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

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 $\frac{\lambda}{3}$.

Consider a two-slit interference arrangement (see figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of $D$ in terms of $\lambda$ such that the first minima on the screen falls at a distance $D$ from the centre $O$.

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

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: