In interference experiment the path difference between two interfering waves at a point \(A\) on the screen is \(\lambda/3\), where \(\lambda\) is the wavelength of these waves, and at another point \(B\) the path difference is \(\lambda/6\). The ratio of intensities at points \(A\) and \(B\) is _______.

The phase difference \(\Delta \phi\) is related to the path difference \(\Delta x\) by the relation \(\Delta \phi = \dfrac{2\pi}{\lambda} \Delta x\).

For point \(A\), the path difference is \(\Delta x_A = \dfrac{\lambda}{3}\).

\(\Delta \phi_A = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{3} = \dfrac{2\pi}{3}\)

The intensity at point \(A\) is given by \(I_A = I_{max} \cos^2\left(\dfrac{\Delta \phi_A}{2}\right)\).

\(I_A = I_{max} \cos^2\left(\dfrac{\pi}{3}\right) = I_{max} \left(\dfrac{1}{2}\right)^2 = \dfrac{I_{max}}{4}\)

For point \(B\), the path difference is \(\Delta x_B = \dfrac{\lambda}{6}\).

\(\Delta \phi_B = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{6} = \dfrac{\pi}{3}\)

The intensity at point \(B\) is given by \(I_B = I_{max} \cos^2\left(\dfrac{\Delta \phi_B}{2}\right)\).

\(I_B = I_{max} \cos^2\left(\dfrac{\pi}{6}\right) = I_{max} \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3I_{max}}{4}\)

The ratio of intensities at points \(A\) and \(B\) is:

\(\dfrac{I_A}{I_B} = \dfrac{\dfrac{I_{max}}{4}}{\dfrac{3I_{max}}{4}} = \dfrac{1}{3}\)

Answer: \(1/3\)