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CBSE Class 12-science Answered

Please give a detailed answer to this question .
 
Asked by Varsneya Srinivas | 25 Oct, 2017, 08:51: PM
answered-by-expert Expert Answer
 
Due to total internal reflection light from point source escapes only through area A as illustrated in the first figure.
 
The circular area has to be calculated in spherical coordinates as shown in figure 2 by integrating the small area dA subtended by small solid angle dΩ.
 
begin mathsize 12px style A space equals space integral subscript 0 superscript theta subscript C end superscript r squared sin theta d theta integral subscript 0 superscript 2 pi end superscript d ϕ space equals space 2 pi r squared open parentheses 1 minus cos theta subscript C close parentheses end style 
Here θC  is the critical angle for total internal reflection and is given by 
 
begin mathsize 12px style mu sin theta subscript C space equals space mu subscript 0 space sin space 90 space sin c e space space mu subscript 0 space equals space 1 comma space sin theta subscript C space equals space 1 over mu end style
Fraction of energy escape is the fraction of area A to the total spherical area 4πr2 
 
begin mathsize 12px style fraction numerator A over denominator 4 pi r squared end fraction equals space fraction numerator 2 pi r squared open parentheses 1 minus cos theta subscript C close parentheses over denominator 4 pi r squared end fraction space equals space fraction numerator open parentheses 1 minus cos theta subscript C close parentheses over denominator 2 end fraction space equals space fraction numerator open parentheses 1 minus square root of 1 minus begin display style 1 over mu squared end style end root close parentheses over denominator 2 end fraction space equals space fraction numerator open parentheses mu minus square root of mu squared minus 1 end root close parentheses over denominator 2 mu end fraction end style
 

Answered by | 11 Dec, 2017, 06:45: PM

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