In Young’s double slit experiment, how many maxima can be seen on a screen (including the central maxima) if $d = \frac{5\lambda}{2}$ (where $\lambda$ is the wavelength of light and $d$ is the distance between the two slits)?

$A$ Young's double-slit experiment is performed using monochromatic light of wavelength $\lambda$. The intensity of light at a point on the screen,where the path difference is $\lambda$,is $K$ units. The intensity of light at a point where the path difference is $\frac{\lambda}{6}$ is given by $\frac{nK}{12}$,where $n$ is an integer. The value of $n$ is $......$

In Young's double slit experiment,if one of the slits is closed fully,then in the interference pattern:

Two slits are made one millimetre apart and the screen is placed one metre away from the slits. The fringe width when light of wavelength $500 \,nm$ is used is

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 $.......... \times 10^{14} \, Hz$.

In a double-slit experiment using light of wavelength $600 \; nm$,the angular width of a fringe formed on a distant screen is $0.1^{\circ}$. What is the spacing between the two slits?