In Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ is $I$. If $I_0$ denotes the maximum intensity,find the ratio $\frac{I}{I_0}$.

In the Young's double-slit experiment, the spacing between two slits is $0.1 \, mm$. If the screen is kept at a distance of $1.0 \, m$ from the slits and the wavelength of light is $5000 \, \mathring{A}$, then the fringe width is ........ $cm$.

Light consisting of plane waves of wavelengths $\lambda_1 = 8 \times 10^{-5} \ cm$ and $\lambda_2 = 6 \times 10^{-5} \ cm$ generates an interference pattern in Young's double-slit experiment. If $n_1$ denotes the $n_1^{\text{th}}$ dark fringe due to light of wavelength $\lambda_1$ which coincides with the $n_2^{\text{th}}$ bright fringe due to light of wavelength $\lambda_2$,then:

In an interference pattern,if the ratio of slit widths is $1:9$,find the ratio of maximum to minimum intensity.

Two thin parallel slits are made in an opaque screen. When a monochromatic beam of light passes through them at normal incidence,the first bright fringe in the transmitted light occurs at $\pm 45^{\circ}$ with the original direction of the light beam on a distant screen when the apparatus is in air. When the apparatus is immersed in a liquid,the same bright fringe now occurs at $\pm 30^{\circ}$. The refractive index of the liquid is:

In a Young's double-slit experiment with light of wavelength $\lambda$,the separation of slits is $d$ and the distance of the screen is $D$ such that $D >> d >> \lambda$. If the fringe width is $\beta$,the distance from the point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is: