TCs, DOF and Moose Peterson's Wildlife Photography

[marc schneider dc metro]

In Moose Peterson's Guide to Wildlife Photography, he states that with

a 1.4x TC you retain 40% of the depth of field of the effective f/stop.

 

I not exactly sure what he means by this and it doesn't make much

sense to me any way that I interpret this.

 

I would assume that you have the same DOF of the a lens 1.4 time

longer in focal length and one stop slower. Is this not true?

 

For example:

Using a 400mm lens at f/5.6 focused at 3.5m the DOF is 41.4 inches.

Put an 1.4x TC on this lens results in a 560mm lens at f/8 still

focused at 3.5m the DOF should be 29.9 inches. Is this not correct?

 

What would the DOF be according to Moose's rule? 23.7 inches? 12

inches? Something else? Moose's explaination just doesn't make

physical sense to me.

 

I know of other technical mistakes in this book regarding TCs (esp.

why a TC make a lens one stop slower, Moose claims it is due to more

glass, it is not), so another one wouldn't suprise me.

[marc schneider dc metro]

I just noticed that I calculated the depth of field for 22.6m not 3.5m focus distance. But the questions about Moose's book and DOF with TC remains.

[alex_lofquist]

I believe that the DOF would be 40% less if the image were taken at the same APERTURE SETTING (not relative aperture), and from the same distance. This would be based entirely on ths increased magnification of the image.

 

The important rule to which I refer is that DOF depends (almost) entirely on RELATIVE APERTURE and IMAGE MAGNIFICATION! The effect of focal length is miniscule.

[sergey_oboguev]

DOF is proportional to aperture (F-number) and inversly proportional to the square of magnification (i.e. size of object image in the film plane divided by the size of original object).

 

M = object-image-size / object-size

 

also M ~ lens-focal-distance / distance-to-object

 

DOF ~ F / M^2

 

If you put TC 1.4 on a lens and stay at the same distance from the object being photograhed:

 

magnification would increase by x1.4

 

whereas minimum possible F (assuming you shoot "wide open" in both cases) would increase by x1.4

 

and thus "wide open" DOF will decrease by x1.4.

 

That's the impact on DOF in absolute terms. "Apparent" DOF with respect to dimensions of framed view would decrease twofold.

[klix]

No, it doesn't make sense. My calcs show:

 

Total DOF of 560/5.6 is roughly 51% total DOF of 400/5.6

 

Total DOF of 560/8 is roughly 73% of total DOF of 400/5.6

[brian_kennedy]

Who knows what Moose Peterson was saying? Without seeing the book, it doesn't make a lot of sense and perhaps it was just a poor wording choice or a misunderstanding on his part. The point is, however, that you are thinking of it correctly, Marc. Add a 1.4x TC to any lens and the DOF is the same as a lens one stop slower and 1.4 times longer.