In Fresnel's biprism experiment,on increasing the prism angle,the fringe width will

In a biprism experiment,the slit separation is $1 \,mm$. Using monochromatic light of wavelength $5000 \text{ Å}$,an interference pattern is obtained on the screen. Where should the screen be moved,so that the change in fringe width is $12.5 \times 10^{-5} \,m$?

Dark and colour patterns on thin soap films are due to

If a biprism experiment is immersed in water,and the fringe width in air is $\beta_a$,find the new fringe width if the refractive index of the biprism material is $1.5$ and the refractive index of water is $1.33$.

In a biprism experiment,a wavelength $\lambda$ is used to obtain an interference pattern. The fringe width is $W_1$ at a distance $D_1$ from the screen. When the screen is moved towards the biprism,the fringe width becomes $W_2$ at a distance $D_2$. The distance between the two virtual images of the slit is:

If prism angle $\alpha = 1^\circ$,refractive index $\mu = 1.54$,distance between screen and prism $b = 0.7\,m$,distance between prism and source $a = 0.3\,m$,and wavelength $\lambda = 180\pi\,nm$,then in a Fresnel biprism,find the value of fringe width $\beta$.