Question

Two coherent light sources, each of wavelength $\lambda$, are separated by a distance of $3\lambda$. The maximum number of minima formed on the line $\text{AB}$, which runs from $-\infty$ to $+\infty$, is

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is C $6$

The condition of path difference for minima is given by

$\Delta \text{x}=\left(\text{2n+1}\right)\frac{\lambda }{2}$

From given diagram in question,

$\Delta x=d\sin \theta =3\lambda \sin \theta$

$\left(\text{2n+1}\right)\frac{\lambda }{2}=3\lambda \sin \theta$

$\sin \theta =\left(\text{2n+1}\right)\frac{1}{6}$

As we know that, $-1\le \sin \theta \le 1$

$-1\le \left(\text{2n+1}\right)\frac{1}{6}\le 1$

After solving above inequality,

$\therefore n=-3,-2,-1,0,1,2$

So, Total minima obtained along $\text{AB}=6$

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