Question

Two waves $Y_1 = asin\omega t$ and $Y_2 =asin(\omega t+\delta)$ are producing interference, then resultant intensity is proportional to -

• A. $acos^2 \delta/2$
• B. $a^2cos \delta/2$
• C. $a^2cos \delta$
• D. $a^2cos^2 \delta/2$

Answer 1

Answer: (D)

$a^2cos^2 \delta/2$

Answer 2

The two waves are given by $Y_1 = a \sin(\omega t)$ and $Y_2 = a \sin(\omega t + \delta)$. The resultant displacement $Y_{res}$ is the sum of the individual displacements: $Y_{res} = Y_1 + Y_2 = a \sin(\omega t) + a \sin(\omega t + \delta)$ Using the trigonometric identity $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$: $Y_{res} = a \left[ 2 \sin\left(\frac{\omega t + \omega t + \delta}{2}\right) \cos\left(\frac{\omega t - (\omega t + \delta)}{2}\right) \right]$ $Y_{res} = 2a \sin\left(\omega t + \frac{\delta}{2}\right) \cos\left(-\frac{\delta}{2}\right)$ Since $\cos(-\theta) = \cos(\theta)$: $Y_{res} = 2a \cos\left(\frac{\delta}{2}\right) \sin\left(\omega t + \frac{\delta}{2}\right)$ The resultant amplitude $A_{res}$ is the coefficient of the sine term: $A_{res} = 2a \cos\left(\frac{\delta}{2}\right)$ The intensity ($I$) of a wave is directly proportional to the square of its amplitude ($A$): $I \propto A^2$ Therefore, the resultant intensity $I_{res}$ is proportional to the square of the resultant amplitude $A_{res}$: $I_{res} \propto A_{res}^2$ $I_{res} \propto \left(2a \cos\left(\frac{\delta}{2}\right)\right)^2$ $I_{res} \propto 4a^2 \cos^2\left(\frac{\delta}{2}\right)$ Thus, the resultant intensity is proportional to $a^2 \cos^2\left(\frac{\delta}{2}\right)$.