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 $\beta$.

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,monochromatic light of wavelength $\lambda$ is used. The distance between the two coherent sources $d$ is kept constant. If the distance between the slit and the eyepiece $D$ is varied as $D_1, D_2, D_3, D_4$ and the corresponding measured fringe widths are $W_1, W_2, W_3, W_4$,then:

It is necessary to coat a glass lens with a non-reflecting layer. If the wavelength of the light in the coating is $\lambda$,the best choice is a layer of material having an index of refraction between those of glass and air and a thickness of

$A$ parallel beam of light of intensity $I_0$ is incident on a coated glass plate. If $25 \%$ of the incident light is reflected from the upper surface and $50 \%$ of the light that enters the glass is reflected from the lower surface of the glass plate, the ratio of maximum to minimum intensity in the interference region of the reflected light is

Dark and colour patterns on thin soap films are due to