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pls help......................................................................

pls help......................................................................

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Grade:12

3 Answers

Parth
58 Points
2 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
2 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
25758 Points
2 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
 

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