In a Young's double slit experiment, the maximum intensity of the...

The intensity $I$ at any point in the interference pattern is given by $I = I_0 \cos^2\left(\dfrac{\phi}{2}\right)$, where $\phi$ is the phase difference.

Given $I = \dfrac{I_0}{2}$, we have:

$\dfrac{I_0}{2} = I_0 \cos^2\left(\dfrac{\phi}{2}\right)$

$\cos\left(\dfrac{\phi}{2}\right) = \dfrac{1}{\sqrt{2}}$

For the minimum non-zero path difference, we take the smallest positive angle:

$\dfrac{\phi}{2} = \dfrac{\pi}{4} \Rightarrow \phi = \dfrac{\pi}{2}$

The relationship between phase difference $\phi$ and path difference $\Delta x$ is $\phi = \dfrac{2\pi}{\lambda} \Delta x$.

Substituting the value of $\phi$:

$\dfrac{\pi}{2} = \dfrac{2\pi}{\lambda} \Delta x \Rightarrow \Delta x = \dfrac{\lambda}{4}$

Answer: $\dfrac{\lambda}{4}$