$A$ light of wavelength $5890 \; \mathring{A}$ falls normally on a thin air film. What is the minimum thickness of the film such that the film appears dark in reflected light?

When a mica sheet of thickness $7 \ \mu m$ and refractive index $\mu = 1.6$ is placed in the path of one of the interfering beams in a biprism experiment, the central fringe shifts to the position of the seventh bright fringe. The wavelength of the light used is $............ \ \mathring{A}$.

To ensure almost $100$ per cent transmittivity,photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediate between that of air and glass (which makes the optical element of the lens). $A$ typically used dielectric film is $MgF_2$ $(n = 1.38)$. What should the thickness of the film be so that at the center of the visible spectrum $5500 \, \mathring{A}$ there is maximum transmission?

In Fresnel's biprism $(\mu = 1.5)$ experiment,the distance between the source and the biprism is $0.3\, m$,the distance between the biprism and the screen is $0.7\, m$,and the angle of the prism is $1^\circ$. The fringe width with light of wavelength $6000\ \mathring{A}$ will be: (in $, cm$)

$A$ parallel coherent beam of light falls on a Fresnel biprism of refractive index $\mu$ and refracting angle $\alpha$. The fringe width on a screen at a distance $D$ from the biprism will be (wavelength $= \lambda$).

What is the shape of the interference fringes formed on the screen due to a point source $P$,in the case shown here?