In a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $I$. If $I_0$ denotes the maximum intensity,the ratio $\frac{I_0}{I}$ is equal to:

Two beams of monochromatic light with intensities $64 \ mW$ and $4 \ mW$ interfere constructively to produce an intensity of $100 \ mW$. If one of the beams is shifted by a phase angle $\phi$,the intensity is reduced to $84 \ mW$. The magnitude of $\phi$ is

The ratio of maximum to minimum intensity due to the superposition of two waves is $\frac{49}{9}$. Then the ratio of the intensities of the component waves is:

The intensity variation in the interference pattern obtained with the help of two coherent sources is $5\%$ of the average intensity. Find out the ratio of intensities of the two sources.

In a double slit experiment,the separation between the slits is $d = 0.25 \, cm$ and the distance of the screen $D = 100 \, cm$ from the slits. If the wavelength of light used is $\lambda = 6000 \mathring{A}$ and $I_0$ is the intensity of the central bright fringe,the intensity at a distance $x = 4 \times 10^{-5} \, m$ from the central maximum is

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi}{2}$ at point $A$ and $\pi$ at point $B$. Then the difference between the resultant intensities at $A$ and $B$ is (in $I$)