In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :

In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1m away, was found to be 0.2^o. What will be the angular width of the first minima, ($$\mu $$_water = 4/3) if the entire experimental apparatus is immersed in water?

In Young’s double slit experiment the separation d between the slits is 2 mm, the wavelength $$\lambda $$ of the light used is 5896 $$\mathop A\limits^0 $$ and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20^o. To increase the fringe angular width to 0.21^o (with same $$\lambda $$ and D) the separation between the slits needs to be changed to

Unpolarised light is incident from air on a plane surface of a material of refractive index $$\mu $$. At a particular angle of incidence i, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?