In a Young's double slit experiment,the ratio of the slit widths is $4 : 1$. The ratio of the intensity of maxima to minima,close to the central fringe on the screen,will be

In the Young's double slit experiment with sodium light,the slits are $0.589 \ m$ apart. The angular separation of the third maximum from the central maximum will be (given $\lambda = 589 \ nm$):

In the $Young's$ double slit experiment, the intensity produced by each of the individual slits is $I_0$. The distance between the two slits is $2 \ mm$. The distance of the screen from the slits is $10 \ m$. The wavelength of light is $6000 \ \mathring{A}$. What is the intensity of light on the screen in front of one of the slits?

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900 \text{ Å}$ was $1.2 \text{ mm}$ for parallel narrow slits placed $2 \text{ mm}$ apart. In this arrangement, if the slit separation is increased by one-and-half times the previous value, then the fringe width is (in $ \text{ mm}$)

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,$12$ fringes are observed to be formed in a certain segment of the screen when light of wavelength $600 \ nm$ is used. If the wavelength of light is changed to $400 \ nm$,the number of fringes observed in the same segment of the screen is: