Path Difference in Young Double Slit Experiment

To find the value of $x$, we start by using the expression for the resultant intensity $I$ in a Young's double-slit experiment involving two identical coherent sources:

$I = I_{\max} \cos^2\left(\frac{\phi}{2}\right)$

where:

• $I_{\max}$ is the maximum intensity of the interference pattern.
• $\phi$ is the phase difference between the two interfering waves.

According to the given problem, the intensity at the point on the screen is $\frac{3}{4}$ of the maximum intensity:

$I = \frac{3}{4} I_{\max}$

Substituting this value into the intensity equation gives:

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

Dividing both sides by $I_{\max}$:

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

Taking the square root on both sides for the minimum phase difference:

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

Solving for $\frac{\phi}{2}$:

$\frac{\phi}{2} = \frac{\pi}{6}$

$\phi = \frac{\pi}{3}$

The relation between the phase difference $\phi$ and the path difference $\Delta x$ is given by:

$\phi = \frac{2\pi}{\lambda} \Delta x$

Substitute $\phi = \frac{\pi}{3}$ into the equation:

$\frac{\pi}{3} = \frac{2\pi}{\lambda} \Delta x$

Solving for the path difference $\Delta x$:

$\Delta x = \frac{\lambda}{6}$

Given that the path difference is in the form $\frac{\lambda}{x}$, comparing the two expressions yields:

$x = 6$