In a Young's double slit experiment,the ratio of the amplitude of light coming from the slits is $2:1$. The ratio of the maximum to minimum intensity in the interference pattern is:

In $YDSE$ experiment,the $4^{\text{th}}$ dark band is formed opposite to one of the slits. The wavelength of light used is ($d=$ distance between the slits,$D=$ distance between source and the screen)

In Young's double slit experiment, for the $n^{th}$ dark fringe $(n = 1, 2, 3, \dots)$ the phase difference of the interfering waves in radian will be

In a double-slit experiment,the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1 \; m$ away. The wavelength of light used is $600 \; nm$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water (in $^{\circ}$)? Take the refractive index of water to be $4/3$.

Interference was observed in an interference chamber when air was present. Now,the chamber is evacuated and if the same light is used,a careful observer will see:

In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\angle S_1 C S_2 = \theta$ and $\lambda$ is the wavelength,the fringe width is