Given equation of a travelling harmonic wave is $y(x, t) = 2.0 \cos 2\pi(10t - 0.0080x + 0.35)$ ...(i) The standard equation of a travelling harmonic wave is $y(x, t)=A \cos \left[ \frac{2\pi}{T} t - \frac{2\pi}{\lambda} x + \phi_0 \right]$...(ii) Comparing equation(i) and (ii), we get $\frac{2\pi}{\lambda}=2\pi \times 0.0080 \mathrm{cm}^{-1}$...(iii) $\frac{2\pi}{T}=2\pi \times 10$ and $\phi_0=0.35$ We know that, phase difference = $\frac{2\pi}{\lambda} \times$ path difference ...(iv) (a) When path difference = 4m = 400 cm, then from (iv)phasedifference = $\frac{2\pi}{\lambda} \times 400$ $= 6.4\pi$ rad (b) When path difference = 0.5 m = 50 cm, then phase difference $= 2\pi \times 0.0080 \times 50$ $= 0.8\pi$ rad (c) When path difference = $\frac{\lambda}{2}$ , then phase difference = $\frac{2\pi}{\lambda} \times \frac{\lambda}{2} = \pi$ rad (d)When path difference $= \frac{3\pi}{4}$ phase difference = $\frac{2\pi}{\lambda} \times \frac{3\lambda}{4} = \frac{3\pi}{2}$rad $= \left(\pi + \frac{\pi}{2}\right)$ ∴ effective phase difference = $\frac{\pi}{2}$ rad