Two waves having the intensities in the ratio of $9 : 1$ produce interference. The ratio of maximum to the minimum intensity is equal to

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 $O$.

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 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

In Young's double-slit experiment,the intensity at a point where the path difference is $\lambda$ is $k$. What will be the intensity at a point where the path difference is $\lambda/4$? ($\lambda$ = wavelength of light)

The ratio of intensities of two interfering waves is $9:1$. The ratio of the maximum amplitude to the minimum amplitude of the resultant wave is ........