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--------------------------------------------------------- By Bill Claff

Recall that for light throughput
we confirmed conservation of light between the object and image sides of the
lens and that the solid angle incorporates the half-angle of view.

Etendue can be expressed as:

G
= (π / 2* (Y / ef#))^{2} Where ef# is the effective f#.

And buried in ef# is our usual (1 + m / p) factor; which we will see again shortly.

Close-up photographers often
speak of light "loss" and explain it in terms of the inverse-square
law and the increased distance of the exit pupil from the image plane when
focusing closer than infinity.

We can confirm this behavior by inspecting the image circle of a bare lens.

I prefer to think of the optical
system as constrained by Y, the actual image height; and that the image circle
is defined by this constrained image plane.

So rather than light loss I think of it
in terms of less light gathered. This doesn't change the outcome, it's just a
different point of view.

In this regard, it's helpful to revisit the formula for the half-angle of view (which is a function in part of Y):

tan(w_{O}) = (Y / f) * / (1 + m / p)

Note that the angle goes down as
m is increased;
and that the angle goes up as f is decreased.

Viewed this way it is easier to understand how a close-up lens might lose less
light than extension tubes due to the drop in focal length which somewhat
offsets the increase in magnification.

In fact, perhaps no light is lost, or even, light is gained.

When a close-up lens is attached the new focal length is determined by the power formula:

P
= P_{M} + P_{C} - d * P_{M} * P_{C}

where P is power in diopters,
subscript M indicates main
lens, subscript C indicates
close-up lens, and d is the
distance between principal planes in meters.

Although it is seldom true we will assume that d is zero. This is a very common assumption which is rarely stated.

Since P = 1000mm / f we can convert from P = P_{M} + P_{C}
to focal lengths:

f
= f_{M} * f_{C} / (f_{M} + f_{C})

Let's call the ratio of the
focal length term R_{f}:

R_{f}
= f / f_{M} = **(**f_{M} * f_{C} / (f_{M} +
f_{C})**)** / f_{M} = f_{C} / (f_{M} + f_{C})

Our assumption implies that the rear principal plane doesn't move resulting magnification is:

m
= (1 + m_{M}) * f_{M} /
f - 1

Now, we have a bit of algebra
ahead of us. Let's first substitute for f,
eliminate f_{M}, and
substitute R_{f}:

m
= (1 + m_{M}) * f_{M} * (f_{M} + f_{C})/ (f_{M}
* f_{C}) - 1 = (1 + m_{M}) * (f_{M} + f_{C})/ f_{C}
- 1 = (1 + m_{M}) / R_{f} - 1

The magnification term in the half-angle formula is:

(1 + m / p) = (p + m) / p

Let's call the ratio of the
magnification term R_{m}:

R_{m} = **(**(p + (1
+ m_{M}) / R_{f} - 1) / p**)** / **(**(p
+ m) / p**)** = (p + (1 + m_{M}) / R_{f} - 1) / (p + m_{M})

Let's call the product of R_{f} and R_{m} the light
throughput ratio, R_{g}:

R_{g}
= R_{f} * R_{m}

So we substitute for R_{m} and
simplify:

R_{g}
= R_{f} * R_{m} = R_{f}
* (p + (1 + m_{M}) / R_{f} - 1) / (p + m_{M}) = (R_{f}
* p + (1 + m_{M}) - R_{f}) / (p + m_{M})

To understand whether light is
lost or gained we want to know how R_{g} compares to 1:

1
≟ R_{g }where ≟ indicates an as yet unknown
relationship

We insert our formula for R_{g} and
cross-multiply:

(p + m_{M}) ≟
R_{f} * p
+ (1 + m_{M}) - R_{f}
= R_{f} * (p - 1) + (1 + m_{M})

And then eliminate m_{M} from each side:

p ≟ R_{f} * (p - 1) + 1

It's not surprising that initial
magnification falls out and that as a sanity check when R_{f} is 1 we get p = p

Finally, rearrange with p on the left hand side:

p
* (1 - R_{f})
≟ (1 - R_{f})

and (whew):

p ≟ 1

Remember that we made an
assumption about the close-up lens mating perfectly with the main lens.

With that caveat we have determined that no light is lost if the main lens p (pupil magnification) is unity,
light is gained if p is >
1, and light is lost if p
< 1.

In practice close-up lenses are
normally used with lenses that have pupil magnifications less than unity so in
these circumstances light is in fact lost.

However, we have seen earlier that less light is lost than using extension to
get to the same magnification with the trade-off being less working distance
with a close-up lens.

For example, let's return to our formula:

R_{g}
= (R_{f} * p
+ (1 + m_{M}) - R_{f}) / (p + m_{M})

And for simplicity assume
infinity focus eg. m_{M} = 0

R_{g}
= R_{f} + (1
- R_{f} )/ p

If R_{f} = 0.5 we get 1x at infinity with a
close-up lens. And with a p
of about 0.80 we lose about 1/6 stop of light and even at a p of 0.66 we only lose
about 1/3 stop.

To achieve 1x with extension, using the usual (1
+ m / p) formula, the values are 7/6 stops and 4/3 stops respectively; 1
stop more light is lost than with the close-up lens.

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