In Young's double slit experiment , one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If $${{\rm I}_m}$$ be the maximum intensity, the resultant intensity $${\rm I}$$ when they interfere at phase difference $$\phi $$ is given by :

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 plane-convex lens over a plane glass plane. 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 center of the interference pattern is dark.

An initially parallel cylindrical beam travels in a medium of refractive index $$\mu \left( I \right) = {\mu _0} + {\mu _2}\,I,$$ where $${\mu _0}$$ and $${\mu _2}$$ are positive constants and $$I$$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.

The initial shape of the wavefront of the beam is

A mixture of light, consisting of wavelength $$590$$ $$nm$$ and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the $$4$$th bright fringe of the unknown light. From this data, the wavelength of the unknown light is :