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 $\,nm$)

$A$ 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.....$nm$

$A$ biprism experiment is set up as shown. When the upper half area of the biprism is covered with an opaque material,then:

This question has a paragraph followed by two statements,Statement $- 1$ and Statement $- 2$. Of the given four alternatives after the statements,choose the one that describes the statements.
$A$ thin air film is formed by putting the convex surface of a plano-convex lens over a plane glass plate. With monochromatic light,this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film.
Statement $- 1$: When light reflects from the air-glass plate interface,the reflected wave suffers a phase change of $\pi$.
Statement $- 2$: The centre of the interference pattern is dark.

$A$ ray of light of intensity $I$ is incident on a parallel glass slab at a point $A$ as shown in the figure. It undergoes partial reflection and refraction. At each reflection,$25\%$ of the incident energy is reflected. The rays $AB$ and $A^1B^1$ undergo interference. The ratio $I_{\text{max}} / I_{\text{min}}$ is (in $: 1$)

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 .......$mm$.