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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Radiometry
of point sources – Inverse square law
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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.
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Radiance vs. angle for
Lambertian source.
Cosine to the 4th “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
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 !!!!
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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
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Power emitted by a Lambertian
source and captured by a lens
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Imaging extended sources
Constant brightness theorem again
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T Transmission of system 0 < T < 1
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 = TπL sin θ
2
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/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.
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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.
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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 with
phosphor. 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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Example
Radiometry of projector
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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/m2)
2. Calculate H of condensor from power, lamp brightness
2
⎛π ⎞
φ = ⎜ ⎟ LH 2
⎝n⎠
3. Given slide radius, this gives NA of condensor (H ~rNA)
4. Can now design projection imaging system with this H
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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.
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Point source imaging
Point 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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Field Stop
Aperture Stop
Vignetting
Exit pupil projected onto exit window
Exit window
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.
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Vignetting
calculating overlap factor
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 = dtanθ
The half angles of the unvignetted areas are given by
r22 + Δ2 − r1'2
r1'2 + Δ2 − r22
cos φ2 =
cos φ1 =
2r2 Δ
2r1' Δ
The resulting overlap is given by
'2
'2
'
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A(θ ) = φ r + φ r − r r sin(φ + φ2 )
z
r1
r =
d+z
'
1
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
20
f1=50 f =-45
2
D=20 D=20
Project AS into
image space with
thin lens to find
exit pupil
To=-20 F=-45
-> t1=-13.85
M=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 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°) = 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
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25 degree point has ~50% less intensity than on axis with
roughly equal losses for cos4 and vignetting.
Reading
W. Smith “Modern Optical Engineering”
Chapter 12
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