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# The luminous intensity of a light source is 500cd. The illuminance of a surface distant 10m from it, will be if light falls normally on it A. 5 lux B. 10 lux C. 20 lux D. 40 lux

Last updated date: 17th Sep 2024
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Hint: In this question, we are given the luminous intensity of a light source and the distance of the light source from the screen. We know that the illuminance of a light source is defined as the luminous intensity falling on per unit square area. We just need to substitute the given values in this equation.

Complete Step by step solution:
Luminous intensity is the amount of visible light emitted by any light source in unit time per unit solid angle. It is one of the seven fundamental physical quantities in physics. The SI unit for luminous intensity is Candela, cd.
Illuminance is the total luminous flux incident on any surface per unit area of the surface. It is measured in lux.
Illuminance, \[E = \dfrac{I}{{{R^2}}}lux\], where I is the luminous intensity and R is the distance between the surface and the light source
Given that \[I = 500cd\,and\,R = 10m\]
By substituting the values, we get that

\[ \Rightarrow E = \dfrac{{500}}{{{{10}^2}}} = 5lux\]

Note: We know that Illuminance, \[E = \dfrac{I}{{{R^2}}}lux\], is different from Luminous Intensity that is I. Illuminance is different from brightness as illuminance is the measure of intensity of light falling onto a surface, while brightness is the visual perceptions of light.