How to convert between the luminous intensity Iv of the LED and the illuminance E

First understand the definition of the following illuminance: illuminance refers to the ratio of the luminous flux dφ on the bin on a light receiving surface to the area ds of the bin. The illuminance is expressed in lux and is represented by the symbol lux, which can be expressed as:
E=dφ/ds (71-1)
Obviously, under the same luminous flux, the larger the area of ​​the illuminated bin, the smaller the illumination, and vice versa.
If the luminous flux φ of the LED and the area to be irradiated are known, the illuminance E can be converted. If the luminous intensity Iv and the emission angle θ of the LED are known, the illuminance of the surface irradiated on the surface area S can be converted. .
For example, an LED with an emission angle of 60° and a light intensity of Iv=1cd can be obtained from the following steps when it is illuminated on a plane with a normal distance of 0.1M:
From the above-mentioned conversion of Iv and φ, it can be known that the equivalent luminous flux φ=4π×(60°/360°) ≈21m of the LED light source having an emission angle of 60° and an emission intensity of 1 cd, and the surface irradiated to a distance of 0.1 M. In the case of Yuan, the area S of the illuminated face element is:
S=π(dtan30°) 2≈3.14×(0.1×0.58) 2≈0.0105M 2
Then there are: E = φ / S = 21m / 0.0105 ≈ 190lux. If the distance is 1 M, the illumination at the illumination angle is only 1.92 lux.
From the above interchange of these parameters, it is conditional. The comparison with the use conditions is not related to the luminous flux φ. Therefore, when the LED is used in the field of illumination, the optical flux is often used to express its optical parameters. Can understand.

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