In Young's double slit experiment,the amplitudes of two sources are $3a$ and $a$ respectively. The ratio of intensities of bright and dark fringes will be (in $:1$)

In Young's double-slit experiment,if the separation between two narrow slits is doubled,then in order to maintain the same fringe spacing,the distance $D$ of the screen from the slits must be changed to:

In a double-slit experiment, the angular width of a fringe for sodium light $(\lambda = 5890 \, \mathring{A})$ is $0.20^o$. If the fringe width increases by $10\%$, what is the change in the wavelength?

The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

In a double slit experiment,when the distance between slits is increased $10$ times,while their distance from the screen is halved,then the fringe width

In a Young's double slit experiment, the slits are separated by $0.3 \, mm$ and the screen is $1.5 \, m$ away from the plane of slits. The distance between the fourth bright fringes on both sides of the central bright fringe is $2.4 \, cm$. The frequency of light used is $.......... \times 10^{14} \, Hz$.