In a double slit experiment,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. The fringe width:

White light is used to illuminate two slits in Young's double-slit experiment. The separation between the slits is $b$ and the screen is at a distance $d (d >> b)$ from the slits. The wavelengths missing at a point on the screen directly in front of one of the slits are

In $YDSE$, the distance of the slits from the screen is increased by $25 \%$ and the separation between the slits is halved. If $W$ represents the original fringe width, the new fringe width is (in $\,W$)

Interference fringes are observed on a screen by illuminating two thin slits $1 \, mm$ apart with a light source $(\lambda = 632.8 \, nm)$. The distance between the screen and the slits is $100 \, cm$. If a bright fringe is observed on a screen at a distance of $1.27 \, mm$ from the central bright fringe,then the path difference between the waves,which are reaching this point from the slits,is close to $.... \mu m$.

Young's double-slit experiment is performed using green light,red light,and blue light,one at a time. The fringe widths recorded are $\beta_G, \beta_R,$ and $\beta_B$ respectively. Then:

The distance between two coherent sources is $0.1 \, mm$. The fringe-width on a screen $1.2 \, m$ away from the source is $6.0 \, mm$. The wavelength of light used is ......... $\mathring{A}$.