The equation of two interfering light waves are y_1 = b ω t and y_2...

The given equations of the light waves are:

y_1 = b \cos \omega t = b \sin \left(\omega t + \dfrac{\pi}{2}\right)<br><br> y_2 = b \sin($\omega$t + \Phi)

The phase difference between the $two waves is:<br><br>\Delta \phi = (\omega$ t + \Phi) - \left(\omega$t +$\dfrac{\pi}{2}\right) = \Phi - \dfrac{\pi}{2}

For destructive interference,$the$phase difference must be an odd multiple of \pi:

\Delta \phi = (2n - 1)\pi

$\Phi -$\dfrac{\pi}{2} = (2n - 1)\pi

$\Phi = (2n - 1)$\pi + \dfrac{\pi}{2} = \dfrac{(4n - 1)\pi}{2}

Substituting $n = 1, 2, 3, \ldots$, we get the possible values of $\Phi$:

$\Phi =$\dfrac{3\pi}{2}, \dfrac{7\pi}{2}, \dfrac{11\pi}{2}, \ldots

Answer: \dfrac{3\pi}{2}, \dfrac{7\pi}{2}, \dfrac{11\pi}{2}, \ldots\ldots