In Young's double slit experiment,the angular width of fringes is $0.20^o$ for sodium light of wavelength $5890 \ \mathring A$. If the complete system is dipped in water,then the angular width of the fringes becomes....$^o$

$A$ fringe width of a certain interference pattern is $\beta = 0.002 \text{ cm}$. What is the distance of the $5^{\text{th}}$ dark fringe from the center?

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

In Young's double slit experiment,the two slits are $0.6 \; mm$ apart. The interference pattern is observed on a screen at a distance of $80 \; cm$ from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light is $\dots \; nm$.

In a certain double-slit experimental arrangement,interference fringes of width $1.0 \ mm$ each are observed when light of wavelength $5000 \ \mathring{A}$ is used. Keeping the setup unaltered,if the source is replaced by another source of wavelength $6000 \ \mathring{A}$,the fringe width will be $...... \ mm$.

The fringe width in Young's double-slit experiment can be increased if we decrease