If Fresnel's biprism experiment is performed in water instead of air,what will be the effect on the fringe width?

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?

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

In a biprism experiment,monochromatic light of wavelength $\lambda$ is used. The distance between two coherent sources $d$ is kept constant. If the distance between the slit and the eyepiece $D$ is varied as $D_1, D_2, D_3$ and $D_4$,and the corresponding measured fringe widths are $Z_1, Z_2, Z_3$ and $Z_4$,then:

$A$ thin oil film of refractive index $1.2$ floats on the surface of water $(\mu = \frac{4}{3})$. When light of wavelength $\lambda = 9.6 \times 10^{-7} \ m$ falls normally on the film from air,it appears dark when seen normally. What is the minimum change in its thickness for which it will appear bright in normally reflected light by the same light?

$A$ biprism experiment is set up as shown. When the upper half area of the biprism is covered with an opaque material,then: