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Question 1: 100 = 100% = light at surface 1 = 1% = 1% of surface light dx = depth where light is 1% of surface light (loge 100 – loge 1) k d0 = surface=zero. dx – d0 - Calculate 1% light depth for both the following attenuation co-efficient (K) values: a) K= -0.1098 b) K = -0.0844

Question 1: 100 = 100% = light at surface 1 = 1% = 1% of surface light dx = depth where light is 1% of surface light (loge 100 – loge 1) k d0 = surface=zero. dx – d0 - Calculate 1% light depth for both the following attenuation co-efficient (K) values: a) K= -0.1098 b) K = -0.0844

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Question 1: 100 = 100% = light at surface 1 = 1% = 1% of surface light dx = depth where light is 1% of surface light (loge 100 – loge 1) k d0 = surface=zero. dx – d0 Calculate 1% light depth for both the following attenuation co-efficient (K) values: a) K= -0.1098 b) K= -0.0844
Transcribed Image Text:Question 1: 100 = 100% = light at surface 1 = 1% = 1% of surface light dx = depth where light is 1% of surface light (loge 100 – loge 1) k d0 = surface=zero. dx – d0 Calculate 1% light depth for both the following attenuation co-efficient (K) values: a) K= -0.1098 b) K= -0.0844
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