Calculate the minimum thickness of a soap film $(n=1.33)$ that results in constructive interference in reflected light, if the film is illuminated with light whose wavelength in free space is $532 \,nm$. (in $\,nm$)

In a Fresnel's biprism experiment,the distance between the source and the screen is $1 \, m$ and the distance between the source and the biprism is $10 \, cm$. The wavelength of light is $6000 \, \mathring{A}$. The fringe width is $0.03 \, cm$ and the refracting angle of the biprism is $1^\circ$. Find the refractive index of the material of the biprism.

An oil film flowing on water appears coloured due to interference. For observing this effect,the approximate thickness of the oil film should be:

Light of wavelength $6000 \text{ Å}$ is incident on a thin glass plate of refractive index $\mu = 1.5$ such that the angle of refraction into the plate is $60^{\circ}$. Calculate the smallest thickness of the plate which will make a dark fringe by reflected beam interference.

In a biprism experiment, a steady interference pattern is observed on the screen kept at a distance of $100 \text{ cm}$ using a light of wavelength $5000 \text{ Å}$. Without changing the distance between the virtual sources of the slit, the source of light is replaced by a source of wavelength $6400 \text{ Å}$. Now, to reduce the fringe width by $20\%$ of its initial value, the screen should be moved

In a biprism experiment using sodium light $\lambda_1 = 6000\, \mathring{A}$,an interference pattern is obtained in which $20$ fringes occupy $2\, cm$. On replacing sodium light by another source of wavelength $\lambda_2$ without making any other change,$30$ fringes occupy $2.7\, cm$ on the screen. What is the value of $\lambda_2$ in $\mathring{A}$?