TE Enriched Sample (E)

24 Lighting 1 C Inverse-square law Point light source Consider a small light source giving out light evenly in all directions with a luminous flux Φ . At a distance r from the source, the light spreads to a spherical surface of area 4 r r 2 (Fig. 1.24). The illuminance E 0 on that surface is ( MC: DSE 2020 Q3.1) inverse-square law ̻˙ˀˢ֛ ܛ E A r 4 0 0 2 r U U = = The illuminance on the spherical surface decreases with r 2 , the square of the distance (Fig. 1.25). For example, when the distance r is doubled, the illuminance decreases to 1/2 2 = 1/4 of the original value. The above formula is an example of the inverse-square law . r light source 2 r 3 r r S Fig. 1.24 Light power spreads to a spherical surface of area 4 r r 2 . Assumptions The inverse-square law applies only when • the source is small (similar to a point source), and • the surface is perpendicular to the incident light. The case of oblique （傾斜） incidence will be discussed on the next page. Verifying inverse-square law ( V81-e13) Fig. 1.25 Inverse-square law The law does not apply to a parallel beam of light. Teaching notes As long as the size of the source is much smaller than the distance r and reflection is not significant, the illuminance is proportional to / r 1 2 . The inverse-square law holds even if light is not emitted uniformly in all directions from a small source. Extra information For a non-isotropic source, the illuminance should depend on an orientation factor ( , ) i z Y : ( , ) E r 4 0 2 $ r i z U Y = Extra information In more advanced study, the quantity /( ) r 4 2 r U is called the luminous intensity and carries the unit of candela (Cd). Q&A Q: Refer to Fig. 1.25. If the luminous flux of S and the distance are both doubled, will the illuminance remain unchanged? A: No, it will decrease by half. (See DSE 2022 Q3.1) Q&A Q: Refer to Fig. 1.24. If a reflector is installed so that it makes the light become a uniform distribution over a hemisphere, how would the normal illuminance change? A: E 0 ➞ 2 E 0 ( Ö U unchanged, but A halved) Q: If the reflector concentrates the light into a cone whose base area is 25% of the entire light sphere, what would be the normal illuminance now? A: E 0 ➞ 4 E 0 ( Ö U unchanged, but A ➞ A /4) Q: If the reflector produces a thick parallel beam with radius = 10% of the sphere, how would the illuminance change with distance? A: unchanged ( Ö no longer divergent) (See DSE 2016 Q3.1 option 3) Sample © United Prime Educational Publishing (HK) Limited, Pearson Education Asia Limited 2023 All rights reserved; no part of this publication may be reproduced, photocopied, recorded or otherwise, without the prior written permission of the Publishers.