MATHEMATICS IN PHOTOGRAPHY 175

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MATHEMATICS IN ELEMENTARY PHOTOGRAPHY

ROBERT A. ATKINSBay Ridge High School, Brooklyn 20, N. Y.

It is not often that a teacher is privileged to teach a course inphotography since it is not often included in the high school curricu-lum, nor is it easy to get a teacher with the proper training. However,photography offers a very good medium for instruction in certainprinciples of chemistry, and a marvelous opportunity for creativeteaching of topics in physics and mathematics. It is the purpose ofthis paper to show some examples of how photography makes useof applied mathematics.

I. Box CAMERA MATHEMATICS

The student who shoots a roll of film with his box camera, takesthe exposed roll to the corner drugstore, and gets the prints back aday or two later is a good subject for a lesson in "cost accounting."A roll of film usually contains 8 pictures. Let us assume they all comeout. The problem: how much do you pay per picture?

Discussion: A roll of 120 film costs about $.46. Developing is usually$.20 with a charge of $.07 per print. The total cost is then $.46+.20+8(.07) =$1.22. The cost per print is $1.22�8 or about $.15.

Problem: Devise a formula for the cost of developing and printnigany roll of film.

Solution:

C=P+.20+.07iV

where

C= total cost

P== price of film

N== number of prints

Problem: Draw a graph which relates the cost of purchase, develop-ing, and printing a roll of 120 film.

Solution: See Figure 1. Incidentally, it should be noted that thestraight line connecting the points is actually meaningless (sinceyou cannot have a fraction of a print) but it does serve to show thelinearity of the relationship. The same applies to Figure 2.

Problem: Devise a formula which gives the cost per print. Graph

176 SCHOOL SCIENCE AND MATHEMATICS

this formula.Solution:

P+.20+.WNC=�������

N

See Figure 2.The foregoing examples can be varied in many ways and extended

to problems in costs of paper and chemicals, and discount problems oncameras, enlargers, and other equipment. An interesting study mightbe made on the amount of depreciation of new equipment whentraded in or bought used. Information for these problems can beobtained from catalogs, photo magazines, or the local camera store.

II. ELEMENTARY PHOTOGRAPHIC OPTICSFor a given level of illumination the amount of light entering a

camera depends on the area of the lens opening. The amount of lightfalling on the film (which is actually responsible for the image for-mation) is related directly to the lens opening and inversely to thedistance from lens to film. In the interests of uniformity the f-numbersystem of lens marking has been adopted to relate the factors ofamount of light falling on the film, lens opening, and lens-to-filmdistance.

lens-to-film distancef-number=��������������� �

diameter of lens opening

The lens-to-film distance, which is nearly constant for most cameras,is for most photographic work the same as the focal length of thelens. In all but the simplest cameras the diameter of the lens openingcan be varied by means of an iris diaphragm. Since the focal lengthof the lens is fixed the f-number is inversely proportional to the diame-ter, a small lens opening being indicated by a large number. Forexample, f/8 admits twice as much light as f/11.The calculation of f-numbers is an interesting exercise in squares,

square roots, and elementary geometry. It was noted above that f/8gives double the light transmission of f/11, yet 8 and 11 are neitherhalves nor doubles of each other. Let us use the formula to find theactual diameters of the openings (assume any focal length�say 5inches).

s" yf/8=� f/ll=��

D D’5 5

D=�=.631/ D’^�^AS^8 11

MATHEMATICS IN PHOTOGRAPHY 177

But D is not double Df. How then can we say we have doubledthe light transmission? The answer lies in the fact that the lighttransmission depends on the area of the lens opening, not its diameter.If we use the theorem that areas of circles are proportional to thesquares of the diameters:

A .63 .40 2

A’~ AS^~20~~1 ’

We see, then, that one area is actually double the other. If one f-number of a lens is known, the others can be calculated by simplysquaring the f-number, halving or doubling it, and taking the squareroot; or else by multiplying or dividing the f-number by the squareroot of 2 or 1.415.

III. PROJECTION SYSTEMS

While the design of projection equipment is a matter for experts,the basic theory of image projection is relatively simple. A beam oflight is passed through a transparent film or slide, then through alens and onto a screen. The basic lens formula

1 1 1

Focal length Object distance Image distance

applies in this case, as it does also to camera lenses. In practice acondensing system is added to concentrate the light and the lens-to-film distance may be varied to allow for various projection distances(or "throws’7). As the projector is moved closer to the screen theimage size will of course decrease, and the lens-to-film distance in-creases. This can be shown by the above formula. Since the focallength is constant, if the image distance decreases, the object distancemust increase.We can compute the size of a projected image by similar triangles.

In essence, a projection system resembles Figure 3. Note that theObject (film or slide), Image, and the light rays form a pair of similartriangles. Therefore,

Object Image

Object distance Image distance

For example, in a 15 foot living room a 35 mm. slide (film dimen-sions about 1 by 1^ inches) projected with a 4 inch projection lenswill give an image about 45 by 67 inches.

IV. FLASH PHOTOGRAPHY

When a subject is close to a flashbulb it is illuminated by an intense

178 SCHOOL SCIENCE AND MATHEMATICS

light as the bulb is fired. As the distance from flashbulb to subjectincreases, the light intensity becomes less. The lens must be openedor closed to compensate for these differences. To calculate the properlens opening we use a guide number which can be obtained from thetable which appears on each package of flashbulbs. This guide num-ber depends on the type of flashbulb and film and to some extent bythe shape and finish of the flashgun reflector. For example, if a num-ber 5 flashbulb is used with ordinary black-and-white film, the guidenumber is about 160. The lens setting is obtained by dividing theguide number by the distance from flashbulb to subject. The quotientis the lens setting.

Example:

for 10 feet, 160-10=f/16for 20 feet, 160-20=f/8.

V. DARKROOM MATHEMATICS

If students are given an opportunity to work in the darkroom, therange of uses of mathematics is extended.

Problem: How long should a given type of film be developed, giventhe developer and its temperature?

Solution: This will involve reading a time-temperature develop-ment graph such as the one in Figure 4. A series of these graphs formany types of film is given in the Eastman Kodak Data Book onFilms.Of course many problems can be devised on dilution of solutions,

conversion of units from metric to English, making only half (or one-fourth) a given formula, etc.

VI. SUMMARYIt is hoped that the examples given in this discussion will stimulate

teachers of photography to include some mathematics in their courses,since teaching mathematics with photography has a double ad-vantage: it is easily motivated, and the subject is recalled wheneverthe camera is used.The author wishes to thank the Sales Service Division, Eastman

Kodak Company, for their help in the preparation of this article.

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