In Young's double-slit experiment,if the distance between the slits is twice the wavelength,the number of possible interference maxima is:

The Young's double slit experiment is performed in succession using blue light of wavelength $4360 \, \mathring{A}$ and green light of wavelength $5460 \, \mathring{A}$. If the distance of the fourth maximum from the central maximum is $x$,then:

In Young's double-slit experiment,the intensity at a point is $1/4^{th}$ of the maximum intensity. The angular position of this point is

In a Young's double slit experiment,$I_0$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ at a distance $x$ from the central maximum will be

In a Young's double-slit experiment,white light is used. The distance between the two slits is $b$ and the distance between the slits and the screen is $d$ (where $d >> b$). At the point on the screen directly in front of one of the slits,certain wavelengths are missing. Some of the missing wavelengths are: $(1) \lambda = b^2/d$,$(2) \lambda = 2b^2/d$,$(3) \lambda = b^2/3d$,$(4) \lambda = 2b^2/3d$.

Two slits are made $1\, mm$ apart and the screen is placed $1\, m$ away. The fringe separation when monochromatic light of wavelength $500\, nm$ is used is