91Ó°ÊÓ

For destructive interference to occur, the path difference between the waves reaching the point on the x-axis must be an odd multiple of half the wavelength i.e., \((2n+1)\frac{\lambda}{2}\) where n is an integer.

Let us consider two waves originating from sources P and Q respectively. Let the point on the x-axis where destructive interference occurs be X. Then, we can express the path difference as: Path Difference = \(QX - PX\) Since Q is at a distance 3λ from P and the path difference is an odd multiple of half the wavelength, we can write: \(QX - PX = (2n +1) \frac{\lambda}{2}\)

Since we need to calculate distances, we can use the Pythagorean theorem to describe the distances PX and QX. \(PX^2 + PQ^2 = QX^2\) Substituting the given spacing between P and Q as well as the value of the path difference in the equation: \((PX)^2 + (3\lambda)^2 = (PX + (2n+1)\frac{\lambda}{2})^2\)

Let's simplify the equation and solve for PX: \(PX^2 + 9\lambda^2 = PX^2 + (2n+1)\lambda PX + (2n+1)^2\frac{\lambda^2}{4}\) As the terms \(PX^2\) cancel out, we get: \(9\lambda^2= (2n+1)\lambda PX + (2n+1)^2\frac{\lambda^2}{4}\) Now, we want to get maximum PX when n is minimum. The smallest value n can take is 0 (since it is an integer). When n = 0: \(9\lambda^2 = (2*0+1)\lambda PX + (2*0+1)^2\frac{\lambda^2}{4}\) \(9\lambda^2 = \lambda PX + \frac{\lambda^2}{4}\) Rearranging and solving for PX: \(PX = \frac{(9\lambda^2-\frac{\lambda^2}{4})}{\lambda} = \frac{35}{4}\lambda\) Hence, the maximum distance along the x-axis from P where a minimum intensity occurs is: PX = \(8.75\lambda\) The correct option is (C) \(8.75\lambda\).