Four identical monochromatic source A, B, C, D as shown in figure produce waves of the same wavelength `lambda` and are coherent. Two receivers `R_(1)` and `R_(2)` are at great but equal distance from B. The choose the correct statements.

Explanation

In Fig. A, B, C, D are four idential monochromatic coherent sources producing waves of same wavelength `lambda`.
`AB = BC = BD = lambda//2`.
`R_(1)` and `R_(2)` are two receivers held as shown such that `R_(1)B = R_(2)B = d (gt gt lambda)`.
Let the wave at `R_(1)` from `B and A = lambda//2` so that phase diff. is `pi`.
`:.` The wave at `R_(1)` because of source `B` is
`y_(B) = a cos (omega t - pi) = - a cos omega t`
Similarly, path diff. of singal at `R_(1)` from C and A `= lambda` so that phase diff. `= 2pi`
`:.` The wave at `R_(1)` because of source `C` is
`y_(C) = a cos (omega t - 2pi) = a cos omega t`
Again, path diff. of signal at `R_(1)` From `D` that of `A` is
`= sqrt(d^(2) + ((lambda)/(2))^(2)) - d(d - lambda//2) = d (1 + (lambda^(2))/(4 d^(2)))^(1//2) - d + (lambda)/(2) = d(1 + (lambda^(2))/(8 d^(2))) - d + (lambda)/(2) ~= (lambda)/(2)` (`because d gt gt lambda`)
Therefore, phase diff. is `pi`
`y_(D) = a cos (omega t - pi) = - a cos omega t`
Hence the signal picked up at `R_(1)` from all the four sources is
`y_(R_(1)) = y_(A) + y_(B) + y_(C) + y_(D) = a cos omega t - a cos omega t + a cos omega t - a cos omega t = 0`. ...(i)
Let the the signal picked up at `R_(2)` from to be `y_(B) = a_(1) cos omega t`
Path differecne of signals at `R_(2)` from `D` to that from `B` is `lambda//2`. Therefore, `y_(D) = a_(1) cos (omega t - pi)`
`= - a_(1) cos omega t`.
Again, path diff. of signals at `R_(2)` from `B` and `A` is
`= sqrt(d^(2) + (lambda//2)^(2)) - d = d (1 + (lambda^(2))/(4 d^(2)))^(1//2) -d = (1)/(8)(lambda^(2))/(d^(2))`
As `d gt gt lambda`, therefore this path diff. `rarr 0`, and phase diff. `phi rarr 0 :. y_(A) = a_(1) cos (omega t - phi)`
Similarly, `y_(C) = a_(1) cos (omega t - phi)`
Hence the signal picked up at `R_(2)` from all the four sources is
`y_(R_(2)) = y_(A) + y_(B) + y_(C) + y_(D) = a_(1) cos (omega t - phi) + a_(1) cos omega t + a_(1) cos (omega t - phi) - a_(1) cos omega t`
`y_(R_(2)) = 2 a_(1) cos (omega t - phi)` ...(ii)
Hence receiver `R_(2)` picks up the larger signal.
(ii) If source `B` is swithched off, then from (i),
`y_(R_(1)) = a cos omega t lt I_(R_(1)) gt = a^(2) lt cos^(2) omega t gt = (a^(2))/(2)`
From (ii), `y_(R_(2)) = a_(1) cos (omega t - phi) = a_(1) cos omega t, because phi rarr 0 lt I_(R_(2)) gt = a_(1)^(2) lt cos^(2) omega t gt = (a_(1)^(2))/(2) = (a^(2))/(2)`.
Hence `R_(1)` and `R_(2)` pick up the same signal.
(iii) When source `D` is swithched off, `Y_(R_(1)) = a cos omega t`
`:. lt I_(R_(1)) gt = (a^(2))/(2)` and `y_(R_(2)) = 3a cos omega t` (Taking `phi rarr 0`)
`lt I_(R_(2)) gt = 9 a^(2) lt cos^(2) omega t gt = (9)/(2)a^(2)` `:. R_(2)` picks up larger signal compared to `R_(1)`.
(iv) Hence a signal at `R_(1)` indicates that source `B` has been switched off and an anhanced signal at `R_(2)` indicates that source `D` has been switched off.