In a Young's double slit experiment, the intensity at some point on the screen is found to be \(\dfrac{3}{4}\) times of the maximum of the interference pattern. The path difference between the interfering waves at this point is \(\dfrac{\lambda}{x}\) where \(\lambda\) is wavelength of the incident light. The value of \(x\) is _______.

The intensity at a point in Young's double slit experiment is given by \(I = I_{max} \cos^2\left(\dfrac{\phi}{2}\right)\)

Given \(I = \dfrac{3}{4} I_{max}\)

\(\dfrac{3}{4} I_{max} = I_{max} \cos^2\left(\dfrac{\phi}{2}\right)\)

\(\cos\left(\dfrac{\phi}{2}\right) = \dfrac{\sqrt{3}}{2}\)

\(\dfrac{\phi}{2} = \dfrac{\pi}{6}\)

\(\phi = \dfrac{\pi}{3}\)

The phase difference \(\phi\) is related to the path difference \(\Delta x\) by \(\phi = \dfrac{2\pi}{\lambda} \Delta x\)

\(\dfrac{2\pi}{\lambda} \Delta x = \dfrac{\pi}{3}\)

\(\Delta x = \dfrac{\lambda}{6}\)

Comparing with \(\Delta x = \dfrac{\lambda}{x}\), we get \(x = 6\)

Answer: \(6\)