Q.Four identical, mutually coherent monochromatic point sources $A$, $B$, $C$, $D$ emit waves of the same wavelength $\lambda$, all in phase. They are arranged as follows: $B$ is at the origin; $A$ lies a distance $\lambda/2$ to the left of $B$ and $C$ lies a distance $\lambda/2$ to the right of $B$, so that $A$, $B$, $C$ are collinear on a horizontal line with $AB = BC = \lambda/2$; $D$ lies a distance $\lambda/2$ directly below $B$ ($BD = \lambda/2$). Two receivers sit at great but equal distances $d$ from $B$ (with $d \gg \lambda$): receiver $R_1$ lies far out along the horizontal line to the left, on the same line as $A$–$B$–$C$ and beyond $A$, with $R_1B = d$; receiver $R_2$ lies far out along the vertical directly above $B$, with $R_2B = d$.

Method

Let every source have amplitude $a$ and be in phase at emission. At a receiver the contribution of a source is $a\cos(\omega t-kr)$ with $k=2\pi/\lambda$; only the relative path $r$ matters. A path difference of $\lambda/2$ corresponds to a phase of $\pi$ (a sign reversal). Because $d\gg\lambda$, a source displaced perpendicular to the line joining $B$ to a receiver is essentially equidistant with $B$ (path change $\sim(\lambda/2)^2/2d\to0$).

Receiver $R_1$ (far along the horizontal source line, to the left)

Distances (relative to $B$'s path $d$): $A$ is nearer by $\lambda/2$; $C$ is farther by $\lambda/2$; $D$ is perpendicular, so $\approx d$.

• $A$: path $-\lambda/2\Rightarrow$ phase $+\pi\Rightarrow$ phasor $-a$.
• $C$: path $+\lambda/2\Rightarrow$ phase $-\pi\equiv\pi\Rightarrow$ phasor $-a$.
• $B$: phasor $+a$; $D$: phase $0\Rightarrow$ phasor $+a$.

Receiver $R_2$ (far along the vertical, above $B$)

Now $A$ and $C$ are perpendicular to $BR_2$, so $\approx d$; $D$ is farther by $\lambda/2$.

• $A,B,C$: phase $0\Rightarrow$ phasors $+a$ each.
• $D$: path $+\lambda/2\Rightarrow$ phase $\pi\Rightarrow$ phasor $-a$.

(i) $R_1=0$, $R_2=2a$, so $R_2$ picks up the larger signal.

(ii) $B$ turned off

• $R_1=(-a)+(-a)+a=-a\Rightarrow$ amplitude $a$.
• $R_2=a+a-a=a\Rightarrow$ amplitude $a$. …

Method: Phasor (Vector) Addition of Several Coherent Sources at a Distant Receiver

Use this method whenever multiple coherent, in-phase point sources are combined

and you need the resultant signal at one or more distant observation points —

this is a direct generalisation of two-source interference to N sources.

Steps

Step 1: Convert each source's path difference into a phase

For a receiver far away in a given direction, work out the path-length

difference of each source relative to one reference source. Convert that path

difference to a phase using

A path difference of exactly $\lambda/2$ corresponds to a phase of $\pi$ — i.e.

a sign flip relative to the reference source's phasor.

Step 2: Identify which sources are effectively equidistant

For a receiver very far away ($\text{distance} \gg$ source spacing), a source

displaced perpendicular to the line joining the reference source to that

receiver contributes essentially zero extra path (the correction is second

order and negligible) — treat such a source as in phase with the reference.

Only sources displaced along the line to that particular receiver pick up a

genuine path difference.

Step 3: Sum the phasors algebraically

Since every source shares the same amplitude $a$ here, and each phase works out

to either $0$ or $\pi$ (in-phase or exactly opposite), the vector/phasor sum

collapses to simple addition and subtraction:

with a $+a$ for each in-phase source and $-a$ for each source shifted by …