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 _______.

$$\phi = \frac{2\pi}{\lambda} \cdot \Delta x$$

$$I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$$

$$\phi_A = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3}$$

$$I_A = 4I_0 \cos^2\left(\frac{2\pi / 3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right)$$

$$I_A = 4I_0 \left(\frac{1}{2}\right)^2 = 4I_0 \left(\frac{1}{4}\right) = I_0$$

$$\phi_B = \frac{2\pi}{\lambda} \left(\frac{\lambda}{6}\right) = \frac{\pi}{3}$$

$$I_B = 4I_0 \cos^2\left(\frac{\pi / 3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{6}\right)$$

$$I_B = 4I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = 4I_0 \left(\frac{3}{4}\right) = 3I_0$$

$$\frac{I_A}{I_B} = \frac{I_0}{3I_0} = \frac{1}{3}$$