1、Chapter The Photometry of Optical Systems光学系统的光度学 6.1 Introduction 节 引言 6.2 Photometric definitions光度学的定义 6.3 The flux emitted by a plane source面光源发射的光通量 6.4 Illumination due to a circular source环形光源的光照度 6.5 Illumination in an optical image像面的光照度 6.6 Photometric properties of plane surface面光源的光度学性质
2、6.7 Calculation of intensity loss计算能量损失Basic equationLight source: radiationBrightness (luminance)Basic: B ( ) （ ）2cdmIBAIntensity: oscsNI: Angle of view from normalFlux: F1dSr2803()eKVVisual function6/lmwat(): solid angleFABreceiving detector: illuminance: FE2lmpoint light source : 2AIdSo Or 2cosIE
3、on-axis point : 0inBoff-axis point : 4cs* optical systemtransmittance : 20sinEBu4co* Brightness (luminance) of illuminated surfaceAErB 6.1 IntroductionA quantitative knowledge of the flow of light from object to image is of the greatest importance. It enables us to determine the illumination falling
4、 on the film in a camera, or the brightness of a star seen in a telescope, or an image projected on a screen.* We cannot seelight. We can only see the sourcefrom which the light comes. Eye can detect only the resultant mixture of wavelengths which we refer tocolor of object.* Photometry is concerned
5、 only with the integrated effect of all wavelength to which the eye is sensitive, from about 0.4 to about 0.75 .m! Basic unit is brightness (luminance).* Radiometry is concerned only with the emission and detection of radiant energy, detectors responds to radiant power. Watt6.2 Photometric definitio
6、ns1. Luminance B ( )2cdmBrightness (luminance) is basic to all photometry.The apparent brightness of an object depends on the luminance of an object, the size of the pupil of our eye, and our state of adaptation in a dark place, eye become dark adapted, while our state of adaptation in a bright plac
7、e, eye become bright adapted.* The luminance of a surface is a physical property of the surface. It does not depend on how is viewed. It may not be viewed at all.* Luminance of a surface is quite independent of the size of the source.* The fundamental unit of luminance is the stilb. Nowadays called
8、(candela/ or 2cdmcandle/).2. Intensity I ( )cdThe unit of intensity is the candela.Intensity obeys lamberts cosine law.(6.1)cosIBAA: the area of the sourceB: brightness of the source: the angle of view from the normal of the source area.So the intensity I of an ordinary domestic light bulb varies wi
9、th direction.3. Flux F (lm)Light flux F is a measure of the amount of light radiated by a source into a given solid angle.* The unit of flux is the lumen (lm).(6.2)FI1lmcdSrdefined as the amount of flux radiated by a point source having an intensity of I candela (cd) into a cone having a solid angle
10、 of steradian.4. Illuminance E (lx)When light falls on an object, it causes the object to be illuminance.* The quantity of the illumination is called illuminance (E).(6.3)FEAThe unit of illuminance is Lux (lx) 21lmx1lm flux falls on a of a plane surface.2m* Inverse square lawThe solid angle :Subtend
11、ed by screen A at the source2AdBecause the flux FISo the illuminance 2EA5. The luminance of an aerialWe look at a uniform diffusing surface through a window or lens, the image that we see will have a luminance equal to the luminance of the surface multiplied by the transmittance.(t1) (6.5)Bt* If we
12、look at some object through a microscope or telescope, the image that we see will have the same luminance as the object, only reduced by the transmittance factor of the microscope or telescope.* We look at the moon through a telescope, we get the impression that the image is much brighter that the o
13、bject. (BB)But, if we open the other eye and look at the moon direct, we see that the real moon is brighter than its image in the telescope. (BB)! The telescope accepts more total energy than the unaided eye, which leads to a sensation of dazzle, because the image of the moon is so much larger than
14、the original moon. 6.3 The flux emitted by a plane source1. a small plane source of area A and lumininance B, : semi vertical angleThe solid angle of the shell is 2r2(sin)ddBecause cosIAB2in()rdFd(2sinco)0ABd(6.6a)2sin* If , the flux radiated by a small plane source into a complete hemisphere is 90(
15、Single-side luminescence) (6.6b)FDouble-side luminescence: AB2. The luminance of an illuminated surfacedetermines the relation between the illuminance falling on a surface and the luminance resulting from this illumination. Suppose: a plane receiving surface of area A is under an illuminance of E lu
16、mens per square unit. ( )FEAThe total number of lumens falling on the surface Receiving F: FEIf the surface is perfectly diffusing and has the reflectivity r The luminance of the illuminance surface: rABSo (6.7)3. cosine radiating objectRadiating intensity is cosine distribution cosNIBrightness 2cos
17、(/)sNNAAIIBBcdmdCosine radiating object has a constant brightness.* Diffusing reflecting object is cosine radiating object.* If with ought Flux loss FThen 22sinsinBdudu(For optical system )2isABFrom 222insiJyuy22siBSo 2n 6.4 Illumination due to a circular sourceAn important relation in photometry is
18、 an expression for the illuminance due to a large circular source of a point on its axis.1. illuminance of axial point The intensity of such a zone ( ) in direction of :drOEcos(2)cosIABrdBSince tanrxxeThe illuminance at , by inverse square law is: OE2cosIdE222(tan)e)s/sBxdsicoIntergraded from 0 to (
19、6.8a)2inOE* If the source is of infinite angular extend such as open sky: 902. (oblique) illuminance off-axis point * Asmal source the law4cosThe illuminance on screen will be a maximum on axis and will be decreasing with increasing oblique angle Oblique distance cosxdBy inverse square law, the illu
20、minance at the angle will be inversely proportional to ( )2dThe solid angle: 22(cs)/oBArdx442ssASo (6.8b)4coOEConclusion: on-axis point: Ailluminance: 2sinOB off-axis point: solid angle 4cosA4cosOE* Attention: aperture angle , solid angle , strictly u2u 24sin()Bu* Method 2: cscs/orQQdx 6.5 Illuminat
21、ion in an optical image1. Axial point Illumination with distant object Suppose: We have a lens forming an image of a distant object on a screen. The object is diffusing, with uniform luminance of B, then the exit pupil of lens will be find with light having the same luminance B times the transmittan
22、ce factor of the lens BThe axial point of screen: 22211sin()44ODEBuNfifF-number: So 1DfN 1sin2uNF-number: 2, 2.8, 4, 5.6, 8, 11, 16 Aperture change 12Off-axis point: 4cosOE2. Axial point Illumination with near object : magnification of pupilzsin2()zDux()zff2()zDf1()zN( )224()OzEBfND* If the lens is
23、symmetrical or nearly so, about a central stop, pupil magnification will be approximately 1. ( )1z3. The exposure equationThe American national standard institute (ANSI) established a measure of film speed that become international in scope.The time (T) of exposure required to produce a satisfactory
24、 picture is T-second.2NTBS: F-NumberfNDB: luminance of an object S: ASA film speed: lens transmittance(ASA: American Standard Association)Germany express the film speed as 10DINASSlongASA: 64, 100, 200, 400DIN: 19, 21, 24, 274. Oblique image illuminanceAt points away from axis, the illumination fall
25、s off.* If the exit pupil of the lens were a small circle of constant diameter, the law 4coswould be accurately applicable in the image space.* At higher obliquity, the exit pupil is no longer a circle, and in most lenses it changes its size, shape and location as the obliquity increases.* In the ab
26、sence of vignetting and pupil distortion, the combined effect of image distortion the effect is given by4cos20sincoyEf: Obliquity angle in the object space : Focal lengthF: the rate of change of image height with obliquity angle2y ytanyf2secfWhen : l4oE* If a lens is to have uniform illumination across the field, we must have an amount of distortion such that image height is given by (sin condition)sinyfcofSo is exact.1oE6.6 Photometric properties of plane surface1. Reflection and transmission at normal incidence light is incident perpendicularly to the surface.
