In a Young's double-slit experiment, interference fringes are obtained on a screen at a distance of $1 \, m$ using light of wavelength $6000 \, \mathring{A}$. The distance between the slits is $1 \, mm$. The fringe width is:

In a Young's double-slit experiment,white light is used. The distance between the two slits is $b$ and the distance between the slits and the screen is $d$ (where $d >> b$). At the point on the screen directly in front of one of the slits,certain wavelengths are missing. Some of the missing wavelengths are: $(1) \lambda = b^2/d$,$(2) \lambda = 2b^2/d$,$(3) \lambda = b^2/3d$,$(4) \lambda = 2b^2/3d$.

In Young's double slit experiment, the angular width of a fringe is found to be $0.2^\circ$ on a screen placed $1 \text{ m}$ away. The wavelength of light used is $600 \text{ nm}$. If the entire apparatus is immersed in water of refractive index $4/3$, the angular width of the fringe will be: (in $^\circ$)

Two slits are made one millimetre apart and the screen is placed one metre away. What is the fringe separation (in $mm$) when blue-green light of wavelength $500 \; nm$ is used?

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 Young's double-slit experiment,the fringe width will remain the same if: