Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

Please give a detailed answer to this question .
 

Asked by Varsneya Srinivas 25th October 2017, 8:51 PM
Answered by 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 Expert 11th December 2017, 6:45 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp