In Young's double slit experiment, the intensity on screen at a point, where path difference is…

The intensity at a point in Young's double-slit experiment is given by $I = I_{\max} \cos^2\left(\frac{\phi}{2}\right)$, where the phase difference relates to path difference by $\phi = \frac{2\pi}{\lambda} \Delta x$.

When the path difference is $\Delta x_1 = \frac{\lambda}{4}$, the intensity is $I_1 = \frac{K}{4}$.
The corresponding phase difference is $\phi_1 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
Substituting into the intensity formula yields $I_1 = I_{\max} \cos^2\left(\frac{\pi}{4}\right) = \frac{I_{\max}}{2}$.
Equating with the given intensity gives $\frac{I_{\max}}{2} = \frac{K}{4}$, so $I_{\max} = \frac{K}{2}$.

For a path difference of $\Delta x_2 = \lambda$, the phase difference becomes $\phi_2 = 2\pi$.
The intensity is then $I_2 = I_{\max} \cos^2(\pi) = I_{\max}$.
Substituting the maximum intensity value yields $I_2 = \frac{K}{2}$.

Final answer: $\frac{K}{2}$