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

Let the individual (monochromatic) sources in Young’s experiment have the same intensity $$I_0$$.
For any point on the screen, the resultant intensity is

$$I = 4I_0 \cos^{2}\!\left(\frac{\delta}{2}\right)$$

where $$\delta$$ is the phase difference between the two interfering waves at that point and $$4I_0$$ is the maximum intensity $$I_{\text{max}}$$.

The problem states that the observed intensity is $$\tfrac{3}{4}$$ of the maximum, so

$$I = \frac{3}{4}\,I_{\text{max}} = \frac{3}{4}\,(4I_0) = 3I_0$$

and therefore

$$3I_0 = 4I_0 \cos^{2}\!\left(\frac{\delta}{2}\right)$$

Dividing by $$4I_0$$ gives

$$\cos^{2}\!\left(\frac{\delta}{2}\right) = \frac{3}{4}$$

Taking the positive square root (intensity depends on $$\cos^{2}$$, so the sign of the cosine itself is immaterial):

$$\cos\!\left(\frac{\delta}{2}\right) = \frac{\sqrt{3}}{2}$$

The standard angle that satisfies this is

$$\frac{\delta}{2} = \frac{\pi}{6} \quad\Longrightarrow\quad \delta = \frac{\pi}{3}$$

Phase difference $$\delta$$ and path difference $$\Delta$$ are connected by

$$\delta = \frac{2\pi}{\lambda}\,\Delta$$

Hence

$$\frac{\pi}{3} = \frac{2\pi}{\lambda}\,\Delta \; \Longrightarrow \; \Delta = \frac{\lambda}{6}$$

The statement of the question writes the same path difference as $$\Delta = \frac{\lambda}{x}$$. Equating the two forms:

$$\frac{\lambda}{x} = \frac{\lambda}{6} \;\Longrightarrow\; x = 6$$

Therefore the required value is 6.