The amplitude of the light waves emerging from the two slits in Young's experiment is in the ratio of $2 : 3$. The ratio of the intensity of the minimum to that of the consecutive maximum will be:

Two coherent sources of intensity ratio $\alpha$ interfere. The value of $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ is

Consider an $YDSE$ that has different slit widths. As a result,the amplitudes of waves from the two slits are $A$ and $2A$,respectively. If $I_0$ is the maximum intensity of the interference pattern,then the intensity of the pattern at a point where the phase difference between the waves is $\phi$ is:

Two slits in Young's experiment have widths in the ratio $1 : 25$. The ratio of intensity at the maxima and minima in the interference pattern,$\frac{I_{max}}{I_{min}}$ is

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 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: