I wanted to know a lot more about the black art (or should it be red
art?) of infrared illumination and how to know what results can be
expected at different ranges and angles of lenses. I wanted practical
answers as well as the theory, so I approached Trevor Duffy, Marketing
Director of Derwent Systems to see what he could tell me. The reason
being that Derwent are the acknowledged leaders in the field of
infrared illumination and have probably carried out more research on
this subject than any other company. The result was that he and Brent
Midgley, their Technical Director, gave me a full day seminar on the
subject which of course included their unique approach to the subject
of infrared illumination. Combining this with data gathered for The
Principles and Practice of CCTV has produced the most comprehensive
analysis of the subject ever published.
The main problem in designing a CCTV system incorporating
infrared illumination is that infrared light cannot be measured in lux.
So how do you know how to specify camera sensitivity and how do you
know what will be seen at various distances? Other questions to be
answered are how sensitive are different cameras to infrared light, how
are spectral diagrams to be interpreted and the effect of the lens.
To understand the answers to these and many more questions
requires a basic understanding of three main aspects of physics. One is
the wavelengths of light, another is the inverse square law of
illumination and finally the transmission of light through a lens.
Infrared illumination is used to provide light over scenes that
would otherwise be too dark for a camera to create an image. It is a
compromise because the best results can only be obtained by providing
sufficient white light but of course this is not always possible. In
many cases using powerful floodlights would cause a considerable
nuisance and could be dangerous where there is road traffic moving
towards the lights. It is also difficult to cover a large area when a
pan tilt camera is being used, in this case the illumination is only
required where the camera is directed and infrared lights provide the
answer.
Electromagnetic Radiation
Light is energy in the form of electromagnetic radiation. The
different forms of electromagnetic radiation all share the same
properties of transmission although they behave quite differently when
they interact with matter. Electromagnetic radiation is measured in
nanometres which is the wavelength. One Metre is 1,000,000,000
Nanometres (nm). The frequency is the speed of light divided by the
wavelength, i.e. a wavelength of 830 nm is 3.6 x 1014 Hz.
Light is that part of the electromagnetic spectrum that can be
detected by the human eye. This is a very narrow band within the total
spectrum as shown in diagram 1.
Electromagnetic Waves
The Transmission of light energy can be conveniently described as a wave motion and having the following properties:
Electromagnetic waves require no medium and therefore can travel in a vacuum.
Different types of electromagnetic radiation have different
wavelengths or frequencies. From radio waves through visible light to
gamma rays.
All electromagnetic waves travel at the same velocity, which is
approximately 300,000,000 Metres per second in a vacuum, the speed of
light..
The waves travel in a straight line but can be affected by:
Reflectance. Which is the reversal of direction that occurs at the surface of an object i.e. a mirror.
Refraction. A change of the angle that occurs at the
boundaries of different surfaces. Different wavelengths have different
angles of refraction i.e. a stick apparently bending in the water.
Diffraction. Which is a deflection that occurs at apertures or edges of objects.
Visible Radiation
These are the wavelengths of light that are visible to the human eye and are from approximately 380 nm to 760 nm.
When all these wavelengths are seen simultaneously the eye cannot
distinguish the individual wavelengths and the result is seen as white
light. Therefore, white light is not one wavelength but a combination
of them all. This effect can be demonstrated in reverse by passing
white light through a prism. As stated previously, different
wavelengths have different angles of refraction, therefore when the
light is passed through a prism it is dispersed into its constituent
spectra because each wavelength is refracted differently. The result is
that if a white screen is placed to show the light passing out of the
other side of the prism it will show all the individual colours. This
effect is shown in diagram 2. The result is to show the spectrum of
light and the seven significant colours of the rainbow. In reality
there is a continuous range of hues but the eye sees mainly the main
colours. A real rainbow is created in the same way by the light being
reflected and refracted by droplets of moisture in the atmosphere.
(Remember from school days "Richard Of York Gained Battles In Vain").
Infrared light is considered to be wavelengths Longer than 715
nm. The range of wavelengths that the human eye can see is compared
with the wavelengths of two main types of infrared light used in CCTV
is illustrated in the diagram 3.
Measurement of light energy
As stated light is a form of electromagnetic radiation, its power is
measured in Watts and its intensity measured in Watts per square Metre
(W/M2). This goes for all wavelengths. The visible spectrum is,
however, normally measured in lumens for power and intensity in lux.
The lumen is related to perceived power or brightness and because of
this, the relationship between lumens and Watts is dependant on
wavelength. Lumen values diminish virtually to zero at infrared
wavelengths,. This is why it is not possible to express infrared
radiation in terms of lux values.
In order to measure light radiation in terms of Watts it is
necessary to use a radiometer which normally only exists in the sacred
domain of laboratories and needless to say tend to be very expensive
and beyond the means of normal installation companies.
Camera sensitivity to infrared light.
The sensor in a CCD camera is composed of thousands of tiny
photo-sensitive diodes, a camera with a resolution of 570 lines
incorporates over 400,000 such diodes. There are two main materials
used in the production of sensors, germanium and silicon. These have
very different responses to wavelengths of light as shown in diagram 4.
Therefore for CCTV applications the material used in the chip
has a great bearing on the response to either visible of infrared
light. The development of sensors for CCTV cameras is driven by the
vast Camcorder market and Camcorders are only used in the visible part
of the spectrum. Therefore any response outside the visible is wasted
and reduces the overall sensitivity of the sensor. An enormous amount
of money has been expended by sensor manufacturers to develop a sensor
that accurately follows the spectral range of the human eye, this is
known as a photopic curve. Such a sensor has by definition a low
response in the infrared part of the and therefore cannot be classified
as infrared cameras. This is an important point for CCTV designers to
consider because with new cameras arriving on the market every month it
is essential to keep checking on their suitability for infrared
illumination. Some manufacturers are not very helpful in this respect
and often optimistic. Beware of such comments as ‘suitable for I/R’,
always ask for a spectral response diagram.
Incidentally, all camera sensors are monochrome, colour is
obtained by inserting red, green and blue filters in front. This is why
colour cameras have less resolution than monochrome cameras. Also due
to the filters, colour cameras are not sensitive to infrared light.
Therefore all the discussion on camera sensitivity and suitability for
infrared illumination is confined to monochrome cameras. This is except
for the Dual Mode cameras now becoming available which potentially
offer the best of both worlds. This however is evidence of the fact
that colour cameras cannot be as sensitive as monochrome and the
industry is finally recognising the fact.
Camera spectral response
Cameras have different responses to the spectrum of light. This is
usually shown in diagrammatic form and is known as the relative
spectral sensitivity. Diagram 5 shows the relative response to each
part of the spectrum and is usually in a range from 0 to 1.0. Curves
from two cameras are shown in the following diagram. It can be seen
that camera A covers the visible part of the spectrum very effectively
whereas camera B is sensitive far into the infrared. It could be
thought that camera B would suit all requirements because of the wide
range of wavelengths covered, but not so.. A Further point is that
using an infrared sensitive camera in daylight can produce different
ranges of grey tones because they see a higher content of infrared that
the eye cannot see. Also the infrared sensitive camera can cause the
automatic iris to close due to the amount of infrared light instead of
visible light. This is particularly noticeable if there is foliage in
the scene, chlorophyll reflects at about 715 nm and often appears
bright white instead a shade of grey. In practice the sensors sensitive
to extended infrared light are more sensitive than photopic sensors.
A high power infrared lamp generally uses a tungsten halogen
bulb which due to its operating temperature emits a high proportion if
light in the infrared part of the spectrum as well as covering the
visible spectrum. (Some longer life bulbs now use quartz halogen). In
front of the light source is a filter that blocks wavelengths up to the
required infrared part of the spectrum. There are various filters that
allow different wavelengths to pass, the most common passes light at
715 nm and above. This wavelength is just at the threshold of visible
light and shows as a dull red glow. At 780 nm the light is considerably
reduced and at 830 it is almost invisible to the human eye.
If optimum performance is to be achieved it is necessary to
match the camera to the light source. Reference to diagrams 6 and 7 it
can be seen that the important part of the graph is where the camera
response overlaps the infrared filter response. The camera sensor
effectively integrates all the wavelengths of light falling on it
within the response curve. Therefore the measure of the cameras’
sensitivity to infrared radiation is the AREA UNDER THE OVERLAPPING CURVES,
not just at say 715 or 830 nm. This is why it is essential to know the
shape of both the camera and the infrared filter response curves. It is
not the height or amplitude of the curve but the area enclosed by the
camera curve and the light source curve that determines sensitivity.
Diagrams 6 and 7 superimpose typical infrared light performance
curves over the camera curves in the previous diagram. The resulting
areas under the curves indicate the reduced sensitivity even with the
camera that does have response up to 900 nm.
Diagram 8 shows the area under the curve for an infrared sensitive camera used only in daylight.
The normally undetected radiation @ 2x the area under the curve,
therefore all other things being equal the sensor will be twice as
sensitivity.
Light and illumination
Only natural light provides absolutely even illumination, although
it is of course affected by clouds and shadows. All forms of artificial
light suffer from the fact that as the distance increases from the
light source so the illuminance reduces. This is due to the inverse
square law of illumination where the illuminance falls to a quarter of
its value if the distance is doubled..
Inverse Square Law Of Illumination
As the luminous flux travels away from the light source the area
over which it spreads increases, therefore the illuminance (lux) must
decrease. The relationship is expressed by the inverse square law and
illustrated in diagram 9
The relationship between illuminance and it's effect at a distance is given by:
This factor is particularly important in considering the light
available for a camera. For instance a light source providing a level
of 30 lux at 20 Metres will provide 7.5 lux at 40 Metres and only 3.3
lux at 60 Metres. The other effect of this is that the wide range of
light levels can cause problems with automatic iris lenses. Unless set
up correctly, the foreground light will cause the iris to close and
lose definition in the distance. The reverse is if the iris is set to
the distant light level in which case there will be a lot of flare in
the foreground.
Diagram 10 illustrates the actual light levels over 100 Metres,
the ratio from 100 M to 10M is 100:1. This illustration is for a white
light because infrared light cannot be measured in lux. The effect of
this law though, affects infrared light in exactly the same way. The
effects of this when using infrared illumination on pan tilt zoom units
can be catastrophic. If it is required to zoom in to distant subjects
the power must be sufficient for the distance. The problem is that when
focused on near subjects the picture is frequently flared out.
Cosine Law Of Illumination
Another factor that affects the light level of an area is if the
light is striking the surface at an angle. This is often the case in
CCTV systems when an infrared lamp is located well above ground level.
Without going into too much theory the principle is shown in diagram
12.7.
Diagram 11 shows the effect of light striking a diffuse surface
at an angle. The effective area of the surface is reduced
proportionally to the cosine of the incident light. Avoiding too much
mathematical complexity, this means that the reflected light from a
diffuse target will be approximately proportional to the cosine of the
striking angle and the specific reflectance of target or scene. This is
only true for a diffuse reflector. If a very highly reflective surface
(specula) is encountered, virtually no energy will be reflected back
along the incident light path to the camera. This is why a mirror may
appears black when viewed at an angle.
The maximum illumination is at the centre of the beam of light.
This falls off towards the extremities. As can be seen from diagram 12
there is a greater reduction of light at the farthest distance. The
cosine law formula can be modified to approximate the light levels at
the outside ranges of a lamp. It is necessary to take account of both
the angle of the beam and the angle at which the lamp is directed. The
total angle at the farthest point will be the angle of the lamp from
the vertical plus half the included beam angle.
For example a 300-watt lamp gives 20 lux at a distance of 10
Metres, has a beam angle of 60o and is mounted at 30o. The light level
at the farthest point will then be approximately 10 lux.
Uniform light distribution
The Derwent lamp utilises design principles borrowed from radar
antennas to achieve even levels of illumination for targets at short
and long range and high and low altitude.
As has been seen from the previous text, for targets at say,
100 Metres it is necessary to radiate 100 times the amount of energy
needed to radiate a target at 10 Metres. This is necessary to cancel
out the effects of the inverse square law.
The Derwent system is designed to provide equal illumination
for short and long range targets. Supposing, for instance, a lamp will
radiate targets at 100 Metres and 10 Metres with an illumination level
of 60 mW/M2. This is illustrated in diagram 13 and is more than
sufficient illumination to provide 0.7 volts of video from an average
photopic CCD camera using an average f1.4 lens.
The design concept is that a single lamp will provide the
equivalent of five lux evenly distributed over a scene from 10 to 100
Metres. If twin lamps are used the equivalent light distribution will
be ten lux. The inverse square law may be applied to obtain the
illuminance at other distances. Therefore at two hundred Metres a
single lamp will provide 1.25 lux equivalent, i.e. a quarter of the
light for twice the distance.
The beam power transmitted by the lamp is actually 1,000
milliwatts to provide 10 milliwatts per square Metre at 100 Metres and
10 milliwatts to provide 10 milliwatts per square Metre at 10 Metres.
The structure of the vertical beam is such that a subject at 10 Metres
will be radiated by the same intensity of energy as one at 100 Metres.
Camera sensitivity
For research purposes a comparison was made of five commonly
available cameras to compare the energy required on the sensor to
achieve certain levels of performance. The tests were carried out using
light sources that could put out a calibrated level of light energy
measured in microwatts per square Metre. Two monochromatic light
sources were used, one at 595 nm, the other at 880 nm. These simulate
the wavelengths in the middle of the visible spectrum and well into the
infrared spectrum. The tests were also measured at two output levels of
the sensor. One to provide a signal to noise ratio of 12 dB, the other
to provide a full 0.7 volts video signal. The results are listed in
table 1.
Four of the cameras had similar specifications for sensitivity,
around the 0.1 lux level, one had a sensitivity of 0.05 lux. Note the
greatly different performance at 880 nm! Although as always they are
all qualified by various factors such as; usable picture; acceptable
picture; 50 IRE, etc. In other words they are all the same but
different!
With the 595 nm light source which is in the middle part of the
visible spectrum there is almost no difference in the energy required
at the 12 dB level. For the full video there is more variation with
factor of 4.7:1 between the greatest and the least.
With the 880 nm light source the variation is 5.7:1 for the 12 dB level but for full video the range stretches to 10:1.
It is beyond the scope of this article to analyse these results
further but they are presented to illustrate the variations in light
energy required by different sensors at different wavelengths.
| Light source 595 nm |
Light source 880 nm |
| Camera ref. |
For S/N 12dB |
For 0.7 v video |
For S/N 12dB |
For 0.7 v video |
| A |
10.2 |
32 |
32 |
80 |
| B |
10.7 |
32 |
14 |
32 |
| C |
9.0 |
80 |
80 |
300 |
| D |
9.2 |
80 |
54 |
320 |
| E |
8.9 |
17 |
31.6 |
74 |
Table 1, µ W/M2 required on sensor.
The first component that light from a scene has to pass through is
the lens. If this is wrong then everything after gets progressively
worse. Apart from selecting the focal length for a particular scene,
there is generally very little thought given to this prime element in
the chain to get a picture from a scene to a monitor. There are several
factors in lens selection that will affect the effectiveness of a
system under infrared light. Often they will determine whether anything
at all will be seen.
Spectral response
As has been previously discussed, cameras have varying response to
wavelengths of light but the same also applies to lenses. Just as CCTV
camera sensors are led by the Camcorder market, so CCTV lenses are led
by the photographic market, as well as lenses on Camcorders. They are
all based on the need to create images in visible light. Even in the
dark a flash is used to simulate visible light. Therefore lenses have a
spectral response which is biased towards the visible part of the
spectrum.
An example of a lens spectral response diagram is shown in
diagram 14. In this example the response at 715 nm is only 60% of that
at 550 nm. This can vary greatly between different makes of lens and
can have a significant affect on the infrared performance of a system.
So, get on to your lens manufacturer and obtain diagrams, beware of
those that cannot provide them.
Aperture
The f number of a lens is the ratio of the focal length to the
effective object lens diameter. It is a mechanical ratio and does not
infer the efficiency of a lens. It does affect the amount of light
energy passed to the sensor and will play a significant part in the
resulting picture. Traditionally camera manufacturers have specified
sensitivity with a lens having an aperture of f 1.4. This would be fine
if they all did it the same but they don’t. Some say with 75%
reflectance some say 89% and so on. Then again some will state the
sensitivity with AGC on but not what the AGC gain is. Camera
specmanship is too vast a subject to expand on in this article but
suffice to say the f number of the lens is a most important
consideration. In simple terms the smaller the f number the more light
is passed to the sensor, therefore f1.2 is better than f1.8. The
percentage of light passed by different apertures is listed in table 2.
This shows the percentage of light falling on the lens that is passed
to the sensor. The f stops in bold face are full stops and each number
in the scale halves or doubles the light passed. There are two
intermediate stops shown because they are common stops found in CCTV
lens.
| F number |
f1.0 |
f1.2 |
f1.4 |
f1.8 |
f2.0 |
f2.8 |
f4.0 |
f5.6 |
| % passed |
20% |
15% |
10% |
7.5% |
5% |
2.5% |
1.25% |
0.625% |
Table 2 light passed by f stops
Yes, it is true that with an aperture of f1.4 only 10% of the
light on the lens is passed to the sensor. Some manufacturers specify
camera sensitivity as that on the faceplate or sensor. In these cases
use these ratios to convert to the light required on the lens. i.e. 1
lux faceplate sensitivity requires 10 lux with an f1.4 lens or 20 lux
with an f2.0 lens.
It may seem relatively unimportant to quibble about the
difference between an f1.2 lens and an f1.4 lens, especially when the
latter is much cheaper than the first. It is significant though because
the f1.4 lens needs 50% more light for the same energy on the sensor.
An example taken from a distributors’ catalogue shows two 12mm auto
iris lenses, one f1.4 and one f1.2. The f1.4 is £75.00 the f1.2 is
£152.00! Which lens would a comparatively inexperienced estimator use
in a competitive tender? In a ten camera system it could mean a saving
of £750.00 or about £1,000.00 on the tender price- a considerable
temptation.
The effect of sensor size
As stated earlier, light is energy measured in Watts per square
Metre. Therefore if the area of a sensor is known then the resultant
power in watts can be easily calculated. The nominal areas of the
sensors in common use are listed in table 3.
| Sensor size |
2/3" |
1/2" |
1/3" |
| Nominal area, mm2 |
68 |
36 |
17 |
Table 3, areas of sensors
The power produced by each individual pixel in the sensor is
directly proportional to its area. If three cameras are considered each
with the same resolution of say 500 lines then the number of pixels on
each sensor must be the same. The result of this is that the pixels on
each smaller size of sensor must also be smaller. Therefore the power
produced will be less for the same aperture setting, i.e. the same
amount of light energy.
If for the sake of an example a light source of 1,000
milliwatts per square Metre is passed to the sensors via an f1.8
aperture lens. The amount of light passed by the f1.8 lens will be 7.5%
= 75 mW/M2 . From this the power output can be calculated for each
sensor. This will be the power multiplied by the area of the sensor.
The result is shown in table 4.
| Sensor size |
2/3" |
1/2" |
1/3" |
| Sensor area |
68mm2 |
36 mm2 |
17 mm2 |
| Aperture |
f1.8 |
f1.8 |
f1.8 |
| Power output |
5.1mW |
2.7mW |
1.275 mW |
Table 4 power output of sensors
Therefore the 1/2" and 1/3" sensors will be producing
insufficient power for a full video signal. The answer is to use a lens
with a larger aperture for these sensors so that more energy is passed
to maintain the output power. This is summarised in table 5.
| Sensor size |
2/3" |
1/2" |
1/3" |
| Sensor area |
68mm2 |
36 mm2 |
17 mm2 |
| Aperture |
f1.8 |
f1.2 |
f0.85 |
| % light passed |
7.5% |
15% |
30% |
| Power output |
5.1mW |
5.1mW |
5.1mW |
Table 5 power output corrected by lens f stop
This is the reason that many 1/3" cameras have the sensitivity
specified with an f1.0 or sometimes an f 0.9 aperture. Beware though,
there are only a limited number of lenses made to the 1/3" format. If
the longer focal length lenses must be used they usually have smaller
apertures (higher f numbers) and pass less light energy.
The contra to this argument is that if a sensor of one size has
the same size pixels as a larger one, then the light required will be
the same. However the total number of pixels will be fewer and the
resulting resolution will be proportionally less. There is no such
thing as a free lunch!
Sensor and lens format
The range of lenses available for the 1/2" and particularly the 1/3"
cameras is limited. The largest range of lenses are still 2/3" and 1"
format. It is alright to use a larger format lens on a smaller format
camera but there is another penalty to pay. A 2/3" lens will focus the
scene on the equivalent area around a 1/2" sensor but only the light
energy relative to the area of the sensor will be converted to power.
See diagram 15.
The area of a 1/2" sensor is 53% that of a 2/3" sensor,
therefore only this proportion of the light energy will be converted to
output power and thus a video signal. This would make an f1.4 lens
equivalent to about f2.0. The area of a 1/3" sensor is only 25% that of
a 2/3" which is equivalent to an f2.8 lens! So this is yet another
correction factor that needs to be applied.
Transmission ratio
As stated, the f number is simply a ratio and although it is a
measure of the amount of light that a lens will pass it is not a
measure of the efficiency of the lens. The efficiency of a lens is
measured by its transmission ratio referred to as the 'T' number. A
lens, as stated earlier, is several different glass elements arranged
to provide the correct projection of the image being viewed on to the
sensor of the camera. Every time light is transmitted through a
standard glass to air surface some is lost. Lens designers and
manufacturers use a range of techniques to reduce this factor. The
amount of light that is additionally lost depends on the glass
materials, the thickness and curves of the glass and the coating of
each lens element. Light may be lost by absorption and reflectance in
passing through a lens. It can also be lost by internal reflections.
The mathematics are quite complicated to calculate. The
Transmission Ratio is calculated from a factor known as the
Transparency Ratio. Suffice to say that the effect is to adjust the
effective aperture to a value depending on the quality of the lens. For
example a high quality f1.4 lens with a Transparency Ratio of 0.85
would have an effective aperture of f1.58. A lower quality lens may
have a Transparency Ratio of 0.6 and would produce an effective
aperture of f1.8. This represents a significant loss of available light
to the sensor.
Other lens factors
There are several other factors in lens design that will affect the final picture seen on the monitor.
Flare, caused by some light being lost due to internal reflections.
Astigmatism, caused by the curvature of the lens being different in the vertical and horizontal planes.
Spherical Aberration, Where the surface of a convex lens in not perfectly spherical and all the rays of light do not focus at the same point.
Barrel Distortion, Where the rays from the outer areas of
the lens do not focus at the same point as those from the centre,
usually exaggerated in very wide angle lenses.
Chromatic Aberration, where different frequencies of light refract at different angles causing blurred images.
A high quality lens will minimise these effects, a low quality
lens will exaggerate them. The difference in cost between a high
quality lens and a cheap one may have be a small percentage of a tender
price but it could have a dramatic affect on performance.
Focusing with infrared light
The distance to an object in focus will be different under infrared
light compared to natural light. This is due to the different angle of
refraction and the fact that a lens compensates for refraction mainly
in the visible part of the spectrum. Therefore focusing of cameras
illuminated by infrared light requires that they must be set up and
back focused at night under the infrared light. The focusing is also
more critical under infrared light because the aperture will be at its
maximum with resulting decrease in depth of field. This will generally
be compensated for in daylight with smaller apertures. Another factor
is that the angle of the lamp must be accurately adjusted to the angle
of the camera, just a couple of degrees difference can lose a large
amount of the available light energy.
There are lenses available that will focus visible light and
infrared light at the same focal point, however the price penalty (or
benefit!) is a factor of about four times the cost.
Reflectance
The camera only sees the light energy reflected from the subject, so
this factor must also be considered. Some typical values for
reflectance are given as follows.
- Matt white test card 89%
- Snowy scene 85%
- Glass windows and walls 70%
- White matt paint on concrete 60%
- Unpainted concrete, car park 40%
- Red bricks 35%
- Open country, trees, grass 20% ( This can be 60%-70% for I/R)
- Empty asphalt area 5%
It is interesting to note that photographic cameras with built-in
light meters use a reflectance factor of 18% in the calibration of the
meter.
What about the glass in a camera housing? Try a simple
experiment, Go to the office window and measure the light from the
inside, then measure it from the outside. The loss can be in the order
of 25%-35%, so add this into the equation and see what happens!!
In conclusion a typical example may be as follows using the real world in which we all live.
Supposing a system is designed to provide 60mW/M2 of infrared power, what on earth does that equate to in plain English?
In very approximate terms it can be assumed that 270 Lux equates to 1 Watt/M2 light power over a wide spectrum.
Therefore 60 mW equates to a nominal 16 Lux if it were over the visible part of the spectrum.
This sounds like a lot of light power for a camera with the following specification:
Sensitivity: 0.1 lux for usable video at f1.0 (90% reflectance)
However if the camera specification is examined more closely some surprises transpire.
- 0.1 Lux is at photopic wavelength.
- This refers to a target reflectance of 90%.
- This usually relates to 50% video or ‘usable video’
- Using a non-standard lens with an aperture of f 1.0.
Now come back to planet earth and see how the camera performs at 730 nm!!!
- The camera has about 23% sensitivity at 730 nm.
- It is required to see targets with a reflectance of 10%.
- A full video signal of 0.7 volts is required, not 50%.
- The lens actually fitted will be f1.4.
The infrared power required will actually be the equivalent of 16 Lux! visible light. What a coincidence.
Add just one more factor, supposing that the area from 750 nm
under the lens spectrum curve is 50% of the total. The infrared power
now required will be:
The equivalent Of 32 Lux! of visible light
Is the camera in a housing? Oh dear this could lose another 25%. So now the light power needed is::
The equivalent of 40 Lux! of visible light
It is probably best to forget about a zoom lens where the
effective aperture may reduce at maximum zoom, or that an f1.4 lens may
not be available for the focal length required!
When a system is required to operate under infrared light all the
components are being pushed to their limit of functionality. With all
the factors to be considered it is amazing that any systems function at
all, and of course many do not. More attention given to the many
factors discussed in this article will improve the chances of achieving
a system that will provide optimum performance. These comments equally
apply to any system operating under low light conditions without
infrared light where components are operating towards the limit of
their specification.
The emphasis in system design for low light conditions must be to
assess the products carefully, ask questions and do not shy away from
quality products for the sake of a few percent on the price.
So, how far can your light shine? The answer is, of course FOREVER.
But, how far can your CAMERA see?
See also: Light Transmission Through Lenses
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