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(D) The wavelength $\lambda$ is given by $\lambda = \frac{v}{f} = \frac{350}{350} = 1 m = 100 cm$.The path difference $(\Delta x)$ between the waves a

Vedclass · Accepted Answer

$0.5$

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(D) The wavelength $\lambda$ is given by $\lambda = \frac{v}{f} = \frac{350}{350} = 1 m = 100 cm$.The path difference $(\Delta x)$ between the waves at point $P$ is $\Delta x = AP - BP = 25 cm$.The phase difference $(\Delta \phi)$ is calculated as $\Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{100} 	imes 25 = \frac{\pi}{2} rad$.Since the phase difference is $\frac{\pi}{2}$,the two waves are perpendicular to each other in terms of phase. The resultant amplitude $A$ is given by $A = \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos(\Delta \phi)}$.Substituting the values: $A = \sqrt{(0.3)^2 + (0.4)^2 + 2(0.3)(0.4) \cos(\frac{\pi}{2})}$.Since $\cos(\frac{\pi}{2}) = 0$,we get $A = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 mm$.

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