CCTV: lens is more.
This article examines fundamental lens characteristics and their effect on video system Performance. A basic knowledge of how lenses work makes manufacturers' specifications more meaningful, provides a better foundation for lens and camera selection, and makes system planning more productive.
Focal length. The focal length is the distance from the principal point of the lens to its focal point. The principal point is determined by the number of elements in the lens and how each element is ground. The focal point is the point at which all the light passing through the lens converges. That point is beyond the image plane.
Lens format. The format of a lens is the largest area on which a lens can accurately focus the light passing through it. This plane is always a fixed distance from the principal point and is specified by the lens manufacturer.
In CCTV applications the three standard formats are 1 in., 2/3 in., and 1/2 in. The relationship between the formats is linear. For example, the horizontal and vertical dimensions of a 1/2in. format lens are half the dimensions of a 1-in. format lens. The vertical dimension of each format is always 75 percent of the horizontal dimension.
A smaller format lens does not work on a larger format camera because the lens is designed to focus light on an image plane smaller than the camera's imaging device. Larger format lenses on smaller camera formats produce smaller fields of view because the camera's imaging device is smaller than the maximum image plane of the lens.
Sweet spot. The lens's center is the area of least distortion. That highly focused area is called the sweet spot. To estimate the sweet spot, draw a circle whose diameter is equal to the vertical height of the maximum image plane of the lens and superimpose it over the camera's imaging device area.
For example, for the sharpest image possible across the width of the image area on a 2/3-in. camera, the field of view requires an 8-mm lens. A 1-in. format lens's sweet spot covers almost all of a 2/3-in. camera's imaging device and all the imaging device of a 1/2-in. format camera. Using a 1-in. format 8mm lens on the camera can reduce image distortion around the edges.
A word of caution: Lenses from different manufacturers have different resolving abilities. In CCTV applications resolving ability is generally measured in lines resolved in one horizontal TV line. A horizontal TV line is the vertical dimension of the image format divided by 525 (the number of horizontal scan lines in one video frame).
Another important point when converting image formats concerns magnification or telephoto applications. A lens of any given focal length creates no greater telephoto effect for a 1/2-in. camera than it does for a I -in. camera. Thus an application that required a 200mm lens on a 1-in. or 2/-in. camera to acquire the detail needed will still require a 200-mm lens on a 1/2-in. camera to provide the same quality of detail. The difference between the three formats is the field of view-the smaller the format, the narrower the view.
Back focus. Back focusing is the physical adjustment of the camera's imaging device to match the lens's image plane. The more accurately the camera's imaging device is positioned relative tO the lens's image plane, the clearer and sharper the image created by the camera will be.
Every camera needs to be adjusted every time a lens is installed on it. For optimal performance, back focus should be checked even when a lens with identical specifications and the same manufacturer is installed.
Aperture. The aperture or f-number, also called f-stop, is equal to the distance between the lens and the focal point divided by the effective diameter of the lens. However, that equation does not reflect the actual quantity of light or the quality of the image focused on the camera's imaging device.
You could knock the end out of a soda bottle, grind it down with a coarse grinder to a 16-mm diameter, and shape it to focus most of the light 16 mm behind it. The result would be a 16mm f/1.0 lens. However, it would not work well. The f-number simply is not a valid measure of lens quality.
Iris. The iris mechanically adjusts the f-stop setting of the lens. Positioned between the lens and the imaging device, the iris acts as a throttle to control the amount of light that travels from the back of the lens to the camera; it cannot pass more light than can travel through the lens.
If the lens elements could pass all the light presented to them, the f-number would reflect the ratio of light energy passing through the lens to the light energy that arrives at the image plane. The equation is I divided by the f-number squared. For example, for an aperture of f/2, the fraction of light that reaches the image plane equals 1 divided by 2[sup.2] or .25.
Some of the more common f-number values are as follows: light energy f-number passed 1.0 1 1.4 1/2 2.0 1/4 2.8 1/8 4.0 1/16 5.6 1/32 8.0 1/64 11.0 1/128 16.0 1/256 22.0 1/512
Generally, the larger the f-number a lens can be set to, the better the picture. This is so because light passing through the center of the lens travels in a less refracted path and is less subject to distortion. As the iris closes (a larger f-number), more of the light reflected from an object to the image plane is passing through the center of the lens.
An f/1.4 lens set at f/5.6, assuming adequate light, produces a more desirable image than it does at f/1.4. More of the image area is in focus, and it is easier to see the relationship between objects near and far from the camera.
Lens efficiency. No lens is completely transparent. The more efficiently a lens passes light, the higher its quality.
The R (resistance) factor is a measure of the light-passing efficiency of a lens. R equals the amount of light that reaches the camera from a known source divided by the amount of light that would have reached the camera through a perfectly transparent lens of the same f-number.
The R factor is key to determining lens aperture setting for a given camera and light level. It therefore directly influences the overall quality of the images produced by a system. For general purpose CCTV lenses, the closer the R factor is to 1.0, the more desirable the lens.
The R factor of a higher quality lens may be 0.85, while a lower quality lens may have an R factor of 0. 60. The difference between those two values is significant. (The former is more than 40 percent better than the latter). The results in image quality are equally significant. Transmittance. Transmittance (T) is a measure of the total light attenuation of a lens. For any f-number, T equals [sq.root (f[sup.2] div. by R).
A comparison of two f/1.4 lenses with different R factors, 0. 85 and 0. 60, shows the relationship between the f-number and the R factor.
The value of f/1.4 2 iS 1.96. That is the minimum light attenuation value for all f/1.4 lenses. The square root of (1.96
R) gives the T values: The .85 R lens has a T value of 1.5, and the .60 R lens has a T value of 1.8.
Dividing I by T 2 gives the percentage of light reflected from the scene that actually reaches the camera. With the first lens, 44 percent of the light reaches the camera. With the second lens, only 31 percent of the light makes it through. Thus the first lens is 42 percent more efficient than the second lens and requires 42 percent less scene illumination to provide the same illumination to the camera.
Image distortion. Light does not travel uniformly through a lens. To reduce image distortion, most quality camera optics use several lens elements.
Distortion is most evident outside a circle with a diameter equal to the height of the image area. With the expanding use of virtually distortion-free solid-state imaging devices in cameras, lens distortion is more apparent than ever before.
Color cameras and distortion. A prism causes chromatic distortion that is enjoyable to look at; however, that same distortion occurs naturally in lenses, with a less desirable effect. In color cameras such distortion reduces image quality and causes unreliable color reproduction.
The only defense against chromatic distortion is color-corrected lenses. However, not all color-corrected lenses are equal, and the difference between manufacturers can be significant.
GIVEN THESE OBSERVATIONS, WHAT is the right combination of lens and camera for a given scene? That depends on the light that travels from the scene through the lens to the camera's imaging device. Scene reflectance and maximum and minimum light levels must be examined.
Not all surfaces reflect the same way. A blacktop parking lot may reflect 5 percent of the light that reaches it, while a concrete lot may reflect 70 percent. A warehouse whose walls and ceilings are painted white may reflect twice the light of one not painted.
Light is measured in lumens per square foot, or footcandles (fc). The approximate range of natural light is from 100,000 fc (looking at the sun) to 0.0001 fc overcast night).
To put all the factors together, what camera would be needed for an indoor scene with 50 fc of scene illumination, 80 percent scene reflectance, and a 16mm f/1.2 lens with an R factor of 0.8?
The light footcandles) times the scene reflectance times the R factor equals 50 x 0.8 x 0.8, which equals 32 fc. Looking at a camera's full-video illumination requirements at the imaging device, not the minimum illumination, shows that a vidicon camera provides full video when 0.42 fc re aging device.
The most light that will reach the image plane at f/1.0 is 32 fc, and the camera requires only 0.42 fc to create a good picture. To choose an f-stop for the lens, divide 32 by 0.42 (which equals 76.19) and take the square root of the quotient to find the f-number of the lens (f/8.7). This should be a great picture ! Turning the equation around, you can estimate the performance of a camera for a particular application by using the camera specification sheet. First, never use minimum-video specifications when estimating scene applications-use the full-video specification. Second, multiply the full-video specification by the light attenuation value of the lens's f-stop. Remember that an f/1.2 lens is really designed to operate at approximately f/4.0. Third, divide the result of step two by the R factor. To be safe, if you don't know that factor, divide by 50 percent of the R factor of a similar lens to get the amount of light the camera needs at the face of lens. Fourth, divide the result of step three by the scene reflectance of the application to get the actual scene illumination needed for a good image at the camera's output. good
A typical, general-purpose, solid-state camera might require 0.42 fc at the imaging device. How much light must be at the scene to make a good image?
Multiply 0.42 fc by 16. (The attenuation value of f/4 = 4 x 4 = 16). The product is 6.72 fc, which is the light at the face plate with a lens setting of f/4.
Next divide 6.72 by the R factor of the lens (0. 8). The result is 8.4 fc, which is the amount of light reflected from the scene.
Then divide 8.4 fc by the scene reflectance (.40). The answer is 21 fc, which is the scene illumination needed for a good picture at the camera's output.
Using the same equation for the same lens but with a setting of f/1.2 gives a scene illumination requirement of 1. 89 fc. Had a lens with an R factor of 0.6 at f/1.2 been used, the scene illumination required would be 5.8 fc ! A different example would be a blacktop parking lot. Here we have to calculate the two extremes of lighting: an overcast night with 5 fc of artificial light and 5 percent reflectance, and a clear winter day with snow covering the lot, 100,000 fc of light, and 85 percent reflectance.
The minimum light conditions would equal 5 fc times 0.05 reflectance, which equals 0.25 fc. When that figure is multiplied by the R factor (0.8), the illumination at the camera's imaging device is found to be .2 fc.
A look at specifications shows that one camera needs 0.42 fc for full video, making it unsuitable for the application. Another camera requires only 0.023 fc. Thus the second camera is sensitive enough to handle the low end at an f-stop of sq.rt (0.2 div.by 0.023), or f/2.95.
The maximum light conditions would be 100,000 fc times 0. 85, which equal s 85,000 fc. But here's the tricky part-the maximum attenuation of light by an autoiris lens may depend on transmittance, not the f-number.
Transmittance is the total passing of light through the lens to the image plane, and it includes the R factor. Thus 85,000 fc divided by the T value squared gives the maximum light attenuation capacity of the lens.
Given a T value of 360, the transmittance equals 85,000 fc divided by 36[sup.2] which equals 0.6559 fc at the image plane. That amount is 28.52 times the light the camera needs for full video.
Here a problem arises. The lens does not have enough range to control the light energy over the full spectrum of scene conditions.
The following questions help define the dimensions of the problem:
* How important is the view?
* Exactly what must be viewed?
* Who needs to see it?
* What kind of response is required?
* Does it need to be recorded? What is the budget?
The following are some options:
* Look for a different camera and lens.
* Add an external, servo, neutral-density filter. (Examine its cost, and make sure it will work in the dead of winter.)
* Use two cameras-one for low light and one for extreme bright light.
* Increase the lighting to reduce the sensitivity needed in the camera and add a fixed neutral-density filter.
* Paint the lot white.
* Take what you can get.
The solution to this last problem is less important than understanding how to estimate the dynamic range of an autoiris lens operating in a wide light range. Clearly, you need a basic understanding of the mechanics of video equipment to evaluate and communicate your needs.
Unfortunately, for persons with the problem described, the most common answer is the last one. They don't know how to estimate their equipment needs, and they assume the lowest bidder knows what it's doing.
There is no short cut to quality in the manufacture of camera lenses. High-quality camera optics may cost more, but with good planning the benefits are well worth the difference.
Though important, lenses are only a part of a video system. By themselves they do not ensure good system performance. Adequate system planning requires attention to the details and mechanics of cameras, connectors, cables, power sources, signal handling equipment, monitors, recorders, and recording tape.
The time and effort you invest in system planning and design, along with the quality of the system components, will be apparent throughout the life of your system.
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|Title Annotation:||closed circuit television|
|Date:||May 1, 1990|
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