Two waves have their amplitudes in the ratio $1 : 9$. The maximum and minimum intensities when they interfere are in the ratio

In $YDSE$,the intensity of the central bright fringe is $8 \, mW/m^2$. What will be the intensity at a path difference of $\frac{\lambda}{6}$?

In a $YDSE$ setup,the intensity of two coherent beams differs by $1\%$. If one of the beams has intensity $I$,then the intensity of the minima is:

The intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources of light is $9: 1$. The intensities of the light sources used are in the ratio (in $: 1$)

In a Young's double slit experiment,the intensity at a point is $\left(\frac{1}{4}\right)^{\text{th}}$ of the maximum intensity. The minimum distance of the point from the central maximum is . . . . . . . . $\mu m$.
(Given: $\lambda = 600 \ nm, d = 1.0 \ mm, D = 1.0 \ m$)

In $Y.D.S.E.$ using light of wavelength $\lambda$,the intensity of light at a point on the screen with path difference $\lambda$ is $M$ units. Calculate the intensity of light at a point where the path difference is $\frac{\lambda}{3}$.