In a Young double slit experiment, the wavelength of incident light is \(6000\) Å, the separation between slits \(S_1\) and \(S_2\) is \(5\) cm and the distance between slits plane and screen is \(50\) cm, as shown in the figure below. If the resultant intensity at \(P\) is equal to the intensity due to individual slits, the path difference between interfering waves is __________ Å.

Let the intensity due to each individual slit be \(I_0\).
The resultant intensity \(I_R\) at a point where the phase difference is \(\phi\) is given by:

\(I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi\)

Given that \(I_1 = I_2 = I_0\) and the resultant intensity \(I_R = I_0\), we can substitute these values into the equation:

\(I_0 = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} \cos \phi\)

\(I_0 = 2I_0 + 2I_0 \cos \phi\)

\(I_0 = 2I_0 (1 + \cos \phi)\)

\(1 + \cos \phi = \dfrac{1}{2}\)

\(\cos \phi = -\dfrac{1}{2}\)

The minimum phase difference satisfying this condition is:
\(\phi = \dfrac{2\pi}{3}\)

The relationship between phase difference \(\phi\) and path difference \(\Delta x\) is:
\(\phi = \dfrac{2\pi}{\lambda} \Delta x\)

Substituting the value of \(\phi\):
\(\dfrac{2\pi}{3} = \dfrac{2\pi}{\lambda} \Delta x\)

\(\Delta x = \dfrac{\lambda}{3}\)

Given the wavelength \(\lambda = 6000 \text{ \AA}\):
\(\Delta x = \dfrac{6000}{3} = 2000 \text{ \AA}\)

Answer: \(2000\)