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Super Macro Underwater Photography - The Definitive Guide, Part 5

(Part of The Definitive Guide To Underwater Super Macro Photography)
The following topics are meant to expand on, clarify, or justify some of the statements made in previous sections of this series. They provide “additional” information as opposed to “essential” information. In that sense, you can ignore all of the following and still be a good photographer.

The clearest explanations of why things work the way they do, especially in photography, are usually presented in the form of physics reasoning and mathematical equations. Such explanations can easily become overwhelming, so I will try to keep the math to a minimum and, when appropriate, will cite technical references that can give you more details.

5.1 Teleconverters “stealing” light: The Bellows effect

Photographers often blame their teleconverters and extension tubes for “stealing light”, since there can be noticeable ‘dimming’ seen through the viewfinder when these tools are installed. You may have also noticed the f-stop in your viewfinder increasing/decreasing as you zoom your lens in and out, and wondered to yourself… what’s happening?!

First, consider photographing a subject at 1:1 magnification. Each point on the subject reflects a particular number of photons per unit time toward the camera, through the lens, and then to a particular photosite. The total photon energy deposited per unit time in all of the photosites, divided by the area of the sensor, is the average light intensity at the sensor.

Now, consider installing a 2X teleconverter into the same system, without changing the lens-to-subject distance. The subject area mapped onto the sensor is now one quarter the area of the image taken at 1:1.
Bellows effect
Since the subject is still reflecting the same number of photons per unit area toward the sensor, the photons that were concentrated on ¼ of the sensor’s area are now spread over the ENTIRE area of the sensor! This results in a decrease in intensity by a factor of 4 (i.e. 2 f-stops), which is an example of the bellows effect.
Bellows effect

Here’s a handy formula that you can use to figure how many stops of light will be lost for any particular change in image magnification (when lens-to-subject distances are kept constant):
F-stops lost = 2.88 X ln(M2/M1)

M2 = final magnification ratio
M1 = initial magnification ratio

For example, going from 0.5:1  (M1 = 0.5) to 1.5:1 (M2 = 1.5) means M2/M1 = 3, and you lose 3.16 f-stops.

The same principles apply when a zoom lens is used between its extremes. Zooming in results in an increase in magnification, and therefore a decrease in intensity. Your camera senses this, and displays a higher f-stop to compensate for the “light loss”.

5.2 Light “Gain” by a Diopter?

This is not as crazy as it sounds.

When light reflects from a point on a subject it travels outwards in all directions. The intensity of this light decreases according to the inverse square law, i.e. doubling the distance quarters the intensity, halving the distance quadruples it. Therefore, the lens-to-subject distance is a crucial factor in determining exposure. Aside from the nature of the light source and the lens aperture, the only other important factor determining exposure is the magnification ratio (as discussed above).

The combined effect of these two factors is what determines the light intensity at the sensor. For the various super macro tools that have previously been discussed, the effect can be calculated, but it is too involved to include here. For details, see the Scuba Geek article.

For diopters, it turns out that the increase in intensity caused by the decrease in working distance is actually larger than the decrease in intensity due to the change in magnification. So, overall, the intensity at the sensor INCREASES… albeit slightly!

The corresponding analysis for extension tubes shows that there is a light loss, but not as much as for a teleconverters. In this respect, extension tubes fall between teleconverters and diopters.

5.3 Extensions Tubes: Minimum Working Distance and Magnification

The minimum focus distance with an extension tube is governed by the following equation:
minimum focus distance with an extension tube equation

minimum focus distance with an extension tube equation


f = focal length of primary lens
MWD = minimum working distance without extension tube
MWDET = minimum working distance with extension tube
M = initial magnification ratio at MWD
L = length of extension tube added

The magnification achievable with an extension tube installed is:
magnification achievable with an extension tube installed

A more detailed explanation of this phenomenon can be found here.


5.4 Diopter Power Loss in Water

The magnifying power of a lens in direct contact with water, MW, compared to its power when in contact with air, MA, is given by the equation:


nL = index of refraction of lens material
nM = index of refraction of the medium that the lens is in contact with (1.33 for water)
nA = index of refraction of air (1.00)

According to the above formula, as the index of refraction of the lens approaches that of the medium, the magnifying power of the lens in the medium approaches zero. So, if a magnifying lens has an index of refraction similar to that of water, it will essentially be useless as a magnifier when immersed in water. Even if the lens is made from optical glass (nL approximately 1.5), this formula shows that

In other words, the magnifying power in water is only 34% of the magnifying power in air! There is a loss of 66%.

However, note that if nM = nA in the formula, there is no loss in magnifying power. This can be achieved by keeping the lens in contact with air within its own “housing”. This is precisely the concept behind several of the most powerful underwater close-up lenses, including ReefNet’s SubSee Magnifier.

The derivation of this formula, and relevant background information can be found here

5.5 Diffraction
Because of the wave nature of light, whenever light from a point source passes through a circular opening, there is an interference pattern generated. When projected on a screen, it consists of a central bright spot, and alternating concentric light and dark rings. The rings are at least 50 times dimmer than the central spot.

Since photographic images are the result of focusing a countless number of point sources (points on the subject) on a film/sensor, the potential overlap of central discs from adjacent points can cause a blurring or softening effect. This is what people refer to as “diffraction effects”.

The degree to which diffraction is a problem depends on the size of the central discs (smaller discs will overlap less). Their diameter can be calculated by the formula:


D = diameter of central disc
f = lens’ f-stop
λ = wavelength of light

Note that increasing the f-stop will increase the diameter of the disc, thereby potentially degrading the image. It is also clear that longer wavelengths (red) will produce larger diffraction effects than shorter wavelengths (blue).

When doing super macro photography, it’s often tempting to use the highest possible f-stops to maximize depth of field. However, it should be apparent that the increase in depth of field comes at the cost of increased image degradation due to diffraction. The amount of diffraction effects that one might consider “objectionable” is very subjective. It depends entirely on how the image will be viewed and/or personal preference.

5.6 Depth of Field

When a lens is focused on a subject, the depth of field is the distance between the closest and farthest points in focus in the image. The figure below illustrates this concept.

Point A (the subject) is perfectly focused at the sensor plane. Light rays from a point B, in front of A, focus behind the sensor, intercept the sensor plane as a blurry circle (“circle of confusion”) of diameter “d”. There is also a point C, somewhere behind point A, that will be focused in front of the sensor plane, and will also form a circle of confusion of diameter “d” on the sensor. If “d” is small enough such that the circles do not span more than one of the sensor’s photosites, then all points between B and C will be in focus. This will occur when “d” is smaller than the separation between adjacent photosites. In the following, “d” represents the largest allowable value for a particular sensor/film.

Using basic optics theory, and simple math, the following formula for “geometric” depth of field can be derived:


f = lens f-stop (e.g. 5.6, 11, 22, etc.)
d = diameter of circle of confusion
M = magnification ratio
D = diameter of the lens aperture

For super macro photography (M>=1), the ratio d/DM is negligible, so the formula simplifies to:

Notice that there are only 3 factors that affect the depth of field: the lens f-stop, the magnification ratio, and the diameter of the circle of confusion. For a particular digital camera, the only factor not under your control is “d”, which is determined solely by your sensor’s resolution.

It is clear from the formula that depth of field will decrease as you increase the magnification ratio. This is why super macro photography (large values of M) involves extremely small depth of field. The formula also indicates that the only available corrective action is to increase the f-stop (decrease the size of the aperture).

Please note that this is a purely geometrical expression for depth of field. Observable depth of field depends on many other factors, including viewing perspective, eye resolution, resolution of the viewing medium, etc. Discussion of those aspects is beyond the scope of this article.


Kay Wilson
Apr 7, 2010 2:22 PM
Kay Wilson wrote:
Now I know why I should have tried harder with physics... if only my teacher had been a photographer!
Oct 14, 2012 5:14 AM
L W wrote:
I was wondering about the refractive index formula, should it not be 'relative'? That is of glass to water, which would be 1.5/1.33, which would be 1.127, which compared to 1.5, has a loss of around 73%.

From your formula, it means a +4 would become around +1.3, but according to some other formula, a +4 would become a +1

Oct 17, 2012 6:05 PM
L W wrote:
Ignore my prior comment. I see that the formula given is simplified. It combined the diopter loss of 75% and magnification gain from a flat lens port of 33%, giving it a net magnification loss of 66%.
Keri  Wilk
Oct 23, 2012 12:16 AM
Keri Wilk wrote:
Neither calculation (mine or my father's from scubageek.com) take into account the magnification from a mask or port, since they simply aren't relevant in a discussion about how water immersion affects a lens.
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