How to Select a Stock Lens

Let's say you are put in charge of selecting the lens for your optical application. If you are reading this, that's probably not too far from the truth.

O.k, so how to you go about selecting a stock lens ? You'll need no more than some basic numbers and that old trusty calculator (o.k., the calculator is optional).

There are several different requirements that must be examined (see Machine Vision Basics), however, we can approach the problem from the view of magnification which will lead to the focal length directly. The question of aberrational correction is beyond the scope of this discussion. If you have concerns that your lens be perfectly designed for your application, we encourage you to take a look at our custom design capabilities.

Two main driving requirements behind any lens/camera combination is that the system meet a basic resolution requirement (i.e. it be able to provide enough level of detail) and/or that an object of a certain size be simultaneously captured (field of view). Both of these requirements can be restated in terms of magnification.

You can calculate the necessary magnification via two different approaches. The first method is simply to take the ratio of the object/image size. So for a rectangular image, the longest dimension is taken (for example, a book 64 mm on a side). This is then compared to the camera plane size. A list of common camera planes is given below :

Format	1 inch	2/3 inch	1/2 inch	1/3 inch
Vertical	9.6 mm	6.6 mm	4.8 mm	3.6 mm
Horizontal	12.8 mm	8.8 mm	6.4 mm	4.8 mm

Let us assume that we are interested in a 1/2 inch CCD. Then the longest dimension is 6.4 mm so that the magnification is given by :

Magnification = Object size/Camera plane size = 64/6.4 = 10

Another method of calculating the magnification is to state it in terms of resolution. Let us assume that the smallest feature size that we require is 100 microns in our object. Further assume that our camera pixel size (the "pixels" or picture elements of your camera have a certain size which can be provided by the manufacturer) is 10 microns. We then require that our magnification be :

Magnification = Min. Resolvable Object Size / Pixel Size = 100/10 = 10

In this contrived example the two numbers worked out perfectly (hey we wrote this article), however often you will discover that they don't agree and will have to decide which is more important, resolution or overall object size. In other words a magnification of 10 allows us to "see" 64 mm. Ultimately, if we wanted to see say 640 mm then we would need a magnification of 100 instead of 10. This would decrease our minimum resolution from 100 microns to just 1 millimeter (.010 mm x 100). This basic physical tradeoff is called the "space bandwidth product" and is always something the optical designer has to balance.

So now that you have your magnification the focal length can be derived directly by manipulating the lens makers formula as

Focal length = Dtotal/A

where

A= M+1/M+2

where Dtotal is the total distance from the image plane of the camera (where the detector is) to the object plane and M is the magnification. For those optical types out there we have ignored the small distance between the nodal points for this calculation but it gives an excellent approximation.

So keeping with our example, let us assume that our object is 300 mm away then,

Focal length = 300/(10+0.1+2) = 24.75 = 25 mm

So you would then go to our lens catalog and look up a 25 mm lens with a 1/2" format. That could be for example our V2513 lens system.

Talk to one of our Application Specialists to assist you in selecting a stock lens.

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