Most radiometric measurements do not require an accurate calculation of the spherical surface area to convert between units. Flat area estimates can be substituted for spherical area when the solid angle is less than 0.03 steradians, resulting in an error of less than one percent. This roughly translates to a distance at least 5 times greater than the largest dimension of the detector. In general, if you follow the “five times rule” for approximating a point source, you can safely estimate using planar surface area.

Luminous flux is a measure of the power of visible light. Photopic flux, expressed in lumens, is weighted to match the responsivity of the human eye, which is most sensitive to yellow-green. Scotopic flux is weighted to the sensitivity of the human eye in the dark adapted state.

- RADIANT FLUX:

- 1 W (watt)

- = 683.0 lm at 555 nm

= 1700.0 scotopic lm at 507 nm

- = 1 W*s (watt * second)

= 10

= 0.2388 gram * calories

- 1 lm (lumen)

- = 1.464 x 10

= 1/(4p) candela (only if isotropic)

- = 1 talbot (T)

= 1.464 x 10

l
nm |
Photopic
Luminous Efficiency |
Photopic
lm / W Conversion |
Scotopic
Luminous Efficiency |
Scotopic
lm / W Conversion |

380
390 400 410 420 430 440 450 460 470 480 490 500 507
510 520 530 540 550 555
560 570 580 590 600 610 620 630 640 650 660 670 680 690 700 710 720 730 740 750 760 770 |
0.000039
.000120 .000396 .001210 .004000 .011600 .023000 .038000 .060000 .090980 .139020 .208020 .323000 .444310 .503000 .710000 .862000 .954000 .994950 1.000000
.995000 .952000 .870000 .757000 .631000 .503000 .381000 .265000 .175000 .107000 .061000 .032000 .017000 .008210 .004102 .002091 .001047 .000520 .000249 .000120 .000060 .000030 |
0.027
0.082 0.270 0.826 2.732 7.923 15.709 25.954 40.980 62.139 94.951 142.078 220.609 303.464 343.549 484.930 588.746 651.582 679.551 683.000
679.585 650.216 594.210 517.031 430.973 343.549 260.223 180.995 119.525 73.081 41.663 21.856 11.611 5.607 2.802 1.428 0.715 0.355 0.170 0.082 0.041 0.020 |
0.000589
.002209 .009290 .034840 .096600 .199800 .328100 .455000 .567000 .676000 .793000 .904000 .982000 1.000000
.997000 .935000 .811000 .650000 .481000 .402000 .328800 .207600 .121200 .065500 .033150 .015930 .007370 .003335 .001497 .000677 .000313 .000148 .000072 .000035 .000018 .000009 .000005 .000003 .000001 .000001 |
1.001
3.755 15.793 59.228 164.220 339.660 557.770 773.500 963.900 1149.200 1348.100 1536.800 1669.400 1700.000
1694.900 1589.500 1378.700 1105.000 817.700 683.000 558.960 352.920 206.040 111.350 56.355 27.081 12.529 5.670 2.545 1.151 0.532 0.252 0.122 .060 .030 .016 .008 .004 .002 .001 |

Illuminance is a measure of photometric flux per unit area, or visible flux density. Illuminance is typically expressed in lux (lumens per square meter) or foot-candles (lumens per square foot).

So, 1 steradian has a projected area of 1 square meter at a distance
of 1 meter. Therefore, a 1 candela (1 lm/sr) light source will similarly
produce 1 lumen per square foot at a distance of 1 foot, and 1 lumen per
square meter at 1 meter. Note that as the beam of light projects farther
from the source, it expands, becoming less dense. In fig. 7.4, for example,
the light expanded from 1 lm/ft^{2} at 1 foot to 0.0929 lm/ft^{2}
(1 lux) at 3.28 feet (1 m).

Figure 7.5 illustrates a typical setup to determine the location of an LED’s virtual point source (which is behind the LED due to the built-in lens). Two irradiance measurements at known distances from a reference point are all that is needed to calculate the offset to the virtual point source.

- IRRADIANCE:

- 1 W/cm

- = 104 W/m

= 6.83 x 10

= 14.33 gram*calories/cm

- 1 lm/m

- = 1 lux (lx)

= 10

= 10

= 9.290 x 10

= 9.290 x 10

The radiance, L, of a diffuse (Lambertian) surface is related to the radiant exitance (flux density), M, of a surface by the relationship:

Some luminance units (apostilbs, lamberts, and foot-lamberts) already
contain p in the denominator, allowing simpler
conversion to illuminance units.

- Suppose a diffuse surface with a reflectivity, r,
of 85% is exposed to an illuminance, E, of 100.0 lux (lm/m

- 1.) Calculate the luminous exitance of the surface:

- M = E * rM = 100.0 * 0.85 =
85.0 lm/m

- L = M / p

L = 85.0 / p = 27.1 lm/m

So, for an extended source with a radiance of 1 W/cm^{2}/sr,
and a detector with a viewing angle of 3°, the irradiance at any distance
would be 2.15 x 10^{-3} W/cm^{2}. This assumes, of
course, that the source extends beyond the viewing angle of the detector
input optics.

- RADIANCE:

- 1 W/cm

- = 6.83 x 10

= 683 cd/cm

- 1 lm/m

- = 1 candela/m

= 1 nit

= 10

= 10

= 10

= 9.290 x 10

= 9.290 x 10

= p apostilbs (asb)

= p cd/p/m

= p x 10

= p x 10

= 2.919 x 10

= 2.919 x 10

If you are wondering how the units cancel to get flux/sr from flux/area
times distance squared, remember that steradians are a dimensionless quantity.
The solid angle equals the area divided by the square of the radius, so
d^{2}=A/W, and substitution yields:

The biggest source of confusion regarding intensity measurements involves the difference between Mean Spherical Candela and Beam Candela, both of which use the candela unit (lumens per steradian). Mean spherical measurements are made in an integrating sphere, and represent the total output in lumens divided by 4p sr in a sphere. Thus, a one candela isotropic lamp produces one lumen per steradian.

Suppose that two LED’s each emit 0.1 lm total in a narrow beam: One has a 10° solid angle and the other a 5° angle. The 10° LED has an intensity of 4.2 cd, and the 5° LED an intensity of 16.7 cd. They both output the same total amount of light, however -- 0.1 lm.

A flashlight with a million candela beam sounds very bright, but if
its beam is only as wide as a laser beam, then it won’t be of much use.
Be wary of specifications given in beam candela, because they often misrepresent
the total output power of a lamp.

- RADIANT INTENSITY:

- 1 W/sr (watts per steradian)

- = 12.566 watts (isotropic)

= 4*p W

= 683 candela at 555 nm

- 1 lm/sr (lumens per steradian)

- = 1 candela (cd)

= 4*p lumens (isotropic)

= 1.464 x 10

- You measure 22.0 lux from a light bulb at a distance of 3.162 meters.
How much light, in lumens, is the bulb producing? Assume that the
clear enveloped lamp is an isotropic point source, with the exception that
the base blocks a 30° solid angle.

- 1.) Calculate the irradiance at 1.0 meter:

- E

E

- 220 lm/m

- W = A / r

- W = 2p[1 - cos(330
/ 2)] = 12.35 sr

- 220 lm/sr * 12.35 sr = 2717 lm