Radiometry. Basics Extended Sources Blackbody Radiation Cos4 th power Lasers and lamps Throughput. ECE 5616 Curtis

Transcription

1 Radiometry Basics Extended Sources Blackbody Radiation Cos4 th power Lasers and lamps Throughput

2 Radiometry Terms Note: Power is sometimes in units of Lumens. This is the same as power in watts (J/s) except that it is spectrally weighted by the sensitivity of the human eye. Other photometric terms illuminance in units of lumins/m 2, Luminance (brightness) in units of L/(sr m 2 )

3 Radiometry of point sources Inverse square law

4 Extended sources Lambert s law Irradiance of plane by point source as function of angle Radiance of extended source Radiance vs. angle for isotropic source Lambert s Law. Intensity falls off like cos away from normal. Radiance vs. angle for Lambertian source.

5 Cosine to the 4 th Law The solid angle subtended by the pupil from point A is the area of the exit pupil divided by the square of the distance OA. From point H, the solid angle is the projected area divided by OH which is greater the OA by 1/cosθ -> This gives a cos 2 θ factor. The exit pupil is viewed obliquely from point H, and its projected area is reduced by a factor that is ~ cos θ. (For high speed lens not correct). The reduction above is true for illumination on a plane normal to the line OH, but we want illumination on plane AH. The reduction factor is also cos θ. Total reduced illumination can be cos 4 θ for point H compared to point A!!!!

6 Power emitted by a Lambertian source and captured by a lens Calculate incremental power dφ radiated from tilted area A into cone of solid angle dω

7 Power emitted by a Lambertian source and captured by a lens

8 Imaging extended sources Constant brightness theorem again T Transmission of system 0 < T < 1

9 Example Problem Source is 10 W ster -1 m -2, T = 80% and angle of collection of system s exit pupil (FOV) is 60 degrees total angle. E = TπLsin 2 θ See Smith Ch 12 integration of small source in cone angle θ Is half angle subtended by exit pupil of system, T transmission, L is the object radience E =.8 *3.14*10*(.5) 2 = 7.85W/m 2 Note; It the source is small and for off-axis image points are subject to loss by factor of cos 4 (θ) in addition to any vignetting.

10 Blackbody sources Ideal incoherent sources Radiate energy, but unless T > 700 ok, emit very little visible radiation and thus appear black. Planck s explanation of the blackbody spectrum in 1900 was the beginning of the development of quantum mechanics. The radiance of a blackbody, L, does not depend on angle, and they are thus ideal Lambertian sources.

11 Blackbody Radiation So the surface of the sun is roughly 5500 o K and humans radiate in the infrared at about 9.5 μm.

12 Blackbody Radiation

13 Lasers vs. lamps

14 Light Emitting Diodes (LED) Low power Longer life More robust Inexpensive White light from Voilet/UV diode with phosphor. Can also be 3 separate diodes.

15 Lasers vs. lamps

16 Lasers vs. lamps

17 Nichia Laser Diode Spec

18 Nichia Laser Diode Spec

19 LD Spec

20 Example Radiometry of projector

21 Example Radiometry of projector If Hc < Hp edges of image will appear dark (projection lens is stopped down). If Hc > Hp light is lost on entering the projector. Typically design for Hc just a bit larger than Hp. 1. Calculate power on screen from area and spec. irradiance (W/m 2 ) 2. Calculate H of condensor from power, lamp brightness 2 2 LH n φ = π 3. Given slide radius, this gives NA of condensor (H ~rna) 4. Can now design projection imaging system with this H

22 Calculating lumens Question: Ar laser puts out 1.5W at 488nm and 2W at 514.5nm, what is the photometric power of the laser? Answer: 680 lumins/w x ((1.5x.19)+(2x.6)) = 1010lumins Relative sensitivity of eye with 1W=680lumins at 550nm.

23 Point source imaging Point source with intensity of I e is located t o from thin lens. What is the intensity at t 1 the image plane? Power collected by lens is I e A/t o2 =I A/t 1 2 So I =I e (t 1 /t o ) 2 = I e M 2 For other points in image space this can be thought of as a point source so point a distance R from image plane at angle θ have E = I e M 2 cos 3 θ/r 2,if inside solid angle of lens (zero if not)

24 Scatter into Detector with Lens A laser is focused onto a screen that radiated uniformly in 2π sr. If lens images the spot onto a detector with M=1, F=8cm, and D lens =3cm, what fraction of power makes it to the detector? Solid Angle Ω = π(3/2) 2 /R 2 If M=1 and then t=t =2f=16cm. So Solid angle is Ω =π(3/2) 2 1/16 2 sr Total solid angle of scatter is 2π so fraction is Ω/2π = So a little less than 0.5% is directed to the detector.

25 Aperture Stop Field Stop Vignetting Exit pupil projected onto exit window Exit window V(θ) = A(θ)/πr 2 Where A is the overlap of the exit pupil on the exit window, and r is the radius of the small of the two.

26 Vignetting calculating overlap factor Projecting pupil onto window it s radius is reduced from r 1 to r 1 => The angle θ is given by tanθ = y k+1 /(d+z) The separation between the 2 circles is given by Δ = dtanθ The half angles of the unvignetted areas are given by ' '2 r1 + Δ r2 r2 + Δ r1 cosφ1 = cosφ ' 2 = 2r1 Δ 2r2 Δ The resulting overlap is given by '2 '2 ' A( θ ) = φ1r1 + φ2r2 r1 r2 sin( φ1 + φ2) r ' 1 = r1 d z + z

27 Vignetting Example What is the loss in power for a point located 25 to the optic axis for the system below with object distance 300 AS FS&EW Project AS into 20 image space with thin lens to find exit pupil To=-20 F=-45 -> t 1 = M=0.692 so that r 1 = 6.92 f 1 =50 D=20 f 2 =-45 D=20 Project System with exit pupil and windows Size of separation and angles can now be calculated

28 Vignetting Example What is the loss in power for a point located 25 to the optic axis for the system Now we can calculate the overlap area at 25 degrees A(25 ) = The vignetted area is π(6.66) 2 = Thus the vignetting factor is We are NOT done. Still have cos4th loss which is cos4(25) = So total reduction is 0.76*0.675 = degree point has ~50% less intensity than on axis with roughly equal losses for cos4 and vignetting.

29 Reading W. Smith Modern Optical Engineering Chapter 12