Collimated backlighting to image transparent parts as solid black

Put a clear resin part in front of a diffuse LED backlight and it all but disappears: rays reach the lens from every direction, the part transmits most of them, and the gauge hunts for an edge that is a soft grey gradient. Collimated backlighting flips that. Nearly parallel rays refract at every sloped or curved surface, miss the lens aperture, and the part images as a solid black silhouette — what push pins, clear caps, connector bodies and cannulas need for dimensional checks. Below you get the refraction arithmetic, the aperture angle that decides whether the silhouette goes black, a worked set-up around a 30 mm part, and the cases where collimation still fails. For the broader opaque-part comparison, start with collimated backlight or diffuse backlight for edge gauging; for other geometry options, see machine vision illumination types.

Why does a clear part disappear under a normal backlight?

Because a diffuse backlight fills the lens with rays from a wide cone — about 87 % of peak brightness still arrives at ±30° from a Lambertian emitter — whatever the part refracts away is replaced by other rays that pass straight into the aperture.

A diffuse panel is an extended source with roughly cosine falloff. Your lens, in contrast, collects rays inside a cone of one or two degrees at the working distance; the arithmetic comes two sections down. Intersect the two and the conclusion follows: for every incoming ray the part bends out of the pupil, another incoming angle, bent differently, lands inside it.

The part transmits around 92 % of the light even counting surface reflections, so its interior reads nearly as bright as the background. Opaque parts never face this problem, because they absorb the beam outright. Transparency makes the source geometry, not the part, the deciding factor — and a threshold on an image like that wanders with every grey bump.

How does collimated backlighting turn the same part black?

By replacing the wide cone with nearly parallel rays — one published collimated backlight series is down to 5 % of peak brightness by ±30° — so a single refraction at a sloped surface steers the whole beam past the lens and the pixel receives no light.

Every ray now crosses the part from almost the same direction. Wherever a surface is tilted, Snell's law bends the beam; if the bend exceeds the aperture half-angle, the camera collects nothing from that region and it reads black. The black is an absence of collected light, not absorption — the resin is still transparent.

Curved surfaces work in your favour: a dome or cylinder limb behaves like a steep wedge over a continuum of angles, so its edges darken strongly. Only rays that meet both faces at exact normal incidence pass undeviated, which is why the exceptions near the end of this piece matter. Reflection alone cannot make black: an uncoated resin surface reflects about 4 %, so an undeviated ray still carries over 90 % of its brightness.

How much ray deviation does your part produce?

Roughly (n − 1) times its wedge angle: 5° of mould draft in PMMA (n ≈ 1.49) deviates a ray by about 2.5°, and the same draft in polycarbonate (n ≈ 1.59) by about 3.0° — small numbers that only have to beat a one-to-two-degree lens aperture.

Measure the shallowest angle anywhere on the silhouette — mould draft on side walls, the taper on a conical seat, the local tangent on a curve near the limb — and design to that value, not the average. Between n = 1.45 and n = 1.60 a 5° wedge spans 2.25° to 3.0° of deviation, so the index matters mainly for marginal, shallow-draft parts.

How small a lens aperture keeps the silhouette black?

Small: the lens accepts a cone of half-angle atan(f / (2 · N · WD)) — 1.6° for a 35 mm lens at f/4 seen from 155 mm — and the part goes solidly black once the ray deviation reaches about twice that acceptance angle.

The three lines interact: sensor and field of view fix the magnification, the focal length then fixes the working distance, and the f-number fixes the aperture cone. The lens focal length calculator runs the geometry for you, and the object-space pixel size that follows comes straight from the magnification — calculate object space pixel size from magnification works it through.

There is a floor to the trick. Halving the acceptance angle means halving the aperture diameter, and eventually diffraction dominates: the black-to-white transition spreads across many pixels and the edge location loses repeatability. The worked example puts numbers on both sides of that trade.

What does a working set-up look like with real numbers?

A 30 mm clear PMMA part with 5° draft walls, a 2/3″ 5 MP monochrome camera with 3.45 µm pixels and a 35 mm lens at f/5.6 from about 155 mm: the walls deviate rays 2.5° against a 1.2° acceptance cone and read solid black.

The arithmetic, step by step:

1. Magnification: m = 8.8 mm sensor width / 30 mm field of view = 0.29; the 6.6 mm sensor height then covers 6.6 / 0.29 = 22.5 mm of part.
2. Working distance: WD = 35 × (1 + 0.29) / 0.29 = 155 mm.
3. Aperture cone at f/5.6: diameter 35 / 5.6 = 6.25 mm, half-angle atan(6.25 / (2 × 155)) = 1.2°.
4. Deviation: (1.49 − 1) × 5° = 2.5°, a 2.1× margin over the cone, so the drafted walls go fully black.
5. Object pixel: 3.45 µm / 0.29 = 11.9 µm; the part spans 30 mm / 11.9 µm ≈ 2,520 px.
6. Edge blur at f/5.6: 2.44 × 0.55 µm × 5.6 = 7.5 µm on the sensor, 7.5 / 0.29 = 26 µm in object space, about 2.2 px — inside the 2-3 px target.

The LS-FFL-0679 is a 35 mm, 2/3″, 6 MP C-mount machine vision lens. That resolution class leaves headroom over 11.9 µm object pixels, and C-mount keeps the camera choice open.

Size the lit area to the field of view plus placement tolerance: 1.25 × is a workable margin, so a 30 × 22.5 mm field of view wants at least 38 × 28 mm of emitting surface. The LS-BKL-0830 is a 120 × 120 mm collimated back light on 24 V:

Change one input and the decision flips. A 1° draft deviates rays only 0.49°; holding the 2× margin needs an acceptance cone near 0.25°, which at this working distance means an aperture under 1.4 mm — about f/25 on a 35 mm lens. Diffraction there spreads the edge over roughly 10 px, so the honest answer is not a smaller aperture but a telecentric lens.

When do you need a telecentric lens with the collimated light?

When the shallowest surface angle on the part falls under about 2°, or when part height varies by more than a few tenths of a millimetre: a telecentric lens accepts only rays within a few hundredths of a degree, and magnification stops depending on distance.

An ordinary lens sees a size change of Δz / WD when the part sits Δz closer or further — 0.5 mm of carrier height at 155 mm is a 0.3 % size error, fatal to a tight gauge. A telecentric lens removes it, and its acceptance cone, centred on the axis to better than about 0.1° on published specs, is cleared fivefold by the 0.49° deviation of a 1° draft. The mechanics are the price: the front aperture must be at least as large as the field of view, and the collimated beam must fill it. The reasoning behind the switch is laid out in when you need a telecentric lens for gauging.

The LS-TCL-0574 is a 0.4x fixed-magnification C-mount lens for a 1.2″ sensor with a 227 mm working distance.

Match the light to the lens, not the part: the beam diameter must cover the front aperture with margin, and these collimated lights are built for exactly that pairing:

When is collimated backlighting the wrong tool?

When the part has two parallel flat faces: a window or flat film deviates rays by a net zero and transmits about 92 % of the background (near 3.9 % Fresnel loss per uncoated surface at n = 1.49), so it stays bright however parallel the beam.

Four cases account for most failures. Flat windows and films — use crossed polarized lighting for transparent film defects instead, where stress and thickness modulation do the work refraction cannot. Frosted or textured surfaces, which scatter light into every angle including the aperture; the silhouette goes grey and soft, and no aperture closes that gap.

Cylinders and rods carry a third trap: the axial ray meets both surfaces at normal incidence and passes undeviated, leaving a bright core down the middle of an otherwise black silhouette. Gauge the diameter from the dark limbs — rod work under backlight has its own write-up in coated rod inspection for diameter and core eccentricity.

The fourth is ambient light, which refills the aperture from outside the collimated cone. Shroud the optical path, or run the light strobed with a short exposure so room light cannot accumulate. And for parts that are flat and thin as well as clear, how to detect small transparent parts with machine vision collects the geometries that beat a silhouette.

What should you check before ordering?

That the deviation beats the aperture with 2× margin, that the lit area beats the field of view by 25 %, and that nothing — ambient light, a colour sensor, a dirty cover plate — refills the black after the light has done its job.

• Measure the shallowest wedge angle α on the part and its resin index n; compute δ = (n − 1) · α.
• Compute the acceptance half-angle at your real working distance and f-number; require δ ≥ 2 · θ.
• Check the diffraction cost of that f-number: 2.44 · λ · N / m in object space, aimed at 2-3 px.
• Size the lit area to at least 1.25 × the field of view; with a telecentric lens, the beam must also cover the front aperture.
• Run a monochrome camera — a Bayer colour sensor smears the black-white edge across extra pixels.
• Shroud the path against room light, or strobe the light with a short exposure.
• On the sample image, measure: part interior under 20 of 255, background over 230, edge transition inside 3 px.
• Hold repeatability to one tenth of the tolerance — vision metrology: the 10 to 1 accuracy and repeatability rule — before release.

To turn the numbers into a parts list, the light finder filters collimated models by size and colour, and the configurator walks the camera, lens and controller around the light you pick. Sample-imaging questions go to the contact page.