Enlarger Optics: Condenser, Diffusion, Lens, Alignment and Falloff

You already own a working printer. The contact frame puts the negative against the paper, presses them together and shines an even light through both, and its uniformity is measured. Everything that follows on this page comes of one change: putting a lens between the negative and the paper, so that the picture can be bigger than the film.

That change buys the whole of enlarging and it costs four things. It costs light, because the same picture now covers more paper. It costs contrast in one direction and gains it in another, depending on how the negative is lit. It costs tolerances, because three planes now have to agree with each other instead of two surfaces being clamped together. And it opens the path to stray light, because there is now a volume of air, a bellows and a lens barrel between the source and the sheet. This page is those four things, in that order, so that the head you build next is designed against them and the commissioning measures them.

The optical path, and the two distances that set everything

Five things sit on the axis: a source, a condenser or a mixing chamber that turns that source into illumination of the negative, the negative, the lens, and the easel. The lens obeys the ordinary thin-lens relation, and every number an enlarger user cares about falls out of it.

1 ÷ u + 1 ÷ v = 1 ÷ f

The conjugates

u is the distance from the negative to the lens, v the distance from the lens to the paper, and f the lens’s focal length. The magnification is the ratio of the two distances, and it is what the enlarger’s column is really adjusting.

m = v ÷ u

Magnification

Combine them and each distance is fixed once you have chosen f and m, which is why a head at a given height on a given column can only make one size of print with a given lens:

u = f (1 + 1 ÷ m) and v = f (1 + m)

Where the two stages have to sit

A 50 mm lens at 4× puts the negative 62.5 mm below the lens and the paper 250 mm below it. At 8× the negative comes up to 56.25 mm and the paper drops to 450 mm. The negative barely moves and the head travels a long way, which is why the focusing movement on an enlarger is short and fine while the column is long and coarse.

The projection path at 4x, drawn from the numbers

Drawn from the thin-lens relation, so the drawing cannot show a magnification the distances deny. The image is inverted, which is why a negative goes into the carrier upside down and back to front.

The exposure consequence, derived rather than asserted

Enlarging further spreads the picture over more paper, so it needs more exposure. That much is obvious; the size of the effect is not, and the obvious guess is wrong.

The guess is that the exposure goes as the area, which grows as m². It does not, and the reason is in the diagram above: as m rises the negative comes closer to the lens, so the lens subtends a larger solid angle at the negative and collects more of its light. Work it through and the two effects combine into a single factor.

Let the lens’s aperture diameter be D = f/N, where N is the marked f-number. A small patch of negative of area A and luminance L sends into the lens a flux of L A × (πD²/4)/u², and that flux lands on an image patch of area m²A. Divide, substitute u = f(1 + 1/m) and D = f/N, and the focal length cancels completely:

E = π L ÷ [4 N² (1 + m)²]

Illuminance at the easel

E is the illuminance on the paper, L the luminance of the negative as the lens sees it, N the marked f-number and m the magnification. The factor (1 + m) is the whole story of enlarging exposure, and since exposure time is inversely proportional to illuminance:

t₂ ÷ t₁ = [(1 + m₂) ÷ (1 + m₁)]²

Exposure factor between two magnifications

Worked example. A negative prints well at 8 seconds at f/8 and 4×. Crop harder and go to 6×. The factor is (1 + 6)² ÷ (1 + 4)² = 49 ÷ 25 = 1.96, so the new time is 15.7 seconds — call it 16 — and the change is log₂(1.96) = 0.97, which is one stop to within three per cent. Going from 2× to 4× is a factor of (5/3)² = 2.78, or 1.47 stops.

🧮The same (1 + m) turns up as the effective aperture, and that is not a coincidence

Write the illuminance relation as E = πL ÷ [4(N(1 + m))²] and the bracket is an f-number: the effective f-number at the image side is N(1 + m). It is the enlarger’s version of the bellows factor Part VI met inside a camera, and it is the same piece of geometry seen from the other end.

Rodenstock state the consequence at one specific magnification, and it is the check on the algebra: for a lens focused at a scale of about 1:1 — that is, m = 1 — the effective aperture is approximately two f-stops smaller than the nominal aperture. Two stops is a factor of two in f-number, and 1 + m = 2. The manufacturer’s sentence and the derivation agree exactly.

This matters twice more on this page: it sets the aperture at which diffraction starts to cost you resolution, and it sets the depth of focus at the easel, which is where the alignment tolerance comes from.

Condenser against diffusion, which is the same negative rendered two ways

A silver image does not simply absorb light. It scatters it, because the developed silver is a mass of irregular filaments comparable in size with the wavelength. What happens to the scattered light depends entirely on the illumination geometry, and that is the whole difference between the two kinds of head.

The same negative, the same grain, two lamphouses

1. 1Compact source and condensers — a narrow, near-parallel beam through the negative
2. 2Scattered rays that miss the lens — counted as absorbed, so the density reads higher
3. 3Mixing chamber and opal exit — light arrives from every direction within a wide cone
4. 4Scatter replaced by scatter — what leaves one direction is made up from another
5. 5The lens, accepting a cone set by its aperture — the cone is what decides which rays count

Nothing about the negative differs between the panels. Only the illumination geometry does, and it is enough to move the print about a grade.

This is the Callier effect, and Part XV owns its definition, including the reasons that page prints no value for the Callier coefficient. What matters here is not the coefficient but its shape: because scattering grows with the amount of silver, the effect is bigger in the dense parts of the negative than in the thin parts. It therefore stretches the tonal scale rather than shifting it, which is exactly what a change of contrast is.

How big is it? ILFORD state it plainly and this course takes their figure: a condenser enlarger gives about an extra grade of contrast compared with a diffuser, and the difference depends on how much silver is left in the negative. Kodak say the same thing from the other side without quoting a grade: T-Max’s published starting-point development times are aimed at a diffusion enlarger, and to print with a condenser you shift one column to the left in their adjustment table — which is to say, develop less.

What that difference actually costs and buys, at the easel

• Dust, scratches and fingerprints. A condenser renders every one of them crisply, for the same reason it renders grain crisply: a scratch is a scattering feature, and specular light punishes scattering. A diffusion head fills the scatter back in and most small marks all but vanish. Printers who have used both put a large part of a condenser session into spotting.
• Grain. Sharper and more evident under a condenser. Whether that is a fault or a virtue is a picture decision, not a technical one, and both answers have made good prints.
• Negative development. This is the expensive one, because it is decided months earlier. A file of negatives developed for a diffusion head is about a grade too contrasty on a condenser, and the whole file is affected at once. It is the clearest demonstration in the course that contrast belongs to a system rather than to a film — the point Part XIII makes when it refuses to publish a contrast-index target without naming the printing system.
• Exposure. A condenser is more efficient. A mixing chamber absorbs a large fraction of its lamp’s output in its own walls while it is evening the field out, which is the trade the glossary records: uniformity is bought with brightness.

Why this course builds a diffusion head

Four reasons, and only the first is about contrast.

It is scatter-insensitive, so it is measurable. A diffuse head reads the negative in something close to the geometry a diffuse densitometer reads it in, which means the density figures Part XV’s instrument produces and the tones the enlarger produces are talking about the same quantity. Under a condenser they are not, and the difference is a coefficient the course has declined to invent.

It tolerates dust, which matters more in a home darkroom than in a laboratory, and more in a converted bathroom than anywhere.

It has somewhere to put a second colour. ILFORD’s variable-contrast papers are a mixture of three blue-sensitive emulsions carrying different amounts of green sensitising dye, so blue light gives high contrast and green light low contrast. A mixing chamber with two independently driven channels is therefore a contrast control with no filters in the light path at all. Part XVIII owns that theory and this course does not pre-empt it; what the head has to do now is leave room for it, which is a decision about chamber geometry taken before anything is cut.

Its evenness is a property of the chamber rather than of the lamp’s position. A condenser system images the lamp filament into the lens; move the lamp and you move a hot spot. A mixing chamber integrates, so its uniformity is set by geometry that does not drift.

📷What you give up, said honestly

Sharpness of rendering, not of focus. A condenser print of a fine-grained negative has a crispness at the grain level that a diffusion print does not, and printers who prize it are not imagining it. You also give up light: expect longer exposures for the same negative and aperture, which is a real cost in a session and the reason the head build sizes its emitter from an exposure target rather than from what is in the parts drawer.

The enlarging lens, and what makes it different from a camera lens

A camera lens is corrected to image a distant subject onto a flat film. An enlarging lens does the reverse job at close conjugates: it images a flat subject, 60 to 100 mm away, onto a flat easel, and it must hold its correction across the whole field at that near distance. The corrections are optimised for a magnification range, and manufacturers publish that range as a specification.

Rodenstock’s own tables are the evidence. Their simplest three-element Rogonar is specified 2× to 8×; the four-element Rogonar-S runs 2× to 10× at the standard focal lengths and 10× to 30× for the very short ones made for sub-miniature formats; the six-element Rodagon 50 mm is rated 2× to 15×. Their Apo-Rodagon-D exists because “even the best enlarging lenses for larger scales begin to show their weak spots” near 1:1, and it is specified 0.8× to 1.2×. A lens has a range, and outside it the same glass is a worse lens.

Focal length is chosen by format, not by taste

The lens must throw an image circle that covers the negative diagonal, and it must do it at the working magnification. Rodenstock’s model tables give the conventional pairings directly:

Negative format	Diagonal	Conventional focal length	Rodenstock’s example
24 × 36 mm	43 mm	50 mm	Rodagon 50 mm f/2.8, 2× to 15×
6 × 6 cm	85 mm	75 to 80 mm	Rogonar-S 75 mm f/4.5, 2× to 10×
6 × 7 cm	92 mm	80 to 90 mm	Rogonar-S 90 mm f/4.5, 2× to 8×
6 × 9 cm	108 mm	105 mm	Rogonar-S 105 mm f/4.5, 2× to 8×
4 × 5 in	162 mm	135 to 150 mm	Rodagon 150 mm f/5.6, 2× to 10×

The pattern is that the focal length is close to the negative’s diagonal, which is the same rule of thumb a “normal” camera lens follows, and for the same reason: it puts the field angle in the range the design is corrected for.

Too short a lens for the format and the image circle runs out before the corners of the negative do. What you see is a print that is soft, dim and finally black towards the corners, with the transition smooth rather than abrupt. Too long a lens costs you nothing optically and a great deal mechanically: v = f(1 + m) means a 105 mm lens needs twice the column height of a 50 mm one for the same magnification, and most darkrooms run out of column before they run out of ambition. Rodenstock’s Rodagon-WA exists for exactly that squeeze — a shorter focal length with a wider field, giving a 70 per cent larger projection area at a shorter projection distance, so a small room can make a big print without projecting onto the floor.

Mechanical vignetting is a different fault and looks different

Coverage is an optical limit. A mechanical vignette is a piece of the enlarger physically clipping the cone of light: a carrier aperture cut for 24 × 36 mm with a 6 × 6 negative in it, a filter drawer left half open, a bellows fold sagging into the path, a lens board recess too deep for a short lens. The two are separable by one test, and it is the most useful discriminator in this part:

Stop down two stops and look again. Optical falloff improves, because stopping down narrows the cone about the axis. A piece of metal is in the way at every aperture. The troubleshooting entry sets the same test out as a differential across four causes.

Choosing the working aperture

Two effects run in opposite directions as the diaphragm closes, and the best aperture is where they cross.

Aberrations fall. A lens’s residual aberrations are worst at the margins of the aperture, so stopping down throws away the badly behaved rays. This is why every manufacturer’s recommendation is to stop down at least a little, and why the recommendation is smaller for a simpler lens: Rodenstock ask two to three stops from the four-element Rogonar-S, two from the six-element Rodagon, and only one to two from the apochromatic Apo-Rodagon-N. A better-corrected lens needs less help.

Diffraction rises. Beyond some aperture the finite width of the hole limits resolution on its own, exactly as it does for a pinhole, and no amount of correction can recover it. Rodenstock name the point for one of their lenses and give the reason in the same breath: the Apo-Rodagon-D’s optimum working aperture is between f/5.6 and f/8, and stopping down beyond nominal f/8 “would result in visible blur because of diffraction” — because at 1:1 the effective aperture is already two stops smaller than the marked one.

!The effective aperture is what diffracts, and at 4x it is five times the marked number

This is the practical sting of N(1 + m). Set f/16 on a lens printing at 4× and the beam converging on the paper has an effective f-number of 16 × 5 = f/80. That is deep into the region where diffraction, not the lens, decides the resolution, and it is why the received advice to “stop right down for sharpness” is wrong for enlarging in a way it is not wrong for a camera.

The working range this course uses is two stops down from full aperture, and then whatever the test says. ILFORD’s own beginner sheet says the same thing in the simplest possible terms: turn the ring from full aperture to f/8, to increase edge sharpness and give more even illumination, counting the clicks so you can do it in the dark. They add the second criterion, which is a working one rather than an optical one — aim for an exposure of about ten seconds, because shorter times are hard to time accurately and longer ones are tedious.

Where those two criteria fight, the aperture wins and the head’s brightness has to change. That is a design input for the next page.

Buying a used lens, which is what almost everybody does

Enlarging lenses are among the few genuinely cheap things left in a darkroom, and the reason is that millions were made and the demand collapsed. Three faults are worth knowing by name, because all three destroy contrast rather than resolution, and a contrast fault in the lens is diagnosed as a fault in the negative or the paper nine times out of ten.

• Haze, a fine deposit on internal surfaces, usually from outgassed lubricant. It scatters, which is flare: shadows lift, maximum black falls, and the print looks flat in a way no grade change quite fixes.
• Fungus, a branching etch on the coating and eventually the glass, from storage in damp. Early fungus scatters like haze; late fungus is permanent.
• Separation, the cement between a cemented pair failing, seen as an irregular oily-looking edge creeping inward. It scatters and it is not repairable at hobby cost.

How to look. Hold the lens up to a bright diffuse light and look through it — that shows haze and fungus. Then shine a small torch in from the side, off-axis, and look at the glass rather than through it: scattering surfaces light up. Dust is unimportant and everyone will tell you so; the reason they are right is that a few specks intercept a negligible fraction of the aperture and are far outside any plane the lens images. Haze is different because it is everywhere.

The negative carrier, and the heat that reaches the negative

The carrier’s job is to hold the negative flat, in the same plane every time, with nothing in the light path but film.

Glassless carriers clamp the negative by its rebate and let the middle look after itself. They add no surfaces, cannot make Newton’s rings and cannot trap dust between glass and film — and they do not guarantee flatness, which is a problem that grows with format. A 35 mm frame held at its edges is fairly flat; a 6 × 9 frame is not.

Glass carriers sandwich the negative between two sheets and make it flat by force. They buy flatness and they cost four surfaces to keep clean, plus the risk of Newton’s rings where the film base and the glass are separated by a fraction of a wavelength. That is the same interference the contact frame’s second test hunts for, and the same fix — anti-Newton glass, textured on the face that touches the film — with the same cost, which is that the texture is now in the light path.

Dust is the tax on both. Every particle on the negative or on a carrier glass is imaged onto the paper at m times its own size, so a 30 μm speck prints as a 120 μm white spot at 4×. Compressed air and a soft brush before every negative is not fussiness; it is the difference between a printing session and a spotting session.

Film popping, and why an LED head is not automatically exempt

Film popping is the negative moving during the exposure as it warms in the beam, so the print carries a sharp image and a displaced soft one superimposed. The glossary entry files it as a movement fault belonging to the holder, not an optical one, and the mechanism is straightforward: the film absorbs energy, expands against whatever is gripping it, stores the strain until friction gives, and then jumps. A gradual expansion would be harmless; the suddenness is the whole problem.

Old tungsten heads were bad at this because a filament lamp puts out far more infrared than visible light, and almost all of it went through the negative. It is the historical reason for heat-absorbing glass in a condenser stack, and the historical reason large-format printers pre-warmed a negative in the beam before focusing.

An LED head reduces the problem and does not remove it, for two reasons worth stating plainly. A white LED emits very little infrared, so far less energy reaches the negative for the same amount of visible light. But the emitter’s own waste heat has to go somewhere, and a head that dumps it into the chamber is a head that warms the negative by convection instead of by radiation. That is why the next page puts the heatsink’s fins outside the chamber, and why the same heatsink is a photometric component as well as a thermal one: Part XIV established that an LED’s output falls as its junction warms, so an unheatsinked head drifts dimmer through a session.

📓What the course could not source about popping

No manufacturer or conservation document in this course’s corpus gives a temperature rise at which a negative pops, or a tested comparison of remedies. The mechanism above is thermal expansion against a constraint, which is ordinary physics; the remedies — a glass carrier, letting the negative reach the lamphouse’s temperature before final focusing, and a source that puts little infrared through the film — are stated as engineering reasoning rather than as a sourced list. If you meet a figure elsewhere, check where it came from before you plan around it.

Alignment: three planes that have to agree

The negative stage, the lens board and the easel must lie in parallel planes. When they do not, one part of the paper sits at a different optical distance from the lens than another, and one edge or corner cannot be focused at the same time as the centre. The secondary symptom is a keystoned image: a rectangle printed as a trapezium, because the magnification is different at the two ends.

A tilted negative stage, and the two things it does to the print

1. 1Negative stage, tilted — the plane whose tolerance is m times tighter
2. 2Lens board, level — tilting it moves both the focus and the image shape
3. 3Easel, level — only a band across it is in focus
4. 4Best-focus surface, tilted the other way — the image of a tilted object plane is a tilted image plane
5. 5The print: soft at two edges, and keystoned — unequal magnification at the two ends

The tilt is drawn at about eight degrees so that the effect is visible. The tolerance derived below is a quarter of a degree at the negative stage, which no drawing can show.

Derivation: how much tilt is too much

Nobody publishes an alignment tolerance for a home enlarger — see the honesty note below — so the course derives one, from a quantity the reader can compute for their own equipment.

Step 1: depth of focus at the easel. The cone converging on the paper has an effective f-number N(1 + m). Move the paper a distance δ from best focus and the point spreads into a circle of diameter δ ÷ [N(1 + m)]. Turn that round: if the largest blur you will accept on the print is c, then the paper may sit anywhere within ±cN(1 + m) of best focus, and the total depth of focus is twice that.

T = 2 c N (1 + m)

Depth of focus at the easel

c is a judgement, not a constant: it is the blur you are prepared to see on the finished print. This course uses 0.05 mm, which is about the finest detail a good eye resolves on a print held at arm’s length, and it says so rather than pretending the number is standard. Take N = 5.6 and m = 4: T = 2 × 0.05 × 5.6 × 5 = 2.8 mm. At f/2.8 it halves to 1.4 mm; at f/11 it doubles to 5.5 mm.

Step 2: the same tolerance at the negative. Longitudinal distances scale as m² between object and image space, so the corresponding depth on the negative side is T ÷ m² = 2.8 ÷ 16 = 0.175 mm. The negative stage’s tolerance is far tighter than the easel’s, and that is the result worth carrying away.

Step 3: turn each depth into an angle. A plane tilted by θ about its centre displaces its own edge by θ × a, where a is the half-width. Allow the edge to reach the limit of its depth of focus and:

θₙ = T ÷ (2 m² a)

Allowable tilt of the negative stage

θₑ = T ÷ (2 m a_image) = T ÷ (2 m² a) × m = m θₙ

Allowable tilt of the easel

For a 24 × 36 mm negative, a = 18 mm across the long axis, at m = 4 and T = 2.8 mm: θₙ = 2.8 ÷ (2 × 16 × 18) = 0.0049 rad = 0.28°, and θₑ = 4 × 0.28° = 1.1°.

🧮The one sentence to remember from all that algebra

A tilt at the negative stage costs m times what the same tilt at the easel costs. At 4× the negative stage is four times as critical; at 10× it is ten times. So when time is short, level the carrier first.

And 0.28° is not a bubble you can see. Over a 100 mm level base it is a rise of 0.49 mm — less than the width of a bubble vial’s line. A spirit level checks that a plane is horizontal; it does not check that two planes are parallel to a quarter of a degree. That is why the commissioning page uses a null method with a mirror instead, where the measurement is a coincidence rather than a reading.

Why you focus at full aperture and print stopped down

Depth of focus is proportional to N, so it is narrowest wide open. Focus there and you are working at the most demanding setting you have, which makes the point of best focus easiest to find; then stop down, and the depth of focus widens around the setting you already found. ILFORD’s beginner sheet gives exactly this sequence: focus at full aperture with a focus finder on the grain, then turn the ring to f/8.

Focusing on the grain rather than on the image is the second half of the trick. Grain is a property of the negative’s own plane, so a grain focuser cannot be fooled by a subject that was slightly out of focus in the camera. It is the one tool in this part that is worth buying rather than making.

The filter under the lens, and whether it shifts the focus

A contrast filter may be used above the lens or below it — ILFORD state both, and list a mounted below-lens kit of twelve filters, a safelight filter and a holder — and a flat plate in a converging beam does shift the focus. The paraxial result for a plane-parallel plate of thickness t and refractive index n is a shift away from the lens of:

Δ = t (1 − 1 ÷ n)

Focus shift from a flat plate in the converging beam

No manufacturer figure for this shift was found in this course’s corpus, so the arithmetic below is the course’s own paraxial calculation, offered as an order of magnitude and as a reason to run the test rather than as a specification.

• A thin gelatin or acetate filter, t ≈ 0.1 mm, n ≈ 1.5: Δ ≈ 0.033 mm. Against a depth of focus of 2.8 mm at f/5.6 and 4×, that is about one per cent. Negligible.
• A glass-mounted filter, t = 2 mm: Δ ≈ 0.67 mm. That is a quarter of the depth of focus at f/5.6 and half of it at f/2.8 — which is where you focus.

The rule that follows costs nothing: focus with the filter in place. Then the shift is inside the setting rather than added to it, whatever its size. There is a second reason to do it, and it is the flare argument from the end of this page rather than a focus argument: a scratched or greasy filter scatters exactly as a hazed lens does, and a filter you have just focused through is a filter whose surface you have looked at.

Illumination falloff, and how to state it honestly

Three separate things darken the corners of a projected image, and only one of them is a fault.

The lens’s own falloff. Off-axis, light reaches the paper through an aperture that is foreshortened, over a longer slant distance, and arriving obliquely. Each is a cosine, and the classical result is the cosine-fourth law the course already derives for a pinhole. Rodenstock treat it exactly as a reference rather than a prediction: every one of their published performance charts plots measured fall-off in illumination, in f-stops, against relative image height, with the 1 − cos⁴ curve drawn on the same axes. A real lens may sit above or below that line; the line is what it is judged against.

The condenser’s hot spot. A condenser system forms an image of the source. Get the lamp off-centre or the condenser spacing wrong for the focal length in use and the negative is lit unevenly before the lens has done anything at all.

The mixing chamber’s own profile. A chamber that is too shallow for its exit port does not integrate; it shows the emitter pattern through the diffuser. The contact printer’s light box settled the geometry that fixes it, and the enlarger head inherits it.

What the geometric reference alone predicts at the corner of a print

• cos⁴ reference, 50 mm at 4x on 24 x 36 mm

Show the numbers behind this plot

A plot of relative illuminance against relative image height, image height running from zero at the centre of the print to one at the corner. A single curve starts at 1.00 at the centre and falls smoothly and increasingly steeply to about 0.80 at the corner, following the fourth power of the cosine of the field angle for a fifty millimetre lens projecting a twenty-four by thirty-six millimetre negative at four times enlargement, where the corner sits about nineteen degrees off the axis. A horizontal guide line is drawn at 0.95, marked as the five per cent difference that corresponds to about 0.04 in density on a mid-grey, and the curve crosses it at a relative image height of about 0.47, which is to say just inside halfway out towards the corner. A second guide at 0.80 is marked as the corner value, about one third of a stop down. The teaching point is that the geometric floor at the corner is several times larger than the density difference a printer can see, so an enlarger cannot be held to the uniformity tolerance a contact printer meets.

Series	Relative image height, centre to corner	Relative illuminance
cos⁴ reference, 50 mm at 4x on 24 x 36 mm	0.00	1.00
cos⁴ reference, 50 mm at 4x on 24 x 36 mm	0.10	1.00
cos⁴ reference, 50 mm at 4x on 24 x 36 mm	0.20	0.99
cos⁴ reference, 50 mm at 4x on 24 x 36 mm	0.30	0.98
cos⁴ reference, 50 mm at 4x on 24 x 36 mm	0.40	0.96
cos⁴ reference, 50 mm at 4x on 24 x 36 mm	0.50	0.94
cos⁴ reference, 50 mm at 4x on 24 x 36 mm	0.60	0.92
cos⁴ reference, 50 mm at 4x on 24 x 36 mm	0.70	0.89
cos⁴ reference, 50 mm at 4x on 24 x 36 mm	0.80	0.86
cos⁴ reference, 50 mm at 4x on 24 x 36 mm	0.90	0.83
cos⁴ reference, 50 mm at 4x on 24 x 36 mm	1.00	0.80

Computed from cos⁴ of the image-side field angle for this geometry, not measured from a lens. It is the reference Rodenstock draw their measured curves against, and a real lens may do better or worse than it. The curve is drawn to show the shape, not measured from a real material. Your own materials will differ, and measuring them is what the sensitometry part of the course is for.

State it as density on the print, not as a percentage of light

A percentage of illuminance means nothing to a reader looking at a print. What they see is a density difference, and the conversion is the paper’s own curve.

ΔD = γ × log₁₀(E_centre ÷ E_corner)

Turning an illuminance difference into a visible one

γ is the local slope of the paper’s characteristic curve at the tone in question. For a middle grade of ILFORD’s Multigrade RC, the published ISO Range is about 110 — that is, the paper takes about 1.10 log exposure units — and its published maximum density is 2.15, so an average slope of about 1.95 follows, and the slope through the mid-tones, where the curve is steepest, is higher still. That division is this course’s own arithmetic across two ILFORD sheets and it is an approximation: an ISO range is defined between particular density points rather than from paper white to maximum black, so the quotient establishes an order of magnitude for γ and not a value. Take γ ≈ 2, measure your own when you can, and the arithmetic is easy to carry in your head:

Illuminance difference	Δlog₁₀ H	ΔD at γ ≈ 2
5 per cent	0.021	0.04
10 per cent	0.041	0.08
25 per cent	0.097	0.19
One third of a stop, a ratio of 1.26	0.100	0.20
One stop	0.301	0.60

Two things follow, and the second is uncomfortable. First, a mid-grey sheet is a remarkably sensitive detector: five per cent of light becomes 0.04 in density, which is the figure ILFORD publish as a detectable change on paper. Second, the cos⁴ reference above puts the corner of a modest 4× enlargement at about one third of a stop down — 0.20 in density — which is five times that detection threshold.

!The consequence for the tolerance this part publishes

The contact printer is held to a worst-point deviation of 0.04 in density, a figure borrowed openly from ILFORD’s safelight sheet because the course had no better one. That figure cannot be carried across to an enlarger, because the lens’s own geometry puts a floor above it before anything is misaligned or badly built.

So this course publishes no pass mark for enlarger uniformity. What it publishes is a method: measure the falloff on your own machine in density units, at two apertures and two magnifications, and write the numbers on the certificate. The commissioning page does that, and it is honest about what it is: a record of what your enlarger does, against which any future change is measured, rather than a test anybody passes or fails.

That is not a retreat. A number you measured is worth more than a limit somebody else’s instrument met.

Flare and stray light, which take away the blacks

Everything so far has been about light that goes where it should. Flare is light that does not, and it does one specific thing: it adds a roughly uniform exposure everywhere, which matters most where the exposure was least. So it lifts the shadows, compresses the foot of the paper’s curve, and takes away maximum black. The print looks flat and the reflex is to reach for a harder grade, which is treating a symptom.

Four sources, in the order they are worth checking:

• The lamphouse leak. Light escaping the head sideways and landing on the easel, or on the ceiling and then on the easel. It is fogging, not flare, strictly — but it presents identically and it is by far the commonest of the four on a converted or home-built head. The atlas entry has the signature.
• The bellows. A pinhole in a folded bellows is a point source inside the light path.
• The lens. Internal reflections between uncoated or hazed surfaces, and reflections off the inside of the barrel. This is what the used-lens inspection above is looking for.
• The room. White walls under and around the enlarger take the light that misses the easel and give some of it back. It is the same argument the course makes about a camera’s interior, one scale up.

📷Flare is already inside the number you print by

ILFORD’s ISO Range figures — the ones used to choose a paper grade — are stated for the image as projected on the enlarger baseboard, not as read from the negative on a light box. That qualification is doing real work: the projected range already has your enlarger’s flare subtracted from it and your enlarger’s Callier effect added to it. It is why Part XIII warns that a densitometer reading of the negative and the range ILFORD mean are not quite the same quantity, and it is why the flare test on the commissioning page is not an optional extra: reduce your flare and your grade choice changes.

Alternative route: reading this page without an enlarger

If you have no enlarger and are not going to build one, this page still earns its place, because three of its five subjects are yours already. The falloff argument, the density-against-percentage conversion and the flare argument all apply unchanged to the contact printer’s light box, which is a light source with a uniformity problem and a stray-light problem of its own. The alignment section does not apply, and that is one of contact printing’s genuine advantages: two surfaces clamped together cannot be out of parallel.

If you print by contact only, the useful transfer is the aperture argument turned inside out. You have no aperture, so you have no way to buy evenness by stopping down and no way to spend resolution on diffraction. Your evenness comes entirely from the box’s geometry, which is why that page’s light-box arithmetic is so insistent.

If you are buying an enlarger rather than converting one, read the lens table and the alignment derivation before you go. They are the two things you cannot judge by looking at a machine in somebody’s garage: whether the lens matches the format and the magnifications you want, and whether the column and stage will hold a quarter of a degree once they are tightened. A twenty-year-old enlarger with an unknown history is not a certificate, and the price of finding out is one sheet of paper.

Recap

Putting a lens between the negative and the paper introduces two conjugate distances, u = f(1 + 1/m) and v = f(1 + m), and the second of them carries almost everything else on the page. Exposure goes as (1 + m)², not as m², because the lens comes closer to the negative as the magnification rises; the effective f-number is N(1 + m), which Rodenstock confirm at 1:1 with their two-stop statement; and that same factor sets both the aperture at which diffraction takes over and the depth of focus at the easel.

Condenser and diffusion heads render the same negative about a grade apart, because a specular beam loses the light the silver scatters and a diffuse one replaces it. That is the Callier effect, Part XIII and Part XV own it, and its consequence is a decision about negative development taken long before the printing session. The course builds a diffusion head because it is scatter-insensitive, dust-tolerant, measurable, and has room for the second colour channel Part XVIII will want.

An enlarging lens is specified by a magnification range and a format, works best one to three stops down depending on how well corrected it is, and is limited by diffraction at an effective aperture that is five times the marked one at 4×. The alignment tolerance follows from depth of focus and comes out at about a quarter of a degree at the negative stage — m times tighter than at the easel, and far below what a spirit level can see.

Falloff is stated in density on the print rather than as a percentage of light, because that is what the reader sees: five per cent of light is 0.04 in density on a mid-grey, and the cos⁴ reference alone puts the corner of a 4× enlargement five times further down than that. So the course publishes a method rather than a pass mark. And flare — from the lamphouse, the bellows, the lens and the room — is the one fault on this page that presents as a chemistry problem and is cured with black paint.

Check your understanding

Question 1. A negative prints correctly at 6 seconds at f/8 and 3x magnification. You raise the head to 5x and change nothing else. What exposure do you need, and why is it not 6 x (5/3)²?

• About 10 seconds, because exposure goes as the magnification squared
• About 13.5 seconds, because exposure goes as (1 + m)² and the lens moves closer to the negative as m rises, collecting more of its light than the area argument alone allows for
• About 17 seconds, because exposure goes as the print area
• Unchanged, because the aperture has not changed

Show the answer and why

Answer: About 13.5 seconds, because exposure goes as (1 + m)² and the lens moves closer to the negative as m rises, collecting more of its light than the area argument alone allows for

6 x (1 + 5)² ÷ (1 + 3)² = 6 x 36 ÷ 16 = 13.5 s, which is 1.17 stops. The pure area argument would give 6 x 25 ÷ 9 = 16.7 s and would overexpose by about a third of a stop, because it ignores the second half of the geometry: u = f(1 + 1/m), so raising the magnification brings the negative closer to the lens and the lens subtends a bigger solid angle at it. At large magnifications the two answers converge, because (1 + m)² approaches m², but in the range a darkroom actually works in the difference is worth having.

Question 2. Two prints are made from one negative, one on a condenser head and one on a diffusion head, both exposed so that the deepest shadow reaches the same density. The condenser print has lighter highlights and a scratch that is invisible on the other. What single mechanism accounts for both?

• The condenser lamp is brighter, so the highlights received less relative exposure
• Light scattered by the silver and by the scratch is lost from the narrow condenser beam and recovered in the diffuse one, and since scattering grows with silver density the effect stretches the tonal scale rather than shifting it
• The condenser head has a different colour temperature and the paper is blue-sensitive
• The diffusion head has more flare, which fills the shadows

Show the answer and why

Answer: Light scattered by the silver and by the scratch is lost from the narrow condenser beam and recovered in the diffuse one, and since scattering grows with silver density the effect stretches the tonal scale rather than shifting it

One mechanism, two symptoms. A scratch and a dense highlight are both scattering features; a condenser collects only the near-axial rays, so both count as denser than a diffuse instrument would say. Because the loss grows with the amount of silver, the highlights move further than the shadows do, which is an increase in contrast rather than a change of exposure — ILFORD put it at about a grade. Colour is a real effect on variable-contrast paper but it is not this one, and it would not explain the scratch.

Question 3. You are printing at 4x and set the lens to f/16 on the reasoning that stopping right down gives the sharpest result. What have you actually done?

• Reached the lens optimum, since most lenses are best at f/16
• Given the paper an effective f/80 beam, which is deep into diffraction-limited territory, having passed the lens optimum several stops earlier
• Increased the depth of focus with no cost at all
• Made the illumination less even

Show the answer and why

Answer: Given the paper an effective f/80 beam, which is deep into diffraction-limited territory, having passed the lens optimum several stops earlier

The effective f-number at the image side is N(1 + m) = 16 x 5 = 80. Diffraction scales with that number, not with the marking on the barrel, which is why Rodenstock warn that stopping their 1:1 lens past nominal f/8 gives visible blur: at m = 1 the effective aperture is already two stops down. The camera habit of stopping well down does not transfer, because a camera lens at infinity has m near zero and its effective and marked apertures are the same. Depth of focus does increase, which is the one true part of the reasoning, and it buys nothing you needed.

Question 4. Your negative stage and your easel are each out of parallel by a quarter of a degree, in the same direction, at 5x magnification. Which matters more, and by how much?

• The easel, because the print is bigger and the error is magnified
• The negative stage, by a factor of five, because the allowable tilt at the negative is the easel tolerance divided by m
• Neither: a quarter of a degree is invisible at any magnification
• They are equivalent, because the light passes through both

Show the answer and why

Answer: The negative stage, by a factor of five, because the allowable tilt at the negative is the easel tolerance divided by m

Depth of focus at the easel is T = 2cN(1 + m), and longitudinal distances scale as m² between image and object space, so the negative side gets T ÷ m². Converting each into an angle over its own half-width gives θn = T ÷ (2m²a) and θe = mθn: the negative stage is m times more critical. At 5x it is five times, at 10x ten. This is why alignment procedures start at the carrier, and why a quarter of a degree — which is beyond the resolution of a spirit level — is a real tolerance rather than a fussy one.

Question 5. A uniformity sheet shows the corner of a 4x enlargement about 0.20 in density lighter than the centre. Your contact printer meets 0.04. What is the correct conclusion?

• The enlarger has a fault that should be found and fixed before printing
• The cos⁴ geometric reference alone accounts for roughly that much at this field angle, so the figure is recorded on the certificate as what this enlarger does, and any pass mark borrowed from the contact printer would be meaningless
• The paper is fogged at the corner
• The uniformity sheet was read wet

Show the answer and why

Answer: The cos⁴ geometric reference alone accounts for roughly that much at this field angle, so the figure is recorded on the certificate as what this enlarger does, and any pass mark borrowed from the contact printer would be meaningless

The two instruments are not comparable, and pretending otherwise would fail every enlarger ever made. A contact printer has no lens and so no off-axis geometry; an enlarger has both, and the fourth-power cosine at the corner of a 24 x 36 negative projected at 4x by a 50 mm lens is about 0.80, which is a third of a stop and about 0.20 in density at a mid-tone slope of 2. What the measurement is for is comparison with itself: stop down two stops and re-measure, and what improves was optical falloff while what does not is mechanical. Reading a print wet is a real error — it reads low — but it would not produce a corner-to-centre gradient.

Question 6. Prints from a newly bought used lens are correctly exposed but flat, with dull blacks, and no grade change quite fixes them. The negatives measure normally on the densitometer. Where do you look first?

• The developer, which is probably exhausted
• Into the lens from the side with a small torch, for haze, fungus or separation — all of which scatter, which is flare, which lifts the shadows and takes away maximum black
• The paper, which is probably out of date
• The alignment, which is probably out

Show the answer and why

Answer: Into the lens from the side with a small torch, for haze, fungus or separation — all of which scatter, which is flare, which lifts the shadows and takes away maximum black

The clue is that the fault survives a grade change and that the negatives measure normally. Flare adds a roughly even exposure, so it does most where there was least: it lifts the foot of the paper curve, which is a loss of maximum black rather than a loss of slope, and a harder grade cannot restore a black the paper never reached. Scattering surfaces show up when you look at the glass off-axis rather than through it. Exhausted developer and stale paper are real causes of weak blacks — and they belong to the print chemistry Part XVIII owns, which is exactly the boundary the break/fix page at the end of this part draws.

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