order to get imaging with no vignetting?

(b) What should be the size of the field stop if the fieldof view is 10 mm?

(c) Find the location of all elements of the system andcalculate the aperture angle and position of the en-trance pupil.

P.1.1.13. In the system of Problem P.1.1.10, find theminimum size of lens L3 which enables one to get imageson the CCD with no vignetting.

Fig. 1.1.24 Field of view and vignetting.

Fig. 1.1.25 Finding the field aperture.

Fig. 1.1.26 Problem P.1.1.12 – Imaging with two lenses.

12


1.1.4. Prisms in optical systems

Prisms serve three main purposes in optical systems: (i)to fold the optical axis; (ii) to invert the image; and (iii)to disperse light of different wavelengths. Here we willconsider the first two purposes. It is quite understand-able that both are achieved due to reflection of rays onone or several faces of the prism. So, it is worth keepingin mind how the system of plane reflectors (plane mir-rors) can be treated (e.g., see Problem P.1.1.14).

A great variety of prisms are commonly used in nu-merous optical architectures. Only a few simple cases aredescribed below.

Right-angle prism. This is intended for changing thedirection of the optical axis through 90�. The cross-sec-tion of this prism is shown in Fig. 1.1.27a. The rayscoming from the object (the arrow 1–2) strike the inputface AB at 90� and after reflection from the hypotenuseside emerge along the normal to the face BC. It can beseen that beyond the prism the object is inverted. Theshortcoming of this prism is revealed if the incoming lightis not normal to the prism face. In this case the anglebetween incoming and outgoing rays differs from 90�.Another issue is concerned with the total reflection ofrays on face AC: it might happen that for some tilted raystotal reflection does not occur. In such a case a reflectingcoating on AC is required.

Penta-prism. This prism has effectively four faces withan angle of 90� between AB and BC and 45� between thetwo other sides (Fig. 1.1.27b). The shortcoming of theright-angle prism does not occur here, i.e., the outgoingbeam is always at 90� to the input beam, independent ofthe angle of incidence. Also, the object is not inverted.This results from a double reflection in the prism and isevidence of the common rule for any prism or systemwith reflectors; namely, the image is not inverted if thenumber of reflections is even.

Dove prism. The angles A and D are of 45� and theinput and output beams are usually parallel to the basisface AD (Fig. 1.1.27c). While traveling through theprism the beams are inverted. Another feature of thisprism is its ability to rotate an image: when the prism is

inserted in an imaging system rotated around the inputbeam with angular speed u the image in the system willbe rotated at a speed 2u.

If it is necessary to invert beams around two axesa combination of prisms, like the Amici prism shown inFig. 1.1.28, can be exploited. This prism is actuallya right-angle prism with an additional ‘‘roof’’ (for thisreason it is also called the roof-prism). As a result thebeams are inverted in both directions: upside-down andleft-right.

In general, any prism inserted in an imaging systemmakes the optical path longer. This effect should betaken into account if a system designed for an unbentconfiguration has to be bent to a more compact size usingprisms and mirrors. With regard to its influence on imagequality and optical aberrations the prism acts as a block ofglass with parallel faces. As was demonstrated earlier (seeProblem P.1.1.6 where the propagation of a divergent–convergent beam through a glass slab of thickness t wasconsidered) the block of glass causes a lengthening of theoptical path by (n � 1)t/n compared to the ray tracing inair. Therefore instead of tracing the rays through the slaband calculating the refraction at the entrance and exitsurfaces one can replace a real plate by a virtual ‘‘air slab’’of reduced thickness, t/n, and perform ray tracing for aironly. To apply this approach to prisms we have to find theslab equivalent to the prism with regard to the ray pathinside the glass. This can be done by the following pro-cedure based on unfolded diagrams (see Fig. 1.1.29). Westart moving along the incident ray until the first re-flection occurs. Then we build the mirror image of theprism and the rays and proceed moving further along theinitial direction until the second reflected surface is met.Then again we build the mirror image of the configura-tion, including the ray path, and proceed further until theinitial ray leaves the last (exit) face of the prism. Detailsof the procedure can be seen in Problem P.1.1.15.

Creating unfolded diagrams is aimed at calculating thethickness, te, of the equivalent glass block. For the casesdepicted in Fig. 1.1.29:

(a) right-angle prism with an entrance face of sizea: te ¼ a;

Fig. 1.1.27 Layout of different prisms: (a) right-angle prism;(b) penta-prism; (c) Dove prism. Fig. 1.1.28 Amici prism.

13


(b) penta-prism with the same size a of the entranceface: te ¼ a(2 þ O2);

(c) Dove prism of height a and 45� angles betweenfaces: te ¼ 3.035a.

Once te is known, the apparent thickness in air is calcu-lated from te/n.

Problems

P.1.1.14. Imaging in systems of plane mirrors. An objectAB is positioned as shown in Fig. 1.1.30 in front ofa mirror corner of 45�. Find the location of the imagebeyond the mirrors.

P.1.1.15. Find the reduced (apparent) thickness ofa 45� rhomboidal prism of 2 cm face length. The prism ismade of BK-7 glass (n ¼ 1.5163).

P.1.1.16. Alens L of 30 mm focal length transfers theimage of an object AB positioned 40 mm in front of L toa screen P. A penta-prism with 10 mm face size isinserted 20 mm beyond the lens. Find the location of thescreen P relative to the prism if it is made of BK-7 glass(n ¼ 1.5163).

P.1.1.17. Dispersive prism at minimum deviation. Findthe minimum deviation angle of a prism with vertexangle b ¼ 60�. The prism is made of SF-5 glass with re-fractive index n ¼ 1. 6727.

1.1.5. Solutions to problems

P.1.1.1. We are looking for a solution in the paraxial rangeand assume the lens is of negligible thickness. To find theimage of point A we use two rays emerging from A: ray 1parallel to the optical axis and ray 2 passing through thecenter of the lens (Fig. 1.1.31). Ray 1 after passing throughthe lens goes through the back focus F0. Ray 2 does notchange its direction and continues beyond the lens alongthe incident line. The intersection of the two rays after thelens creates the image A0 of point A. Once the image A0 isfound, the image O0 of point O is obtained as the in-tersection of the normal from point A0 to the optical axis.

It should be noted that instead of ray 1 or 2 one canuse ray 3 (dotted line) going through the front focus F inthe object space (in front of the lens) and parallel to OXafter the lens. Intersection with the two other raysoccurs, of course, at the same point A0. Also note that inour approximation of the paraxial range the homocentricbeam also remains homocentric in the image space.

P.1.1.1.2. In both cases, Figs. 1.1.32a and b, we drawthe ray (dotted line) parallel to AB and passing throughthe center of the lens. The ray crosses the back focal planeat point C. Since the ray and AB belong to the same par-allel oblique bundle and all rays of such a bundle are col-lected by the lens in a single point of the back focal plane,this must be point C. Therefore, the ray AB after passingthrough the lens goes from B through C to point A0 at theintersection with the axis. This point is the image of A. Inthe case of Fig. 1.1.32b the focus F0 and correspondingback focal plane are located to the left of the lens. Hence,not the ray itself but its continuation passes through pointC. The intersection with OX is still the image of the pointsource A which becomes virtual in this case.

P.1.1.3. First we will derive the ray tracing formulavalid for the paraxial approximation. By multiplying both

Fig. 1.1.29 Unfolded diagram for (a) the right-angle prism, (b) thepenta-prism, and (c) the Dove prism.

Fig. 1.1.30 Problem P.1.1.14 – Imaging in a mirror corner.Fig. 1.1.31 Problem P.1.1.1 – Graphical method of finding theimage.

14


sides of Eq. (1.1.4) by h (see Fig. 1.1.33) and denotingh/S ¼ tan(u) z u and h/S0 ¼ tan(u0) z u0 we get

u0 � u ¼ hF;

which yields for a number of lenses (k ¼ 1,2,.,N):

ukþ1 ¼ uk þ hkFk; (A)

with the additional geometrical relation

hkþ1 ¼ hk � ukþ1dk;kþ1: (B)

Expressions (A) and (B) enable one to calculate the raytrajectory in a system of thin lenses. To start the calcu-lation we need the values u1, h1. Usually these values canbe arbitrarily chosen, as they do not affect the final re-sults. Going back to the numerical data of the problem,we choose u1 ¼ �0.1 and then proceed as follows (seeFig. 1.1.34):

u1 ¼ �0:1; h1 ¼ S1u1 ¼ ð�30Þð�0:1Þ ¼ 3:0

u2 ¼ �0:1þ 3:0

15¼ 0:1;

h2 ¼ 3:0� 0:1� 40 ¼ �1:0

u3 ¼ 0:1� 1:0

15¼ 0:0333;

h3 ¼ �1:0� 0:333� 60 ¼ �3:0

u4 ¼ 0:0333� 3:0

20¼ �0:1167;

S0

3 ¼h3

u4¼ �3:0

�0:1167¼ 25:71 mm:

It can be easily checked that exactly the same result willbe obtained if we choose another initial value of u1 (say,

u1 ¼ �0.2). Of course, this results from the linearity ofthe expressions (A) and (B).

P.1.1.4. Measurement of the optical power of a lens orits focal length is often required in the practice of opticaltesting. The method described here is particularly usefulbecause it is based only on measurements of linear dis-tance (L) and linear displacement (a) which can be easilyand accurately realized.

We start with a derivation of working formulas.Combining Eqs. (1.1.4) and (1.1.5) gives

S0 ¼ VS;1

f 0¼ 1

VS� 1

S¼ 1� V

VS;

S ¼ f 01� V

V; S0 ¼ f 0ð1� VÞ:

Therefore

S0 þ S ¼ f 0ð1�V2ÞV

¼ a; S0 � S ¼ �f 0ð1�VÞ2

V¼ L

(remember that V< 0, S< 0, and S0 > 0 in both positions1 and 2 of the lens). Solving the last equations for V andfor f 0 we get

f 0 ¼ aV

1� V2¼ LV

ð1� VÞ2;

V ¼ �Lþ a

L� a; f 0 ¼ L2 � a2

4L¼ L

4� a2

4L:

It is understood that linear magnifications V1 and V2 inpositions 1 and 2 are reciprocal values (V1 ¼ 1/V2), andthe segments S, S0 are just replacing each other whilemoving from position 1 to 2.

In deriving the above expressions we did not take intoaccount the thickness of the lens, or, more exactly, thedistance A between the principal planes. The rigorousrelation is L ¼ S0 � S þ A. Actually D is unknown andtherefore it is the origin of uncertainty in the value of L.Differentiating the above expression for f 0 with regard toL and denoting dL ¼ D we obtain

df ¼ D

4

�1þ a2

L2

�;

Fig. 1.1.33 Problem P.1.1.3 – Ray tracing through a single lenses.

Fig. 1.1.34 Problem P.1.1.3 – Ray tracing through a system ofthree lenses.

Fig. 1.1.32 Problem P.1.1.2 – Graphical method of finding theimage with (a) a positive lens and (b) a negative lens.

15


Now for the numerical data of the problem we have

f 0 ¼ 135

4� 452

4� 135¼ 30:0 mm:

If the thickness of the lens is about 6 mm then the dis-tance between its principal planes is about 2 mm (ap-proximately one-third of the lens thickness). Hence, forthe uncertainty of the focal length we have

df ¼ 2

4

"1þ

�45

135

�2#¼ 0:56 mm:

P.1.1.5. Consider the layout of the thick lens shown inFig. 1.1.35. We use Eqs. (1.1.7) and (1.1.8) and applythem to two surfaces of the lens. We choose an arbitraryvalue for h1 and start with u1¼ 0, remembering that in ourcase n1¼ n3¼ 1; n2¼ n. Since we are looking for a solutionin the paraxial range, where the heights of all rays aresmall, one can neglect the segments x1, x2 (the latter is notshown in the figure) assuming that the distance betweenh1 and h2 is equal to the lens thickness, t. Then we get

u2 ¼1

nu1 þ

h1

R1nðn� 1Þ ¼ h1

R1nðn� 1Þ;

h2 ¼ h1 � u2t ¼ h1 �h1t

R1nðn� 1Þ;

u3 ¼ nu2 þh2

R2ð1� nÞ ¼ h1

R1ðn� 1Þ

� h1

R2ðn� 1Þ þ h1t

R1R2nðn� 1Þ2;

which enables one to calculate the focal length and SF0

(BFL):

1

f 0¼ u3

h1¼ ðn� 1Þ

�1

R1� 1

R2

�þ tðn� 1Þ2

R1R2n

SF0 ¼h2

u3¼ f 0

�1� tðn� 1Þ

R1n

�:

To find the segment SF (FFL) we should repeat thesame procedure, but to assume that the ray which isparallel to OX is incident on the surface R2 of the lens

from the right. Then the exit ray intersects the opticalaxis in the front focus F (left of the lens) and replacingf 0 by f and R1 by R2 in the above expression for SF0 wefinally get

SF ¼ f

�1þ tðn� 1Þ

R2n

�:

Here we should remember that in our problem f < 0 andR2 < 0, hence the value of SF is negative. Using the nu-merical data of the problem we obtain

F ¼ 1

f 0

¼�0:5163

�1

50þ 1

75

�� 6ð0:5163Þ2

1:5163 � 50� 70

�� 103

¼ 16:92 dioptry

f 0 ¼ 1

F¼ 59:1mm; SF0 ¼ 59:1

�1� 6� 0:5163

50� 1:5163

�

¼ 56:69mm; SF ¼ �59:1

�1� 6� 0:5163

75� 1:5163

�

¼ �57:49mm:

As we see, the principal planes H and H0 are located 1.61mm and 2.41 mm, respectively, inside the lens.

P.1.1.6. Consider the ray incident on the slab ata height h1 along the direction of the angle i (seeFig. 1.1.36). We have h2 ¼ h1 � t � tan(r) where the re-fraction angle r is calculated from Eq. (1.1.1). Since theincident angle at point 2 is also r, the refraction angle here(found again from Eq. (1.1.1)) is equal to i, meaning thatthe exit ray is parallel to the incident one. Then O0A0 ¼ h2/tan(i)¼ h1/ tan(i)� t tan(r)/ tan(i); OA¼ h1/tan(i); andtherefore AA0 ¼ O0A0 � (OA � t) ¼ t � t[tan(r)/tan(i)]

AA0 ¼ t 1�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� sin2ðiÞn2 � sin2ðiÞ

s24

35:

As we see, AA0 depends on i, which means that each rayof the homocentric incident beam intersects the opticalaxis after the slab in another point A0. In other words, thehomocentricity of the beam is violated. As the measure

Fig. 1.1.35 Problem P.1.1.5 - Finding the location of the principal planes in a thick lens.

16


of this violation one can choose the value d ¼ (AA0)i max.Since imax is related to the given solid angle, u, as u ¼2p[1 � cos(i max)], we obtain

cosðimaxÞ ¼ 1� u

2p¼ 1� 1:5

2p¼ 0:761;

sinðimaxÞ ¼ 0:649;

d ¼ 5

"1�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:761

2:25� 0:421

r #¼ 1:775 mm;

O0A0 ¼ 31:775 mm:

[Note: The above expression for AA0 is rigorous, it is validfor any angle i. For small angles i (paraxial approximation)we have sin i z i; sin2(i) 1� n2; AA0z t(1� 1/n); and AA0

does not depend on i. If n¼ 1.5 then AA0 ¼ t/3. This meansthat in the paraxial approximation the center of the incidentbeam is just relocated with regard to initial point A by one-third of the glass slab thickness (1.667 mm in our case).]

P.1.1.7. Considering a two-lens system in general, andreferring to Fig. 1.1.37 one obtains

S2 ¼ f 01 � d;1

S02¼ 1

f 02þ 1

S2¼

f 01 � dþ f 02f 02ðf 01 � dÞ ;

S02 ¼1�F1d

F1 þF2 � F1F2d;

h2 ¼ h1S2

f 01¼ h1ð1� F1dÞ;

f 0e ¼ S02h1

h2¼ 1� F1d

F1 þF2 � F1F2d� h1

h1ð1� F1dÞ

¼ 1

F1 þ F2 �F1F2d¼ 1

Fe: (A)

Now, by substituting the problem data in expression (A)we get

Fe ¼1

100þ 1

75� 30

100� 75

¼ 0:01933 mm-1 ¼ 19:33 diopter;

f 0e ¼1

0:01933¼ 51:72 mm;

S02 ¼1� 0:01� 30

0:01933¼ 36:2 mm:

Thus, the two lenses could be considered as a singlesystem with 51.72 mm focal length and the focal pointpositioned 36.2 mm behind the second component.[Note: Replacing two lenses by a single equivalent lens isuseful only if a parallel beam strikes the system. If im-aging is performed for an object positioned at a finaldistance from the first lens then Eq. (A) above becomesuseless and calculations should be done according to Eqs.(1.1.4) and (1.1.5), first for the first element and then forthe second one.]

P.1.1.8. For a ball lens of radius r the Eqs. (1.1.9) and(1.1.11) are transformed as follows ðd ¼ 2r; r1 ¼ �r2Þ:

SF0 ¼ f 0�1� 2ðn� 1Þ

n

�¼ f 0

2� n

n;

1

f 0¼ 2ðn� 1Þ

r� 2ðn� 12Þ

rn¼ 2ðn� 1Þ

nr(A)

and therefore

a0 ¼ f 0 � SF0 ¼ f 0�

1� 2� n

n

�

¼ nr

2ðn� 1Þ �2ðn� 1Þ

n¼ r: (B)

Thus, the principal plane H0 is located at the center ofthe ball. Due to the symmetry of the lens one can statethat the front principal plane is located at the same point.From the data of the problem, using the glass data fromAppendix 2 (nD ¼ 1.67270), we find

SF ¼2� 1:6727

2� 0:67273 ¼ 0:73mm;

f 0 ¼ 3� 1:6727

2� 0:6727¼ 3:73:Fig. 1.1.37 Problem P.1.1.7 – Ray tracing through a system of

two thin lenses.

Fig. 1.1.36 Problem P.1.1.6 – Ray tracing through a parallel plate.

17


As we see, the focus is distant from the lens surface by0.73 mm.

P.1.1.9. The angle in air between two chief rays di-rected to two separate object points still distinguished bythe eye is 3� 10�4 rad. Taking into account the ‘‘reducedeye’’ properties, in particular the refractive index of themedium between the eye lens and the retina as n¼ 1.336and the back focal length as 22.9 mm, we get that thelimiting angle in the vitreous is 3� 10�4/1. 336¼ 2.25�10�4 rad. The corresponding distance between twoimages on the retina is 2.25 � 10�4 � 22.9 ¼5.15 �10�3 mm and they should fall on two different cells. This

means that the retina cell size is about 5 mm.P.1.1.10. (a) The intermediate image in the branch to

the eye is formed in the plane of the reticle M of size19 mm. As linear magnification of the objective is V1 ¼� 10, it yields

S01 ¼ �10S1;1

�10S1� 1

S1¼ 1

20; S1 ¼ �22 mm

and this is the working distance of the system. Thecorresponding field of view is 1.9 mm.

(b) The eyepiece has visual magnification determinedfrom Eq. (1.1.15): G ¼ 250/25 ¼10 and therefore thetotal magnification in the branch to the eye is Vtot [ V1G¼ 100.

(c) The optimal imaging in the CCD branch is theone which enables one to see the maximum part of theimage created on the circular reticle M. Such a situationshown in Fig. 1.1.38 means that the maximum imagesize on the CCD is y0 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið4:82 þ 5:62Þ

p¼7.4 mm and

therefore the relay lens L3 provides an optical magnifi-cation V3 ¼ 7.4/19 ¼ 0.388. This can be realized in twopossible arrangements. The first is demonstrated inFig. 1.1.38 and the second in Fig. 1.1.39. In both casesS03 ¼ 20 mm, but S3 ¼ S03/V3 ¼ 20/0.388 ¼ 51.5 mm inthe first case and S3 ¼ �51.5 mm in the second case.

Evidently the shortest configuration is that of Fig. 1.1.38.In this case

1

f 03¼ 1

20� 1

51:5; f 03 ¼ 32:7mm;

and the distance between P and the CCD is 22 �10 � 51.5 ¼ 168.5 mm. In the second case

1

f 03¼ 1

20� 1

51:5; f 03 ¼ 14:4 mm;

and the distance between P and the CCD is 220 þ 51.5¼ 271.5 mm.

P.1.1.11. (a) From the numerical data of Fig. 1.1.22we have S1 ¼ �70; S01 ¼ (�70) � (�3) ¼ 210. There-fore, the distance between the CCD and the last beamsplitter is 210 � (10þ 100 þ 70)¼ 30 mm and the focallength of L1 should be

f 02 ¼�

1

210þ 1

70

��1

¼ 52:5 mm:

In the high-magnification branch the image createdby lens L1 (the same size and position as in the low-magnification branch) serves as the virtual object for thesecond lens, L2 (negative relay lens). Magnification of L2

is V2 ¼ 2 � (� 3)/V1 ¼ 2. Since the distance along theoptical axis between L1 and the CCD is l ¼ 10 þ 70 þ 2� (10 þ 15) þ 100 þ 30 ¼ 260 mm and taking intoaccount that S02�S2 ¼ l�S01 ¼ 50 mm and S02 ¼ 2 � S2,we get S2 ¼ 50 mm and S02 ¼ 100 mm and therefore thelocation of L2 is 70 mm below the last beam splitter. Itsfocal length is

f 02 ¼�

1

100� 1

50

��1

¼ �100 mm:

(b) Increasing the high magnification by 10% requires V2

¼ 2.2 (V1 remains the same as before). Hence,

S02 ¼ 2:2S2;1� V2

V2S2¼ 1

f 02;

Fig. 1.1.39 Problem P.1.1.10 – Formation of image on CCD, thesecond case.

Fig. 1.1.38 Problem P.1.1.10 – Formation of an image onto theCCD plane.

18


which yields S2 ¼ 54.54 mm and S02 ¼ 120 mm. In otherwords, the relay lens should be relocated to 90 mmbelow the last beam splitter. Since in this case S02 � S2 ¼120 � 54.54 ¼ 65.46 mm it is necessary to add thelength 15.46 mm to the optical path of the left branch.This is done by moving down the retroreflector by thesegment Dz ¼ 15.46/2 ¼ 7.73 mm.

P.1.1.12. (a) To get an image with no vignetting it isnecessary to position the field stop in the plane of theintermediate image created by the first lens. Referring toFig. 1.1.40 we get

S1 ¼ �200; S01 ¼

1

f0

1

þ 1

S1

!�1

¼ 200 mm;

V1 ¼ �1;

and therefore the field stop should be of the same size asthe field of view, i.e., dab ¼ 10 mm, and positioned 200mm behind L1. The total magnification V ¼ 3 ¼ V1 � V2

requires V2 ¼ �3, which enables one to find the positionof L2 and M:

1� V2

S2V2¼ 1

f 02; S02 ¼ 50

1þ 3

ð�3Þ ¼ �66:67 mm;

S02 ¼ S2V2 ¼ 200 mm:

(b) The field stop size, as we saw above is 10 mm.(c) To find the entrance pupil we should build the

image of all diaphragms (L2 and ab in our case) in theobject space, e.g., to create their images through L1 atreverse illumination (as if radiation propagates from rightto left). Since ab conjugated with the object plane P, onehas to find only the image of L2 through the first lens. Wehave:

S21 ¼ �226:67mm; S021 ¼�

1

100� 1

266:67

��1

¼ 160:0 mm;

d0L2¼ 20� 160

266:67¼ 12 mm:

Calculating the angle of the margin ray coming from theon-axis point of the obje to the side point of the lens L1

gives a1 ¼ 10/200 ¼ 0.05. The corresponding angle of

the image of L2 is a2 ¼ 6/(200 � 160) ¼ 0.15 > a1 andtherefore the entran pupil is the lens L1 and the apertureangle of the system is aap ¼ a1 ¼ 0.05.

P.1.1.13. Referring to the solution of Problem P.1.1.10and Fig. 1.1.41, we first find t size of the lens L1 usingNA ¼ 0.2 and the distance to the object S1 ¼ �22 mm:DL1 ¼ 2 � 22 � tan(arcsin 0.2) ¼ 9.0 mm. Then weproceed with the margin ray originating in the off-axispoint of object A. This ray comes to the side point A’ ofthe intermediate image (O0A0 ¼ 19/2¼ 9.5 mm) and it isthis ray which determines the active size of lens L3.Geometrical consideration of the figure gives

DL3 ¼ 2� ðO01NþNDÞ

¼ 2��DL1=2þ 168:5

220ð9:5� 4:5Þ

�

¼ 16:66 mm:

P.1.1.14. To demonstrate the advantage of the unfoldeddiagram we describe two approaches in solving theproblem: first without the diagram and then usingunfolding. In the first case we start with imaging throughmirror M1 (see Fig. 1.1.42a) and find the image point A0

using the triangle AOiA0, where AO1 ¼ (80 � 20) ¼60 mm ¼ A0Oi. Obviously the second image point, B’, ison the horizontal line passing through A0. Then, referringto A0B0 as a new object we find its image in mirror M2:A00O2 ¼ A0O2 ¼ 80 mm and B00 is again located on thehorizontal line passing through A00.

Fig. 1.1.41 Problem P.1.1.13 – The margin ray tracing throughlenses L1 and L3.

Fig. 1.1.42 Problem P.1.1.14 – Two approaches to finding theimage: (a) without unfolded diagram; (b) with unfolding.Fig. 1.1.40 Problem P.1.1.12 – Imaging system with the field stop.

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In the second case (Fig. 1.1.42b) we create the imageof the mirror corner in mirror M1. Then mirror M2

image, M02, is vertical and A0B0 is parallel to AB anddistant from M02 by 80 mm. Going back to the realmirror M2 we just put A00B00 beneath M2 at the samedistance 80 mm and 20 mm to the right of the vertex. Aswe see, the second approach is significantly shorter andeasier.

P.1.1.15. We build an unfolded diagram for the prism,as demonstrated in Fig. 1.1.43, and consider the principalray MNPQ striking the entrance face AB at the heightAM ¼ (a�sin 45�)/2 ¼ 0.707 cm. This is the center ofa beam passing through the prism. Obviously MN ¼ AM¼ PQ¼ P1Q1¼ 0.707 cm, NP1¼ a¼ 2 cm, and te¼MNþ NP1 þ P1Q1 ¼ 3.414 cm. Hence, the apparent (re-duced) thickness is

ten¼ 3:414

1:5163¼ 2:251cm:

P.1.1.16. We refer to Fig. 1.1.44 and assume that the lensis working in the paraxial range. Without the prism thedistance from the lens to the screen P would be

S0 ¼�

1

30� 1

40

��1

¼ 120 mm:

The thickness of the glass block which is equivalent tothe prism is te¼ 3.41a¼ 34.1 mm. The prism makes theray trajectory longer by the segment D ¼ te(1 � 1/n) ¼34.1(1 � 1/1.5163) ¼ 11.61 mm. Finally, from the ge-ometry of the figure we get for the distance between thescreen P and the exit face of the prism x¼ 120�20�34.1þ 11.61 ¼88.39 mm.

P.1.1.17. The prism ABC of refractive index n hasa vertex angle b and an input ray strikes the side AB atpoint O1 at an incident angle i1 (see Fig. 1.1.45). Thedeviation angle 4 is defined as the angle between theinput direction and the output direction of the ray.Geometrical consideration of the triangles O1BO2 andO1DO2 yields

4 ¼ ði1 � r1Þ þ ðr2 � i2Þ; r1 þ i2 þ g ¼ 180+

¼ bþ g;

r1 þ i2 ¼ b; (A)

4 ¼ i1 þ r2 � b: (B)

Snell’s law gives i1 ¼ arcsin(n sin r1) and r2 ¼ arcsin(nsin i2) ¼ arcsin[n sin(b � r1)]. By substituting these ex-pressions in (B) we get

4 ¼ arcðn sin r1Þ þ arcsin½n sinðb� r1Þ� � b: (C)

To find the minimum deviation angle we calculate thederivative d4=dr1 and find the angle at which it has a zerovalue, as usual:

d4

dr1¼ n cos r1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1� n2 sin2 r1

p � n cosðb� r1Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� n2 sin2ðb� r1Þ

q ¼ 0;

cos r1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� n2 sin2 j

q¼ �cos j

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� n2 sin2 r1

q¼ 0

where the new variable j ¼ b � r1 is introduced. Fromthe last equation we have

cos2r1

cos2j¼ 1� n2sin2r1

1� n2sin2j;

Fig. 1.1.43 Problem P.1.1.15 – Unfolded diagram of a rhomboidalprism.

Fig. 1.1.44 Problem P.1.1.16 – Imaging through a penta-prism.Fig. 1.1.45 Problem P.1.1.17 – Deviation of a ray travelingthrough a prism ABC.

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and denoting z ¼ sin2 r1 we proceed as follows:

1� z

cos2j¼ 1� n2z

1� n2sin2j;

z ¼ 1=cos2j� 1=ð1� n2sin2jÞ1=cos2j� n2=ð1� n2sin2jÞ

¼ sin2j� ð1� n2Þ1� n2

¼ sin2j:

The last equation is satisfied if r1 ¼ j and therefore r1 ¼b � r1; and r1 ¼ b/2. With this value we have from (C):

4 ¼ 2arcsin

�n sin

b

2

�� b (D)

and

i1 ¼ arcsin

�n sin

b

2

�: (E)

The last two expressions allow one to calculate the angleof minimum deviation of the prism and the incidenceangle corresponding to such a deviation. Going back tothe problem, we find 4 ¼ 2 arcsin(1. 6727 � sin 30�) �60� ¼ 53.51� and the incidence angle i1 ¼ arcsin(1.6727� sin 30�) ¼ 56.76�.

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geometrical optics - elseviergeometrical optics menn 1.1.1. ray optics conventions and practical...

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