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Question

(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \cdot \Delta x

Vedclass · Accepted Answer

$\infty: 1$

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(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \cdot \Delta x$.For the first point,$\Delta x_1 = 0$,so $\Delta \phi_1 = 0$. The intensity is $I_1 = I_{max} \cos^2(\frac{\Delta \phi_1}{2}) = I_{max} \cos^2(0) = I_{max}$.For the second point,$\Delta x_2 = \frac{\lambda}{2}$,so $\Delta \phi_2 = \frac{2 \pi}{\lambda} \cdot \frac{\lambda}{2} = \pi$. The intensity is $I_2 = I_{max} \cos^2(\frac{\pi}{2}) = 0$.The ratio of intensities is $\frac{I_1}{I_2} = \frac{I_{max}}{0} = \infty$.Thus,the ratio is $\infty: 1$.

Light waves from two coherent sources arrive at two points on a screen with path difference of $0$ and $\frac{\lambda}{2}$. The ratio of intensities at the points is $(\cos 0^{\circ}=1, \cos \pi=-1)$.

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