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Asked by eknathmote50 | 25 May, 2019, 11:43: AM
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Transmitter-B is lagging a phase difference of π/2 with respect to transmitter-A.
 
if phase difference is  π/2 and  path difference  is x,  then (2π/λ)x = π/2   or x = λ/4
 
where λ is wavelength that equals 20 m
 
Hence to get maximum intensity at C, pathe difference BC-AC = nλ-(λ/4)  = [ n -(1/4) ]λ  ..............(1)
 
where n is integer.
 
eqn.(1) is satisified only for the two cases of path difference i.e, 55 m and 75 m
 
for 55 m, we get path difference = [ 3 -(1/4)]×20 = 55 m
for 75 m, we get path difference = [ 4 - (1/4)]×20 = 75 m
 
for other given path differences, i.e., 60m and 65 m, if we apply the condition given in eqn.(1), we will not get integer value for n.
 

Answered by Thiyagarajan K | 25 May, 2019, 18:04: PM

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