Hint: The signal received by the two receivers depends on the phase of the signals.

Step 1: Find the signal received by ${\mathrm{R}}_1$.

(i) Consider the disturbances at the receiver ${\mathrm{R}}_1$ which is at distance d from B.
Let the wave at ${\mathrm{R}}_1$, because of A be $y_A$ = cos $\omega t$. The path difference of the signal from A with that from B is $\lambda /2$ and hence, the phase difference is $\mathrm{\pi }$.
Thus, the wave at ${\mathrm{R}}_1$, because of B is $y_B=\cos (\omega t-\pi )=-a\cos \omega t$
The path difference of the signal from C with that from A is $\mathrm{\lambda }$ and hence the phase difference
is 2$\pi$.
Thus, the wave at ${\mathrm{R}}_1$, because of C is $y_C= \cos \left(\omega t-2\pi \right)=a\cos \omega t$

The path difference between the signal from D with that of A is;

$P=R_{1}D-R_{1}A=\sqrt{{\left(R_{1}B\right)}^2+{\left(BD\right)}^2}-\left(R_{1}B-AB\right)$
$\begin{array}{rl}=\sqrt{d^2+{\left(\frac{\lambda }{2}\right)}^2}-(d-\lambda /2)& =d{\left(1+\frac{\lambda }{4d^2}\right)}^{1/2}-d+\frac{\lambda }{2}\end{array}$
$=d(1+\frac{{\lambda }^2}{8d^2})-d+\frac{\lambda }{2}=\frac{{\lambda }^2}{8d^2}+\frac{\lambda }{2}=\frac{\lambda }{2} (\because d>>\lambda )$

Therefore, the phase difference is $\pi$.

$y_D=a \cos (\omega t-\pi )=-a \cos \omega t$

Thus, the signal picked up at $R_1$ from all four sources is

$y_{R_1}=y_A+y_B+y_C+y_D$
$=a \cos \omega t -a \cos \omega t+ a \cos \omega t-a \cos \omega t=0$

Step 2: Find the signal received by $R_2$.

Let the signal picked up at $R_2$ from B be $y_B=a_1 \cos \omega t$

The path difference between signal at D and that at B is $\lambda /2$.

$\therefore y_D=-a_1\cos \omega t$

The path difference between signal at A and that at B is 

$\sqrt{{\left(d\right)}^2+{\left(\frac{\lambda }{2}\right)}^2}-d=d{\left(1+\frac{{\lambda }^2}{4d^2}\right)}^{1/2}-d=\frac{{\lambda }^2}{8d}$

As d >> $\lambda$, therefore this path difference $\to 0$

and, phase difference = $\frac{2\pi }{\lambda }\left(\frac{1}{8}\frac{{\lambda }^2}{d}\right)=0$
$$

Hence,    $y_A=a_1 \cos (\omega t-\varphi )$

similarly, $Y_C=a_1\cos (\omega t-\varphi )$

$\therefore$The signal picked up by $R_2$,

$\begin{array}{rl}{y}_A+y_B+y_C+y_D& =y=2a_1\cos (\omega t-\varphi )\\ \therefore |y{|}^2& =4a_1^2{\cos }^2(\omega t-\varphi )\\ \therefore <I>& =2a_1^2\end{array}$

Thus $R_1$ picks up the larger signal.

Step 3: Find the signal received by $R_1$ and $R_2$ when B is switched off.

(ii) lf B switched off, 

$R_1$ picks up y=acos$\omega t$

$\begin{array}{rl}\therefore ⟨I_{R_1}⟩& =\frac{1}{2}{a}^2\\ R_2 picks up y& =a\cos \omega t\\ \therefore ⟨I_{R_2}⟩& =a^2<{\cos }^2\omega t>=\frac{a^2}{2}\end{array}$

Thus $R_1 and R_2$ picks up the same signal.

Step 4: Find the signal received by $R_1$ and $R_2$ when D is switched off.

(iii) lf D is switched off,

$R_1$ picks up y=acos$\omega t$

$\begin{array}{rl}\therefore ⟨I_{R_1}⟩& =\frac{1}{2}{a}^2\\ R_2 picks up y& =3a\cos \omega t\\ \therefore ⟨I_{R_2}⟩& =9a^2<{\cos }^2\omega t>=\frac{9a^2}{2}\end{array}$

(iv) Thus $R_2$ picks up a larger signal compared to $R_1$ indicated B has been switched off and an enhanced signal at $R_2$ indicated D has been switched off.