﻿ Lux

# Lux

Lux is the ( derived SI ) unit of illuminance and its corresponding emitter size, the specific light emission. Your unit symbol is: lx = lm / m². The name is derived from the Latin word lux for light.

## Practical significance

The illuminance in lx is the quotient of the luminous intensity of a point light source in cd and the square of the distance in meters.

Thus, the illuminance gets a quick reference when a single light source is considered, such as under artificial lighting. Conversely, the brightness outdoors has virtually nothing to do with distances, since the light source is virtually infinitely large (for example, cloud cover ).

The luminous intensity is measured with the lux meter or calculated in simulations, eg, to determine if a work area is sufficiently illuminated. There are also applications in photography.

At the Physikalisch-Technische Bundesanstalt ( PTB) illuminance levels between 0,001 and 100,000 lx lx can be realized. This is, inter alia, the calibration of illuminance measuring instruments.

## Worked examples

Example 1 The light intensity of a candle is 1 candela ( cd). At a distance of 2 m, the illuminance ( " brightness " ) is:

Result: White objects lit by a candle at a distance of about 1.8 m, appear about as bright as the light of the full moon (see also Examples of typical illuminance ).

Example 2 The brightness of an isotropic radiating light source amounts to 3 meters 20 lx = 20 lm / m².

• The solid angle of a 1 square meter area at a distance of 3 m is: 1 / ( 3.3) sr = 1/9 sr.
• The light intensity is thus 20 lm / (1 /9) sr = 20.9 lm / sr = 180 cd.
• Integrated over the whole space is the luminous flux 180 lm / sr · 4π sr = 180 · ln 4π = 2260 lm.

Example 3 Converting the units of candelas, lumen and Lux, depending on the radiation angle and distance.

A light emitting diode is illuminated with a light intensity I of 6 cd within a light cone of α = 30 °, corresponding to a solid angle of Ω = 0.2 sr. Outside of this cone it was dark.

• The luminous flux Φ is: Φ = I × Ω, ie: Φ = 6 lm / sr sr × 0.2 = 1.2 lumens
• The area A = illuminates the light emitting diode with a distance r Ω 0.2 sr is: Ω = A · r ². For example, they illuminated at a distance of 20 cm an area of ​​0.2 × 0.04 m = 0.008 m.
• The illuminance E is E = Φ / A, see table: At a distance of 0.6 m, the illuminance is 17 lux

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