Measurements on Lambertian objects:some novel approaches

Henry L. LaMuth

The problem of measuring Lambertian objects at close distances is examined for both a sphere and a coneirradiated by a plane wave. The equations describing the scattered power collected by a receiver are de-rived using the principles of radiometry, and some experimental results are provided.

1. IntroductionA number of practical problems exist in which the

radiance of an object is required. 1 - 4 If the directionof irradiance on the object is defined and if the mo-tion (rotation, tumbling, etc.) of the object is known,the magnitude and fluctuations of radiation from theobject can be calculated for a given receiver.3 '4 Al-most all problems, however, consider (1) the object tobe far away from the receiver, (2) the object to beLambertian, and (3) the object to be smaller than thereceiver field of view.5- 8

When laboratory measurements on Lambertianobjects must be made, the assumption of the receiverbeing far away from the object may no longer hold.This problem in scaling the distance of the receiverwill be examined for two problems. The first case tobe considered is that of a sphere; this problem is sim-ilar to viewing the moon with the eye, where themoon is a good deal larger than the eye's resolutionelement. The second case is that of a cone observedfrom the direction of the cone apex when the receivercontacts the apex.

The two examples were chosen to demonstrate thebasic principles of applying radiometric formalismsto problems of interest. Certain features of the ex-amples made each easier and more interesting for theparticular scaling principle demonstrated. The ap-plications are, however, perfectly general, and the ex-amples could be interchanged.

In Sec. II the basic equation is described. In Secs.III and IV the sphere and cone, respectively, will bedescribed, and in Sec. V the conclusion will be given.

When this work was done, the author was with Martin MariettaCorporation, Orlando, Florida 32805; he is now with Battelle-Co-lumbus Laboratories, Columbus, Ohio 43201.

Received 26 December 1973.

II. Basic Equation

Depending on the specific problem geometry em-ployed, a problem can be made exactly soluble. Themost general approach of the greatest applicability isthat developed by Look in which an arbitrarilyshaped object can be described in a regular coordi-nate system.7 The important equation is shownbelow:

P = Hpo'f[(nno)(nn,)/(np)]dA'dSŽr. (1)

The specific terms in this equation are defined in thefollowing list and refer to Fig. 1 for the case of an ir-regular object:

Pr = power at the receiver;H = irradiance on the object;n = unit normal to an elemental area on the

object;ni= unit vector pointing toward incident radia-

tion;no = unit vector pointing towar receiver;dS = elemental area on object within receiver

field of view;p = unit normal to projected area of dS on the

preferred coordinate plane;dA' = elemental area of dS projected on pre-

ferred basis;d2r = elemental solid angle of receiver as seen

from dA';Po' = reflectivity of object.

The object is assumed Lambertian. Replacing Po' by7r' would specialize the magnitude of the reflectivityto that of a perfect Lambertian object. Also, if theobject surface is described by a function f(xyz) = 0,then n = v vAl. In this paper, Po' will always be aconstant independent of angle, that is, representativeof a Lambertian surface.

1150 APPLIED OPTICS / Vol. 14, No. 5 / May 1975

n0

An

J

Fig. 1. Orientation of an irregular surface in the X- Y-Z (i-j-k)coordinate system.

W1. Lambertian SphereThe first and easier problem of the two to be con-

sidered is that of a sphere irradiated by a plane wave.This problem is of interest because it is in contrastwith the usual example of Lambertian surfaces wherea flat surface is uniformly illuminated and fills a re-ceiver's field of view. The radiance of the flat sur-face is constant, and the distance of the receiver fromthe surface is immaterial. For an object over whichthe radiance is not constant, however, the size andlocation of the surface filling the field of view doesmatter. For the flat plate no receiver input varia-tions occur; for a sphere, as an example, the receiverdoes in fact record intensity changes as the locationof the field of view changes location on the sphere.

The irradiation is chosen to be from the + Y direc-tion, so ni = j. The normal to the sphere n is givenby

n = i sine cos + sine sine + k cos9.

of view of the receiver moves across the object sur-face.

The sphere appears to be 50% illuminated with aterminator in the X-Z plane, as shown in Fig. 2(a).The elemental area dA' is inside the total field ofview area, where the center of the total field is atcoordinates

(X0,Yo) = (ao cospo, a sinO0), (4)

as defined in Fig. 2(b). The area dAl has centroidcoordinates

(3Y) = (a cos , a sinq)= (ao cosqo + a' coso',a, sins + a' sink'), (5)

where (X, Y) are the coordinates of the center of dA'relative to the sphere coordinates, and

(X', Y') = (a' cosO, a' sinp') (6)

are the coordinates of the center of dA' relative to theprimed coordinates, which are parallel to the primarycoordinate system but located at the center of thefield of view area at (XO, Yo). The elemental areadA' inside the field of view is integrated over the fieldof view, given by

dA' = a'da'd/Y, (7)

and Eq. (1) becomes

LAMBERTIANSPHERE

- _ .- H (IRRADIANCE)

a)

(2)

The area dS on the sphere is projected on the x-yplane as dA' with normal p = k (or -k depending onthe location of dS above or below the plane), and

fdA'

represents the size of the receiver field of view on theplane.

The receiver is located along the positive z axis, sono = k. In Eq. (1) the value of dQr is a constant andcan be removed from the integral, giving

fdQ, = D, = A/R 2, (3)

where Ar is the receiver aperture, and R is the dis-tance of the receiver from the sphere. The receiverdistance and aperture are assumed large enough sothat no appreciable variation in no occurs as the field

X

Fig. 2. Geometry of a sphere for evaluating Eq. (1).

May 1975 / Vol. 14, No. 5 / APPLIED OPTICS

H ___

i

p, - SINoSINK . a SINOr

K 0u90.

=1 0o'80'

O D 60°

I i Il I I .- I I I I

I I I I II I I I

50' 60' 70' 80' 90

given size of the field of view, as ao approaches r, partof the field of view will fall outside the boundaries ofthe sphere, causing a rapid decrease in receivedpower. The same condition holds as the values of koand ao cause the field of view to approach the termi-nator.

If the field of view is at the center of the sphereand if the size of the field of view is exactly the sameas the disk of the sphere (a' = r), Eq. (11) yields

0P 40

00=20'

l 0'o5

a)

Fig. 3. Normalized power received from a Lambertian sphere ir-radiated from a direction perpendicular to the receiver direction:

(a) plot; (b) description of angles O, o, and distance ao.

Pr = po'HAr(2r2 /3R2 ) = po'HAQv(2/3T).

This represents the maximum possible power in thegiven geometry and is the same result as given else-where.5-8

In Fig. 3, the plots of Eq. (12) are of

P Pr/PO'HAnFOV = sinb0 sin o,

where ao and ko define the center of the field of view.It is obvious that despite the apparently uniformbrightness of a sphere as seen by the eye, consider-able intensity variation occurs. The main reasonsfor apparent uniformity are (1) the eye never sits atjust one fixed point; (2) the intensity variation is aslowly varying function of position except at the ter-minator or limb; (3) the eye perceives strong gradi-ents in intensity rather than absolute values of inten-sity.

If the irradiance of the sphere is changed so thatthe receiver and irradiance are in the same directionk, Eq. (1) becomes

Pr = (po'HArR 2 )f sine sin'a'da'dP (8)

In order to evaluate Eq. (8), sinO and sinq must begiven in terms of the primed coordinates. From Eq.(5)

a sino = a sin 0 + a' sinp'. (9)

The projection of dS on the X- Y plane is dA'. Thecoordinate of a point in dS projected onto dA' is

a = rsinO (10)

which is the projection onto the plane of a coordinatepoint at 6 on the surface of a sphere of radius r.Thus, Eq. (8) becomes

Pr = po'HAR2)f (a, sino 0 + a' sin0')(a'/r)da'dk,(11)

where 0 < < 2 7r, 0 < a' r', and r' is the radius ofthe circular field of view area on the X- Y plane.Evaluation of Eq. (11) yields

Pr = po'HAr7Tr2 sinpOao/rR2

Pr = (po'HA/r)ao sinOQfov,

(12a)

(12b)

where QFOV = 7rr2 /R2 is the field of view of the re-ceiver.

The plots of Eqs. (12) are shown in Fig. 3. In thisplot no edge effects are considered. That is, for a

Pr = (poHA/R2 )f cos a'da'dp'. (15)

Now, cosO = (1 - sin2 o)1/2 = (r2- a2 )1/2 /r from Eq.

(10). Also, from Eq. (5), a2 = a 2 + a 2 + 2a'aocos(' - O), so that Eq. (15) becomes

Pr = (I'HA R2r f fo 2s [r2 - a02

- a 2

- 2a 0a' cos(o' - )]1 1 2 a'da'dop. (16)

This can only be evaluated numerically for a generalsolution. A closed solution can exist for ao = 0. Inthis case the solution to Eq. (16) is simply

Pr = (po'HAr27r/3rR 2)[r 3- (r 2

- r 2)3/2] (17) ;

If the field of view is centered on the sphere andjust large enough to include the whole sphere, ao = 0and r' = r. In this case Eq. (17) yields

Pr = 'HAr27 Tr 2/3R2 (18)

which is r times larger than Eq. (13).

IV. Lambertian ConeThe second problem to be considered is that of a

Lambertian cone uniformly irradiated from the sideand viewed from the apex direction, as shown in Fig.4. In this case the power from a field of view wan-

1152 APPLIED OPTICS / Vol. 14, No. 5 / May 1975

1.0*

0.5

lo~ '1 I 0 10' 20' 30' 40'

~0(13)

(14)

-

7

Y

LAMBERTIAN CONE

H (IRRADIANCE)

l i

- FULL CONE ANGLE

Figure 5 defines the vectors for the cone just as Fig. 2did for the sphere. The normal to the cone is

n, = i cos¢c~ cosa + cosa sink - k sina, (20)

where Oi is the incidence angle relative to the hori-zontal, fp is the angular position on the cone, and a isthe cone half-angle. The radiance of the cone is

N(0c) = pH (cosa sinO, cosOi - sina sinGi), (21)

and the angular extent of the illuminated area isgiven by

RECEIVER sin(P = tan9i tana. (22)

Each wedge dA' of the cone area projected on theX- Y plane is given by

dA' = (D/2)2 0C, (23)

where each elemental wedge has constant radiance,and D is the diameter of the cone base. With p = k,when the receiver is far enough away so that no isconstant, Eq. (1) becomes

P, = Hpo'Q(D2/8)f (cosa sin4g cosi

- sinOi sina,)dkW, (24)

Fig. 4. Lambertian cone irradiated from an elevation angle O: (a)isometric view; (b) radiance pattern.

dering across the cone is not being sought. Rather,the cone outline just fills the receiver field of view,and the receiver is brought forward toward the coneapex. In this case radiation scatters off the part ofthe cone that is in the receiver field of view but withangles as large as the acceptance angle of the receiv-er.

This problem is of interest because in laboratorymeasurements an object may be so much smallerthan the field of view of a receiver that the only wayto measure its properties is to be very close to it.When the object and receiver are close, however, theformalism used in Sec. III does not hold. The prob-lem occurs because the assumption that no is nearlyconstant must be reworked.

The problem of the power received at large dis-tances from a cone irradiated from any direction hasalready been solved.6 The received power dependson both the irradiance and receiver directions rela-tive to the cone axis. For the case to be discussedhere, the irradiance is either horizontal, above, orbelow the horizontal -plane, and the receiver is alongthe z axis off the cone apex as shown in Fig. 4(a).

Seen from the z axis, the irradiated area of thecone projects on the X-Y plane as a wedge-shapedarea with nonuniform radiance as shown in Fig. 4(b),and the angular extent of the wedge is a function ofthe irradiance direction relative to the horizontal.The irradiance direction is taken to be in the Y-Zplane, and ni is given by

ni = coso + k sin9i. (19)

where the limits on Xc are given by solutions to Eq.(22). Calling 01 and 02 the solutions to Eqs. (22) andD/2 = 1 tana, Eq. (24) yields

P, = (Hp,'&Ž7/2)1 3 tan2 Ca[cosa COSpli(COSO - cos¢p2 )

- (02 - cp1) sinOi sina], (25)

1 2n

Fig. 5. Geometry of a cone for evaluating Eq. (1): (a) elemental

area description; (b) description of R.

May 1975 / Vol. 14, No. 5 / APPLIED OPTICS 1153

-y H

IPC

where is the cone height. As an example, when Oi =0, then ql = 0 and ¢2 = r, giving for Eq. (25)

P = HpOQ2l 2 tan2ca cosa (26)

This agrees with results derived elsewhere. 6

As the receiver approaches the cone apex, the de-tected radiation is no longer scattered into just the Zdirection, but from each cone surface point into asmall solid angle defined by the field of view of thereceiver optics. Equation (1) is still to be solved, butthe formulation of the problem has changed. Theproblem will be solved for the receiver contacting thecone apex. Intermediate ranges are more difficult todeal with.

In order to find the expression for the receiverpower, an intermediate concept of the radiation foot-print on the receiver input aperture must be consid-ered. As shown in Fig. 5(b), a point on the cone withcoordinates (Z,0c) scatters radiation into the receiveraperture at a location with coordinates (rL,OL). Avector R from the cone to the lens is given by

R = i(rL cos - Z tana coso,)

+ j(rL sinOL - Z tana sinkL) - kZ. (27)

R makes an angle to the aperture normal given by

cos4' = -RnL/ I R Z/[Z 2 sec2C + rL2

-2 rLZ tanai cos(OL - kd)], (28)

and 4 < max, where Omax is one half of the full field ofview angle of the receiver. It is useful to solve Eq.(28) for the ratio r/Z. This ratio gives the angularextent of the footprint at the receiver and can besolved for the maximum and minimum rL/Z for =Xc. More important is the angular extent of theratio. Putting L - O = A, rL/Z becomes single val-ued. The value of A is given by

cosA = cota (sec 2a - sec2 p)1/2 . (29)

Equation (29) shows that from a point on the conewith angular coordinate L, radiation is scatteredinto the receiver between aperture angular coordi-nates

the cone is small, the receiver aperture must beclosed down so that the cone does fill the field ofview. The development now proceeds in much thesame fashion as when a plain flat white surface isbrought close to the receiver with no intensity varia-tions being encountered.

The important step occurs in Eq. (1):

P7 = f N((P)dsdQ = fN(bL)d0FOvdAr,

which can befield of view.equation to be

(31)

done when the object fills a receiverThe field of view is constant, so the

solved is

P = O'Harj (cosa sinOL cosi - sini sina)dA,. (32)

The lens area, dAr, is contiguous to the projectedarea, dA', in angular extent. So, Eq. (31) is analo-gous to Eq. (1).9 Also, because of the contiguous-ness, spc = XL. The constancy of radiance as a func-tion of radial distance means that the same formal-ism obtains in the case of the cone as when a planeflat sheet fills a receiver field of view. That is, theprojected area of the radiant surface onto the receiv-er aperture is larger than the aperture for all ranges.The limits on Spc are given by Eq. (22). Putting lr =2 7r(1 - cosOmax) Eq. (32) becomes

P, = pHA,(1 - cosom".) [CosCa Cosei (cos 1 - cos0 2 )

+ (01 - 02)sini sina]. (33)

Equation (33) has been verified with measure-ments, shown in Fig. 6. The power received has beenplotted as a function of the incidence angle Oi andcompared with the theoretical value from Eq. (33).

V. ConclusionsBy application of Eq. (1), two different radiation

measurement conditions were examined. In the firstexample the case of a spherical Lambertian scattererthat is much larger than the field of view of a receiverwas developed. A sphere was used in the example,although any object would work. In the second ex-ample a cone was examined. The radiation reachinga receiver that is in contact with the apex of the cone

L = ±c ± A. (30)

For example, if the cone half-angle is 300 and the re-ceiver half-field angle is 30, A = 40.

The usefulness of the concept of the footprint is toshow that the angular pattern of the detected radia-tion entering the receiver is given by the angular pat-tern of the cone irradiance, which is given by solutionto Eq. (22), projected on the receiver aperture. Eventhough the match is not exact, the radiance near theterminator on the cone is negligible compared to ra-diance at an angular point nearer to the incidence di-rection. The error, then, resulting from the approxi-mation is small.

The receiver is assumed focused at infinity, and thecone is assumed to fill the receiver field of view. If

I.

1., SOLID LINE-THEORY

0- EXPERIMENT

-10° 0'

INCIDENCE ANGLE ( i)

Fig. 6. Power collected by the receiver as a function of the inci-dence irradiation angle and normalized to horizontal (i = 0)

incidence.

1154 APPLIED OPTICS / Vol. 14, No. 5 / May 1975

was examined, and the applicable equation wasfound to give good results. Again, any object couldbe used. Closed solutions do require the properchoice of geometry, however.

The two techniques were examined because theyapproximate laboratory conditions that are encoun-tered. The techniques provide a method for makinglaboratory scale measurements when the distancesthat can make Eq. (1) exactly soluble leave the re-ceived intensities very small. Occasionally, laborato-ry distances are required. It is easier to measure aclose object than a distant object when the totalpower scattered is small. Also, when detailed infor-mation on an object's surface is desired, near proxim-ity of the receiver to the object is necessary.

References1. R. Tousey, J. Opt. Soc. Am. 47, 261 (1957).2. R. A. Schmeltzer, Appl. Opt. 8, 179 (1969).3. E. R. Lanczi, J. Opt. Soc. Am. 56, 873 (1966).4. E. R. Lanczi, J. Opt. Soc. Am. 57, 202 (1967).5. M. Sussman, J. Opt. Soc. Am. 48, 278 (1958).6. W. R. Rambauske and R. R. Gruenzel, J. Opt. Soc. Am. 55, 315

(1965).7. D. C. Look, Jr., J. Opt. Soc. Am. 55, 462 (1965).8. K. W. Brand and F. A. Spagnolo, J. Opt. Soc. Am. 57, 452

(1967).9. Because the projection plane for the cone is identically the en-

trance aperture of the receiver, the projected cone area is con-tiguous (point by point) with the aperture. [See A. Taylor, Ad-vanced Calculus (Ginn and Company, New York, 1955), pp380-389.]

a

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May 1975 / Vol. 14, No. 5 / APPLIED OPTICS 1155