In Young's double-slit experiment,if the amplitudes of the interfering waves are not equal,then . . . . .

In Young's double slit experiment,the amplitudes of the two waves incident on the two slits are $A$ and $2A$. If $I_{0}$ is the maximum intensity,then the intensity at a spot on the screen,where the phase difference between the two interfering waves is $\phi$,is:

In an interference pattern,the $(n + 4)^{th}$ blue bright fringe and $n^{th}$ red bright fringe are formed at the same spot. If red and blue light have wavelengths of $7800\,\mathring{A}$ and $5200\,\mathring{A}$ respectively,then the value of $n$ is:

In Young's double-slit experiment,the distance between the two slits is $2 \times 10^{-3} \, m$ and the distance between the slits and the screen is $2.5 \, m$. The wavelength of the light used ranges from $2000 \, \mathring{A}$ to $9000 \, \mathring{A}$. What wavelength (in $\mathring{A}$) will form a bright fringe at a distance of $10^{-3} \, m$ from the central maximum?

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 the $Young's$ double slit experiment, the intensity produced by each of the individual slits is $I_0$. The distance between the two slits is $2 \ mm$. The distance of the screen from the slits is $10 \ m$. The wavelength of light is $6000 \ \mathring{A}$. What is the intensity of light on the screen in front of one of the slits?