Explanation

(b) In interference phenomenon, when both the waves are in opposite phase, we got destructive interference (or minima). As resultant amplitude of two superimposed wave is
$R = \sqrt{a_{1}^{2} + a_{2}^{2} + 2a_{1}a_{2}\cos\phi}$
And resultat intensity of two superimposed wave is
$I = I_{1} + I_{2} + 2ka_{1}a_{2}\cos\phi$
For minima (destructive interference), phase difference should be odd multiple of $\pi.$
ie, $\phi = (2n - 1)\pi$
where, $n = 1,\ 2,\ 3,\ \ldots.$
As $\ \phi = \pi,3\pi,\ 5\pi,\ \ldots$
Hence, $\cos{\phi = \cos{(2n - 1)\pi = - 1}}$
So, $R_{\min} = \sqrt{a_{1}^{2} + \ a_{2}^{2} - 2a_{1}a_{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{\left( a_{1} - a_{2} \right)^{2}}$
Or $R_{\min} = a_{1} - a_{2}$
Also, $I = I_{1} + I_{2} - 2ka_{1}a_{2}$
Or $\ I = I_{1} + I_{2} - 2a_{1}\ \sqrt{k}a_{2}\sqrt{k}$
$= I_{1} + I_{2} - 2\sqrt{I_{1}I_{2}}$
Hence, $I < I_{1} + I_{2}$
Hence, when the phase difference is odd multiple of $\pi,$ then destructive interference is obtained. The resultant amplitude is equal to the difference of the individual amplitude and the resultant intensity is less than the sum of individual intensities.