Question

Two points sources separated by $2.0\text{m}$ are radiating in phase with $\lambda =0.50\text{m}.$ A detector moves in a circular path around the two sources in a plane containing them. How many maxima are detected ?

• A. $16$
• B. $20$
• C. $24$
• D. $32$

Answer 1

The correct option is A $16$


Condition of path difference for maxima is given by

$\Delta \text{x}=n\lambda$

According to the figure shown, path difference at any point is given by,

$\Delta \text{x}=S_{1}P=d\cos \theta$

$\Rightarrow d\cos \theta =n\lambda .......\left(1\right)$

$\cos \theta =\frac{n\lambda }{d}=n(\frac{0.50}{2.0})=0.25n$

As $\cos \theta$ lies between $-1$ and $1$, so we wish to find all values of $n$ for which $-1\le \left(0.25n\right)\le 1$.

So, corresponding $n$ values are $-4,-3,-2,-1,0,+1,+2,+3,+4$.

For each of these, there are two different values of $\theta$ except for $-4$ and $+4$.

A single value of $\theta$, $-{90}^0$ and $+{90}^0$, is associated with $n=-4$ and $n=+4$, respectively. Thus, there are $16$ different angles in all which satisfies equation $\left(1\right)$, and therefore $16$ maxima will be detected by the detector.

Hence, option $\left(\text{A}\right)$ is correct.

Alternate solution: At the end of vertical diameter, zeroth maxima $(\Delta x=0)$ is obtained, and at the end of horizontal diameter $4\text{th}$ maxima $(\Delta x=4\lambda )$ is obtained. Hence, there are three maxima in each quadrant and another four maxima at the ends of horizontal and vertical diameters. Therefore, total maxima $=(3\times 4)+4=16$.