TE Enriched Sample (E)

23 Measuring illumination 1.3 B Lambert’s cosine law Suppose a parallel light beam shines on a surface (Fig. 1.23). When the beam is perpendicular to the surface, it falls onto a certain area A 0 = ab . However, when the beam shines on the surface at an angle θ , it falls onto a larger area cos cos A a b A 0 $ i i = = . A 0 θ θ θ A 0 A 0 b / cos θ b cos θ a b A 0 Fig. 1.23 A light beam shining on a surface Hence, the illuminance E is reduced by a factor of cos θ as compared with the normal incidence: / cos cos E A A A 0 0 i i U U U = = = Thus, E = E 0 cos θ E 0 is a scalar, but for a mnemonic, this formula looks like taking the vertical component of a vector. If you imagine E 0 is a vector, then E is the vertical component of it. θ E 0 cos θ E 0 Checkpoint 4 1. True or false: (a) Illuminance measures how bright a surface appears to the human eye. (b) The illuminance on a surface is greatest when light from a light source is incident on it perpendicularly. 2. Which of the following is a unit of luminous flux? A. lm W - 1 B. lm C. lm m - 2 D. lx 3. Which of the following is a unit of illuminance? A. lm W - 1 B. lm C. lm m - 1 D. lx 4. The illuminance on a surface is 500 lx when light shines perpendicularly on it. What is the illuminance on the surface if it is rotated by 30°? F T B D 433 lx Lambert ´ s cosine law ࣦЬቱָ֛ ܛ This is known as Lambert’s cosine law . cos θ ≤ 1 (reduced) 1 cos θ ≥ 1 (enlarged) Simulation Illumination on a surface by a torch (See DSE 2015 Q3.1) 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.