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.

Assertion: Thin films such as a soap bubble or a thin layer of oil on water show beautiful colours when illuminated by white light.
Reason: It happens due to the interference of light reflected from the upper surface of the thin film.

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 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 $, cm$)

$A$ very thin transparent film (like a soap bubble) is viewed in white light. The color of the film appears .......

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?