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Radiometry

• Basics• Extended Sources• Blackbody Radiation• Cos 4th power• Lasers and lamps•Throughput

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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/m2, Luminance (brightness) in units of L/(sr m2)

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Radiometryof point sources – Inverse square law

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Extended sourcesLambert’s law

Irradiance of plane by point source as function of angle

Radiance of extended source

Radiance vs. angle forisotropic source

Lambert’s Law.Intensity falls off like cos away from normal.

Radiance vs. angle forLambertian source.

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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 cos2 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 cos4 θ for point H compared to point A !!!!

Cosine to the 4th “Law”

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Power emitted by a Lambertiansource and captured by a lens

Calculate incremental power dΦ radiated from tilted area A intocone of solid angle dΩ

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Power emitted by a Lambertiansource and captured by a lens

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Imaging extended sourcesConstant brightness theorem again

T Transmission of system 0 < T < 1

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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.

θπ 2sinLTE =

E = .8 *3.14*10*(.5)2 = 7.85W/m2

Note; It the source is small and for off-axis image points are subject to loss by factor of cos4(θ) in addition to any vignetting.

θ Is half angle subtended by exit pupil of system, T transmission, L is the object radience

See Smith Ch 12 integration of small source in cone angle

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Blackbody sourcesIdeal 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 thebeginning of the development of quantum mechanics.

• The radiance of a blackbody, L, does not depend on angle, and they are thus ideal Lambertian sources.

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Blackbody Radiation

So the surface of the sun is roughly 5500oK and humans radiate in the infrared at about 9.5 μm.

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Blackbody Radiation

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Lasers vs. lamps

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Light Emitting Diodes (LED)• Low power• Longer life• More robust• Inexpensive

White light from Voilet/UV diode withphosphor. Can also be 3 separate diodes.

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Lasers vs. lamps

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Lasers vs. lamps

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Nichia Laser Diode Spec

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Nichia Laser Diode Spec

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LD Spec

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ExampleRadiometry of projector

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ExampleRadiometry of projector

• If Hc < Hp edges of image will appear dark (projection lens isstopped 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/m2)2. Calculate H of condensor from power, lamp brightness

3. Given slide radius, this gives NA of condensor (H ~rNA)4. Can now design projection imaging system with this H

22

LHn

⎟⎠⎞

⎜⎝⎛=

πφ

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Calculating lumens

Relative sensitivity of eye with 1W=680lumins at 550nm.

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

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Point source imagingPoint source with intensity of Ie

is located to from thin lens. What is the intensity at t1 the image plane ?

Power collected by lens is IeA/to2 =I’ A/t12

So I’=Ie (t1/to)2 = IeM2

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’ = IeM2cos3θ/R2 ,if inside solid angle of lens (zero if not)

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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 Dlens=3cm, what fraction of power makes it to the detector ?

Solid Angle Ω = π(3/2)2/R2

If M=1 and then t=t’=2f=16cm.So Solid angle is Ω =π(3/2)2 1/162 sr

Total solid angle of scatter is 2π so fraction is

Ω/2π = 0.0044

So a little less than 0.5% is directed to the detector.

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VignettingAperture S

top

Field Stop

V(θ) = A(θ)/πr2

Where A is the overlap of the exit pupil on the exit window, and r is the radius of the small of the two.

Exit pupil projected onto exit window

Exit window

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1'

1 rzd

zr+

=Projecting pupil onto window it’s radius is reduced from r1 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 = dtanθThe half angles of the unvignetted areas are given by

The resulting overlap is given by)sin()( 212

'1

2'22

2'11 φφφφθ +−+= rrrrA

Δ−Δ+

= '1

22

22'1

1 2cos

rrrφ

Δ−Δ+

=2

2'1

222

2 2cos

rrrφ

Vignettingcalculating overlap factor

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Vignetting ExampleWhat is the loss in power for a point located 25° to the optic axis for the system below with object distance 300

20

f1=50D=20

f2=-45D=20

AS FS&EWProject AS into image space with thin lens to find exit pupilTo=-20 F=-45 -> t1=-13.85M=0.692 so that r1= 6.92 System with exit pupil and windows

Project

Size of separation and angles can now be calculated

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Vignetting ExampleWhat 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°) = 106.6

The vignetted area is π(6.66)2 = 139.3. Thus the vignetting factor is

We are NOT done. Still have cos4th loss which is cos4(25) = 0.675

So total reduction is 0.76*0.675 = 0.51

25 degree point has ~50% less intensity than on axis with roughly equal losses for cos4 and vignetting.

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Reading

W. Smith “Modern Optical Engineering”

Chapter 12