Ratio of Intensities at Points with Different Path Differences

To find the ratio of intensities at points $A$ and $B$, we use the relationship between the path difference $\Delta x$, the phase difference $\phi$, and the resultant intensity $I$ in an interference pattern.

The resultant intensity $I$ at any point on the screen due to two coherent sources of equal intensity is given by: $I = I_{\max} \cos^2\left(\frac{\phi}{2}\right)$

where $I_{\max}$ is the maximum intensity, and the phase difference $\phi$ is related to the path difference $\Delta x$ by: $\phi = \frac{2\pi}{\lambda} \Delta x$

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1. Intensity at Point $A$:

Given the path difference at point $A$ is $\Delta x_A = \frac{\lambda}{3}$.

The corresponding phase difference $\phi_A$ is: $\phi_A = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3}$

Thus, the intensity at point $A$ is: $I_A = I_{\max} \cos^2\left(\frac{\phi_A}{2}\right) = I_{\max} \cos^2\left(\frac{\pi}{3}\right)$

Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$: $I_A = I_{\max} \left(\frac{1}{2}\right)^2 = \frac{I_{\max}}{4}$

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2. Intensity at Point $B$:

Given the path difference at point $B$ is $\Delta x_B = \frac{\lambda}{6}$.

The corresponding phase difference $\phi_B$ is: $\phi_B = \frac{2\pi}{\lambda} \left(\frac{\lambda}{6}\right) = \frac{\pi}{3}$

Thus, the intensity at point $B$ is: $I_B = I_{\max} \cos^2\left(\frac{\phi_B}{2}\right) = I_{\max} \cos^2\left(\frac{\pi}{6}\right)$

Since $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$: $I_B = I_{\max} \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3 I_{\max}}{4}$

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3. Ratio of Intensities:

Taking the ratio of $I_A$ to $I_B$: $\frac{I_A}{I_B} = \frac{\frac{I_{\max}}{4}}{\frac{3 I_{\max}}{4}} = \frac{1}{3}$

Hence, the correct option is C (or 1/3).