In two separate setups for a Biprism experiment using the same wavelength,fringes of equal width are obtained. If the ratio of slit separation is $2:3$,then the ratio of the distance between the slit and the screen in the two setups is:

White light is incident at an angle of $30^\circ$ on a soap film with a refractive index of $4/3$. The wavelength of the transmitted light is observed to be $6 \times 10^{-5} \, cm$. Find the minimum thickness of the film.

Light of wavelength $6000 \, \mathring{A}$ is incident on a thin glass plate of refractive index $1.5$ such that the angle of refraction in the plate is $60^\circ$. Find the thickness of the plate so that it appears dark in reflection.

What is the minimum thickness of a thin film required for constructive interference in the reflected light from it (in $ nm$)? Given, the refractive index of the film $= 1.5$, wavelength of the light incident on the film $= 600 \, nm$.

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 $12.5 \times 10^{-5} \,m$?

In a Fresnel biprism experiment,the two positions of the lens give separations between the virtual slits as $16 \; cm$ and $9 \; cm$ respectively. What is the actual distance of separation (in $cm$)?