Answer:
Option B
Explanation:
For path difference λ, phase difference = $ 2\pi$ ($Q = \left(\frac{2\pi}{\lambda}x=\frac{2\pi}{\lambda}\lambda=2\pi\right)$)
$I = I_{0}+I_{0}+2I_{0}\cos2\pi$
$I = 4I_{0} (\therefore\cos 2\pi = 1)$
For $x= \frac{\lambda}{4}$, Phase difference = $\frac{\pi}{2}$
$I' = I_{1}+I_{2}+2\sqrt{I_{1}}\sqrt{I_{2}}\cos\pi/2.$
$If = I_{1} = I_{2} = I_{0}$ then I' = $2I_{0}$ = 2.I/4 = I/2