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(D) The intensity of the resultant wave in interference is given by $I = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference and $I_0$ is the in

Vedclass · Accepted Answer

$\infty:1$

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(D) The intensity of the resultant wave in interference is given by $I = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference and $I_0$ is the intensity of each individual source.Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.For path difference $\Delta x = 0$,the phase difference $\phi = 0$. Thus,$I_1 = 4I_0 \cos^2(0) = 4I_0$.For path difference $\Delta x = \lambda/2$,the phase difference $\phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{2} = \pi$. Thus,$I_2 = 4I_0 \cos^2(\pi/2) = 4I_0 \cdot 0 = 0$.The ratio of intensities is $\frac{I_1}{I_2} = \frac{4I_0}{0} = \infty$.Therefore,the ratio is $\infty:1$.

Light waves from two coherent sources arrive at two points on a screen with path difference of zero and $\lambda / 2$. The ratio of the intensities at the points is

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