In a Fresnel's biprism experiment,the fringe width is $1 \,mm$. The distance of the nearest dark fringe from the central bright fringe is .......$mm$.

$A$ thin film of soap solution $(n = 1.4)$ lies on the top of a glass plate $(n = 1.5)$. When visible light is incident almost normal to the plate,two adjacent reflection maxima are observed at two wavelengths $420 \ nm$ and $630 \ nm$. The minimum thickness of the soap solution is.....$nm$

White light reflected from a soap film (Refractive Index $\mu = 1.5$) has a maxima at $600 \ nm$ and a minima at $450 \ nm$ with no minimum in between. Then the thickness of the film in units of $10^{-7} \ m$ is:

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 $= 600 \, nm$.

$A$ Fresnel's biprism experiment is performed in water. If the fringe width in air is $\beta_a$,and the refractive indices of the biprism material and water are $1.5$ and $1.33$ respectively,find the new fringe width.

In a biprism experiment,monochromatic light of wavelength $\lambda$ is used. The distance between two coherent sources is kept constant. If the distance between the slit and the eyepiece $(D)$ is varied as $D_{1}, D_{2}, D_{3}$ and $D_{4}$,and the corresponding measured fringe widths are $z_{1}, z_{2}, z_{3}$ and $z_{4}$,then: