📐 Camera Lens Coverage Calculator
Choose the right lens for the job: enter focal length, sensor size, and distance to see the horizontal field of view and exactly how wide a scene the camera captures.
📐 Field of view & coverage
Field of view = 2·atan(sensor width ÷ 2·focal length); scene width = distance × sensor width ÷ focal length. A wider lens covers more area but shrinks faces and plates at distance; a longer lens identifies detail far away but sees a narrow slice. Real optics — barrel distortion on very wide lenses isn't modelled.
Wide coverage vs. distant detail
Every camera placement is a trade-off between how much you see and how well you see it. A short focal length sweeps in a wide scene — great for watching a whole yard — but a person at the far edge is only a handful of pixels. A longer lens zooms in to capture a recognisable face or plate, at the cost of seeing a much narrower strip.
Use the field-of-view angle to plan how many cameras cover an area without blind spots, and the scene-width figure to confirm a lens actually frames the doorway, gate, or aisle you care about. Then match resolution to the detail you need at that distance.
❓ Frequently Asked Questions
How do I calculate a camera's field of view?
The horizontal angle of view is 2 × atan(sensor width ÷ (2 × focal length)). A 4 mm lens on a common 1/2.8-inch (5.37 mm) sensor gives roughly a 67° horizontal view. Shorter focal lengths widen the angle; longer ones narrow it.
How wide an area does the camera actually see?
At a given distance, the captured scene width equals distance × sensor width ÷ focal length. So that 4 mm lens on a 1/2.8-inch sensor sees about 13.4 m across at 10 m away. Double the distance and you double the width; double the focal length and you halve it.
What focal length should I choose?
For wide area coverage — a yard, car park, or open room — use a short focal length like 2.8 mm. To identify faces or licence plates at distance, use a longer lens (8–12 mm or more) that trades width for detail. Many cameras offer a varifocal lens so you can tune it on site.
What's a sensor 'format' like 1/2.8 inch?
It's a legacy designation for the sensor's optical size, not a literal measurement. What matters for field of view is the sensor's actual imaging width in millimetres, which the calculator stores for each format — for example 1/2.8 inch ≈ 5.37 mm wide.
Is this exact?
It's the standard, real optics formula and accurate for planning coverage. It doesn't model lens distortion (very wide lenses bend straight lines) or the pixel density needed to actually recognise a face, which depends on resolution and the identification standard you're using.