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CBSE Class 12-science Answered

Physic
Asked by balwansungh111334 | 13 Jul, 2022, 07:34: AM
Expert Answer
Resultant intensity IR of two waves of radiation of intensities I1 and I2 is given as
 
begin mathsize 14px style I subscript R space equals space I subscript P space plus space I subscript Q space plus space 2 space square root of I subscript P end root space square root of I subscript Q end root space cos ϕ end style  .............................(1)
where begin mathsize 14px style I subscript P end style is intensity of radiation of source P , begin mathsize 14px style I subscript Q end style is intensity of radiation of source Q and  begin mathsize 14px style ϕ end style is phase difference between sources
At point A , radiation from source P has path difference 5m compare to radiation from source Q.
 
Phase difference for 5 m path difference = (5/20)2π = π/2
 
Since radiation from source P is ahead of source Q by phase (π/2) , additional phase lag (π/2) of source P due to path difference
makes the two sources in phase at point A.
 
Hence if we substitute begin mathsize 14px style ϕ end style = 0 in eqn.(1), we get
begin mathsize 14px style I subscript A space equals open parentheses square root of I subscript P end root space plus space square root of I subscript Q end root close parentheses squared space space space space end style.................................(2)
At point B , there is no path difference for radiations from source P and source Q.
But initially source P is ahead of source Q by phase (π/2)

Hence if we substitute begin mathsize 14px style ϕ end style = (π/2) in eqn.(1), we get
begin mathsize 14px style I subscript B space equals space I subscript P space plus space I subscript Q space end style .........................................(3)
 
At point C , radiation from source P leads a path difference 5m, hence leads additional phase difference (π/2).
Since initially source P is ahead of source Q by phase (π/2), total phase difference becomes π .

Hence if we substitute begin mathsize 14px style ϕ end style = π in eqn.(1), we get
begin mathsize 14px style I subscript C space equals open parentheses square root of I subscript P end root space minus space square root of I subscript Q end root close parentheses squared space space space space end style............................(4)
If two sources have same intensity , then begin mathsize 14px style I subscript A space equals space I subscript B space equals space I end style
Hence using eqn.(2) , eqn.(3) and eqn.(4) , we get the ratio of intensities as
 
begin mathsize 14px style I subscript A space colon space I subscript B space colon space I subscript C space equals space left parenthesis 2 square root of I space right parenthesis squared space colon space 2 I space colon space 0 space space equals space 2 space colon space 1 space colon space 0 end style
Answered by Thiyagarajan K | 13 Jul, 2022, 09:21: AM
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