In a $YDSE$ apparatus,if we use white light,then:

The correct curve between fringe width $\beta$ and the distance between the slits $(d)$ is:

In Young's double slit experiment,the fringe width is $1 \times 10^{-4} \ m$. If the distance between the slit and screen is doubled,the distance between the two slits is reduced to half,and the wavelength is changed from $6.4 \times 10^{-7} \ m$ to $4.0 \times 10^{-7} \ m$,what will be the value of the new fringe width?

The path difference between two interfering light waves meeting at a point on the screen is $\left(\frac{57}{2}\right) \lambda$. The band obtained at that point is

In Young's double slit experiment,the distance between the two slits is $0.1 \,mm$,the distance between the slits and the screen is $1 \,m$ and the wavelength of the light used is $600 \,nm$. The intensity at a point on the screen is $75 \%$ of the maximum intensity. Find the smallest distance in $mm$ of this point from the central fringe.

In a Young's double-slit experiment,two slits are separated by a distance of $3 \, mm$ and are illuminated by light of wavelength $480 \, nm$. The screen is at a distance of $2 \, m$ from the plane of the slits. Find the separation between the $8^{th}$ bright fringe and the $3^{rd}$ dark fringe with respect to the central fringe.