In Young's double slit experiment, the intensity on the screen at a point where path difference is λ is K. What will be the intensity at the point where path difference is λ/4.

$\frac{K}{2}$

By using phase difference $\varphi=\frac{2 \pi}{\lambda}(\Delta)$

For path difference $\lambda$, phase difference $\varphi_1=2 \pi$ and for path difference $\lambda / 4$, phase difference $\phi_2=\pi / 2$

Also by using $I=4 I_0 \cos ^2 \frac{\varphi}{2} \Rightarrow \frac{I_1}{I_2}=\frac{\cos ^2\left(\varphi_1 / 2\right)}{\cos ^2\left(\varphi_2 / 2\right)} \Rightarrow \frac{K}{I_2}=\frac{\cos ^2(2 \pi / 2)}{\cos ^2\left(\frac{\pi / 2}{2}\right)}=\frac{1}{1 / 2} \Rightarrow I_2=\frac{K}{2}$