In interference experiment the path difference between two...

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: $1/3$