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

The maximum intensity in Young's double slit experiment is given as I_\max = K.

The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula:

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

Given the path difference $\Delta x = \dfrac{\lambda}{6}$, the phase difference is:

$\phi = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{6} = \dfrac{\pi}{3}$

The intensity $I$ at any point on the screen is given by:

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

Substituting the values of I_\max and $\phi$:

$I = K \cos^2\left(\dfrac{\pi}{6}\right)$

Since $\cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}$, we get:

$I = K \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3K}{4}$

Answer: $\dfrac{3K}{4}$