Explanation

(d) Let the equation of two waves are
$y_{1} = a\sin(\omega t - kx)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
And $y_{2} = a\sin(\omega t - kx + \phi)\ \ldots(ii)$
When they superpose, the resultant wave is
$y = y_{1} + y_{2}$
$= a\left\lbrack \sin{(\omega t - kx) + \sin(\omega t - kx + \phi)} \right\rbrack$
$= a\left\lbrack 2\sin{\left( \omega t - kx + \frac{\phi}{2} \right)\cos\left( - \frac{\phi}{2} \right)} \right\rbrack$
$= 2a\sin{\left( \omega t - kx + \frac{\phi}{2} \right)\cos\frac{\phi}{2}}$
$= \left( 2a\cos\frac{\phi}{2} \right)\sin\left( \omega t - kx + \frac{\phi}{2} \right)\ \ldots(iii)$
Comparing Eq. (iii) with (i) or (ii), we get
$a = 2a\cos\frac{\phi}{2} \Longrightarrow \cos{\frac{\phi}{2} = \frac{1}{2}}$
$\Longrightarrow \ \cos{\frac{\phi}{2} = \cos\frac{\pi}{3}}\ \ $
$\therefore\ \ \cos{\frac{\phi}{2} = \cos\frac{\pi}{3}}\ \ $
$\therefore\ \ \frac{\phi}{2} = \frac{\pi}{3}$
$or\ \ \ \ \ \ \ \phi = \frac{2\pi}{3}$