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CBSE Class 12-science Answered

Please solve and also explain the reason during the same...
Asked by nitishkrnehu09 | 18 Nov, 2017, 10:32: PM
answered-by-expert Expert Answer
Let two waves from S1 and S2 interfere at P.
 
The two waves are given by
begin mathsize 12px style E subscript S 1 end subscript equals space E subscript 1 sin space omega t E subscript S 2 end subscript equals space E subscript 2 sin space left parenthesis omega t plus delta right parenthesis end style
where δ is the phase difference between two waves
The resultant is given by

begin mathsize 12px style E subscript S 1 end subscript plus E subscript s 2 end subscript space space equals space E subscript 1 sin space omega t space plus space E subscript 2 sin space left parenthesis omega t plus delta right parenthesis space equals space left parenthesis E subscript 1 plus E subscript 2 cos delta right parenthesis sin space omega t space plus space E subscript 2 sin space delta space cos space omega t end style
(student should workout the required trignometry/algebra here )
 
Above equation shows that superposition of two sinusoidal waves with same frequency but with
phase difference gives superposition of two sinusoidal waves of same frequency but different amplitudes.
 
begin mathsize 12px style L e t space E subscript 1 plus E subscript 2 cos delta space equals space E space cos space phi a n d space E subscript 2 sin delta space equals space E space sin space phi end style
Where E is the resultant amplitude and φ is the resultant phase
 
By squaring and adding the above two equations we get
 
begin mathsize 12px style E squared space equals space E subscript 1 superscript 2 space plus space E subscript 2 superscript 2 space plus space 2 E subscript 1 E subscript 2 cos space delta end style
since Intensity I proportional to E2, we can write
 
begin mathsize 12px style I space equals space I subscript 1 plus I subscript 2 plus 2 square root of I subscript 1 I subscript 2 end root cos space delta end style
If I1 = I2 = I0 and δ = 0, we get maximum Imax = 4I0
If I1 = I2 = I0 and δ = 2π, we get minimum Imin = 0
If I1 = I2 = I0 and δ  ranging from 0 to 2π, we get intensity between Imax   and Imin
which is given by I = (1+cosδ)2I0 .
When this intensity is averaged out for δ = 0 to 2π, we get average value 2I0
 
Hence the option (A) in the question is correct answer



Answered by | 04 Dec, 2017, 12:18: PM

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