# pls help......................................................................

Parth
58 Points
3 years ago
in this question you’ve to use the concept of the circle of illuminance
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% or 17 %
Parth
58 Points
3 years ago
in this question you’ve to use the concept of the circle of illuminance
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% or 17 %
pls approve
Arun
25750 Points
3 years ago
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