In Young's double slit experiment, the intensity on screen at a point where path difference is $\frac{\lambda}{4}$ is $\frac{K}{2}$. The intensity at a point when path difference is ' $\lambda$ ' will be

The intensity $I$ in Young's double-slit experiment varies with phase as $I = I_0 \cos^2 \left( \frac{\phi}{2} \right)$, where $I_0$ is the maximum intensity, and $\phi = \frac{2\pi}{\lambda} \Delta x$ relates the phase difference to the path difference $\Delta x$.

Given that the intensity is $\frac{K}{2}$ for a path difference of $\frac{\lambda}{4}$, the corresponding phase difference is $\phi_1 = \frac{\pi}{2}$.
Since $\cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$, the intensity is
$I_1 = I_0 \cos^2 \left( \frac{\pi}{4} \right) = \frac{I_0}{2}$.
Equating $\frac{I_0}{2} = \frac{K}{2}$ yields $I_0 = K$.

For a path difference of $\lambda$, the phase difference becomes $\phi_2 = 2\pi$.
The intensity is then
$I_2 = I_0 \cos^2 (\pi) = I_0$.
Substituting $I_0 = K$, we find that $I_2 = K$.

Final answer: $K$