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pls help......................................................................
in this question you’ve to use the concept of the circle of illuminanceA = Angle at which total internal reflection is occurring, in this case, it is given by sin(A) = 1/mu. mu = 4/3;A = 48.5903 degrees ; in this case percentage of light escaped = [1 – cos(A)]/2 which gives 0.169281 which is nearly 16.92% or 17 %
in this question you’ve to use the concept of the circle of illuminanceA = Angle at which total internal reflection is occurring, in this case, it is given by sin(A) = 1/mu. mu = 4/3;A = 48.5903 degrees ; in this case percentage of light escaped = [1 – cos(A)]/2 which gives 0.169281 which is nearly 16.92% or 17 %pls approve
Dear student A = Angle at which total internal reflection is occurring, in this case, it is given by sin(A) = 1/mu. mu = 4/3;A = 48.5903 degrees ; in this case percentage of light escaped = [1 – cos(A)]/2 which gives 0.169281 which is nearly 16.92% Hope it helps
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