Is it t-stop or f-stop?

[Dj Joofa]

The recent debate between Glazarus and Peter Majtan prompted me to scribble down a few facts that I am sure are known to most people, but still for the record.

Glazarus: t-stop is a measure of transmission loss by the lenses.
Peter Majtan: t-stop > f-stop is a number that limits f-stop.

And, we shall see that, as I mentioned before, both are saying the same things, but in different words.

The exposure equation for a camera can get quite involved as given below:

E = I * B * F^2 * (cos theta)^4 * H * T * time / (4 * v^2 * f^2) ---------------------- (I)

where, the symbols have the usual meaning and I shall not go into them except those that interest us: H is transmission at off-axis points as limited by barrel vignetting, T is the transmission as limited by losses due to reflection and absorption by glass, F is the focal length, and f is the f-stop, time is of course time.

Equation (I) neglects one minor factor, which being the slight decrement of image brightness due to the fact that under some unusual conditions the illumination arising from an object point and incident upon the first surface of the photographic lens is not perfectly uniform.

The interesting thing to note here is the factor F^2 * T / f^2, lets drop the factor F^2, and concentrate on T / f^2. But, before that, I want to mention in passing that the factor (time / f^2) is an important one as it can be used as a measure of the exposure / exposure value.

Now coming back to the factor, T / f^2, it can also be derived in a simpler and a more instructive way. Lets assume that the light passing through the lens is proportional to area of the aperture, A,

A = pi * d^2 / 4, ------------------------------ (II)

where d is the diameter. Lets discount the constant factor pi / 4, and absorb that into the already existing proportionality constant, we can say that the light is proportional to d^2. Now, the f-stop, f, is

f = F / d,

and using equation (II), we see that light is proportional to F^2 / f^2, the same factor we saw in equation (I).

Okay, Glazarus says that t-stop is a measure of transmission loss, which ignoring H, would mean T < 1 in the equation (I). Now consider the factor again:

T / f^2 = 1 / (f^2 / T) = 1 / (f / R)^2,

where I have used R = sqrt (T), and of course if T < 1, then T < R < 1.

Consider the factor f / R, where R < 1, we see that f is multiplied by a factor > 1, (because R < 1, so 1 / R > 1), and then you may consider your t-stop as

t-stop = f-stop * (a number greater than 1),

which essentially boils down to what Peter Majtan was saying:

that t-stop > f-stop is a number that limits f-stop.

 

[Mike Prevette]

F stops assume an optically perfect system, and stand for (F)ractional stops. Ie the iris is a fraction of the total focal length.

(T)ransmission stops take into account the light loss caused by the glass in a system.

Use T-stops for exposure, and F-stops for calculating depth of field.
Don't confuse people with all the math.

 

[Dj Joofa]

Don't confuse people with all the math.

Mike, I sincerely hope that you understand that the intention is not to confuse people but to relate the two quantities t-stop and f-stop as far as exposure is concerned. I am sorry if all the detail has distressed you.

Thanks for the definitions -- we develop highly specialized HD cameras (that are not in (direct) public domain), which are a small part of our overall systems, so I guess you would let me some benefit of doubt.

However, I still apologize.

 

[Mike Prevette]

I didn't mean to complain. The detail of your math is great for me. I just think so many people get so confused over the true meaning and usefulness of the two metrics. I just want everyone to have a basic understanding before they get scared by calculations.

 

[Eryc Tramonn]

F stops assume an optically perfect system, and stand for (F)ractional stops. Ie the iris is a fraction of the total focal length.

(T)ransmission stops take into account the light loss caused by the glass in a system.

Use T-stops for exposure, and F-stops for calculating depth of field.
Don't confuse people with all the math.

Correct. Please allow the law of parsimony to apply. I already feel like film school was enough of a joke without some engineer/physicist confusing the hell out of me - with something I thought I had a handle on. :wacko:

Interesting, but how could that level of calculation be practical in a set environment? Real question. Maybe there is some practical application and I need to bone up on my particle physics.