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: $(1) \lambda = b^2/d$,$(2) \lambda = 2b^2/d$,$(3) \lambda = b^2/3d$,$(4) \lambda = 2b^2/3d$.

In a $YDSE$ apparatus,two identical slits are separated by $1\, mm$ and the distance between the slits and the screen is $1\, m$. The wavelength of light used is $6000\, \mathring{A}$. The minimum distance between two points on the screen having $75\%$ intensity of the maximum intensity is ......$mm$.

For Young's double slit experiment,two statements are given below :
Statement $I:$ If the screen is moved away from the plane of the slits,the angular separation of the fringes remains constant.
Statement $II:$ If the monochromatic source is replaced by another monochromatic source of larger wavelength,the angular separation of the fringes decreases.
In the light of the above statements,choose the correct answer from the options given below :

Two sources produce an interference pattern which is observed on a screen,at a distance $D$ from the sources. The fringe width is $2w$. If the distance $D$ is now doubled,the fringe width will be:

In a Young's double-slit experiment,the slits are placed $0.320 \, mm$ apart. Light of wavelength $\lambda = 500 \, nm$ is incident on the slits. The total number of bright fringes that are observed in the angular range $-30^\circ \le \theta \le 30^\circ$ is:

In Young's double-slit experiment,$\frac{d}{D} = 10^{-4}$. The intensity at point $P$ on the screen is equal to the intensity of one of the sources. If the wavelength of the light used is $\lambda = 6000 \, \mathring{A}$,what is the distance of point $P$ from the central bright fringe in $mm$?