Measurement Of Canon Ef 1.4x Tele Extender Performance

[Photo Dad]

My goal was to extend the focal length of my Canon EF 70-200mm f/4L USM zoom lens. The Canon EF 1.4x Tele Extender appeared the right tool for this, but I wanted to see for myself so I set up tests in my back yard and compared focal lengths with and without the extender for both the EF 70-200mm USM f/4L zoom and the EF 300mm f/4L IS USM lens. (Please let me know if this is of interest to anybody. There should be a drawing attached to this file. I hope it is there.)

I focused the camera without the extender at a target of vertical lines spaced 1/2 inch apart. I measured the distance from the camera to the target and counted the number of lines focused in the viewfinder within the width of the center spot metering circle. Then I mounted the extender and moved the camera back until the same number of lines were focused within the width of the metering circle and measured that distance. I used the formula f2/f1 = q2/q1 where f2 is the focal length with the extender, f1 is the focal length without the extender, q2 is the distance between the camera and the target with the extender and q1 is the distance without the extender. If you’re interested in how I derived this formula see the addendum below. Here are the results:

1.4x Tele Extender with 70-200mm f/4L zoom
Extension at 70mm: 1.34x
Extension at 200mm: 1.50x
Average Extension: 1.42x

1.4x Tele Extender with 300mm f/4L IS: 1.36x

Since Canon doesn’t claim two decimal accuracy the Tele Extender on average meets its specifications with both lenses, although only marginally for the 300mm lens. Unfortunately I was most interested in extending the focal length of the zoom lens as far past its 200mm maximum as possible, but in exactly this region it had the worst performance.

ADDENDUM: DERIVATION OF FORMULA

Refer to the diagram below. Although camera lenses are complicated assemblies of several glass elements, each lens is shown as a single glass element at one position without the extender (f = f1) and at another position with the extender (f = f2). In other words I ignored the length of the lenses (about 8 inches). The distance from the camera to the object was 15 feet to 20 feet. If you do the calculations with and without subtracting the distance the lens stuck out in front of the camera body (I measured the distance from the vertical lines to the camera body), you will find the difference in your results will be negligible. To make my measurements I focused the camera on a series of vertical lines without the extender. Then I attached the extender and moved the camera until the lines appeared the same distance apart in the viewfinder. In other words I place the camera with the extender at a point where the image is the same size as it is at the other point without the extender. The thin lines in the diagram represent light rays diverging from the object (the vertical lines) and converging through the lens to the image plane. The arrows represent the objects and their images. Notice that images for the lens without the extender and the lens with the extender are the same size. I call it w. If you’re not familiar with ray diagrams you need to know that rays of light parallel to the axis of a lens converge to a point on the axis at the focal distance from the lens and that rays through the center of the lens are not bent. I use these facts to determine the location of the images in the diagram. These lines also form sides of the triangles I use for the calculations. In the diagram the lens is focused on an object at a point q1 from the lens without the extender and at a point q2 from the lens with the extender, and the image is located at a point p1 from the lens without extender and at a point p2 from the lens with extender. You can learn in optics courses that object and image locations are related through the formula

(1) 1/q1 + 1/p1 = 1/f1 and 1/q2 + 1/p2 = 1/f2.

Now some calculations are necessary. I used the relations between similar triangles in the diagram to calculate the ratios

(2) w/a = x/f1 and w/b = x/f2.

so that by eliminating x I get

(3) b/a = f2/f1.

Now I notice from the diagram that

(4) p1 = f1 + a and p2 = f2 + b.

Substituting these formulas for p1 and p2 into equation (1) and doing some algebra yields:

(5) (f2/f1)2 = (b/a)[(q2 – f2)/(q1 – f1)].

and using equation (3) to eliminate b/a from equation (5) gives

(6) f2/f1 =(q2 – f2)/(q1 – f1).

In practice the distance from the camera lens to the object can always be made much larger than the focal length so that the focal lengths can be neglected (simply discarded) in these formulas allowing the simplification

(7) f2/f1 = q2/q1.

This is the formula I used to obtain my results.

Here is the diagram.