In Young's double-slit experiment,the ratio of intensities of bright and dark fringes is $9$. What is the ratio of the intensities of the sources?

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

Waves emitted from two identical sources produce an intensity of $K$ units at a point on the screen. If the path difference between these two waves is $\lambda$,calculate the intensity at a point on the screen where the path difference is $\lambda/4$.

In Young's double slit experiment,one of the slits is wider than the other,so that the amplitude of the light from one slit is double that of the other. If $I_m$ is the maximum intensity,the resultant intensity $I$ when they interfere at a phase difference $\phi$ is given by

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: