Explanation

(d) It a is amplitude of each wave,
$I_{0} = k(a + a)^{2} = 4ka^{2}$
Let $\phi$ be the phase difference to obtain the intensity$I_{0}/2$.
$\therefore\frac{I_{0}}{2} = ka_{r}^{2} = k(a^{2} + a^{2} + 2aa\cos\phi)$
$= k2a^{2}(1 + cos\phi) = k4a^{2}\cos^{2}\frac{\phi}{2}$
$= I_{0}\cos^{2}{\phi/2}$
$\cos{\frac{\phi}{\sqrt{2}} = \frac{1}{\sqrt{2}} = \cos{45{^\circ}}}\therefore\phi = 90{^\circ}$.
If $\mathrm{\Delta}x$ is path difference between the two waves, then
$\mathrm{\Delta}x = \frac{\lambda}{2\pi}\phi = \frac{\lambda}{2\pi}\left( \frac{\pi}{2} \right) = \frac{\lambda}{4}$
Therefore, displacement of sliding tube $\frac{1}{2}(\Delta x) = \lambda/8$