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Asked by Prashant DIGHE | 26 Jan, 2020, 09:38: PM
Expert Answer
Figure shows the illumination area appears at the top of surface due to total internal reflection
for the light source placed at the bottom of lake.
Let h be the depth of lake. For a depth h of greater magnitude, illumination area approximately
equals the cap surface area 2πR(R-h) as shown in figure, where R is the radius of sphere and R equals
the length of light path when light incident at critical angle.
Ratio r of light intensity that emerges from the illumination surface shown in figure to the light intensity
that emerges from the complete surface area of sphere of radius R is equal to ratio of respective areas
Hence r = 2πR(R-h) / ( 4πR2 ) = (1/2) (R-h)/R = = (1/2) [ 1 - cos(c) ] ................(1)
where c is critical angle which is related to refractive index μ as, sin(c) = 1/μ = 3/4 .
hence we have cos(c) = [ 1 - sin2c ]1/2 = [ 1 - (3/4)2 ]1/2 = √7/4 = 0.661
Hence the required ratio r using eqn.(1) is given by, r = (1/2) [ 1 - 0.661 ] = 0.169
Hence percentage of light emerging through the illumination surface = 17%
Answered by Thiyagarajan K | 27 Jan, 2020, 02:32: PM
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