$A$ 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?

In a biprism experiment, the $4^{th}$ dark band is formed opposite to one of the slits. The wavelength of light used is (where $D$ = distance between source and screen, $d$ = distance between the slits).

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?

$A$ thin film of water $(\mu = 4/3)$ is $3100 \,\mathring{A}$ thick. If it is illuminated by white light at normal incidence,the colour of the film in the reflected light will be:

In a biprism experiment,if the $5^{\text{th}}$ bright band with wavelength $\lambda_1$ coincides with the $6^{\text{th}}$ dark band with wavelength $\lambda_2$,then the ratio $(\lambda_1 / \lambda_2)$ is:

When exposed to sunlight,thin films of oil on water often exhibit brilliant colours due to the phenomenon of