The distance of the $5^{\text{th}}$ dark fringe from the center is $4 \, mm$. If $D = 2 \, m$ and $\lambda = 600 \, nm$,then the distance between the slits is (in $mm$):

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 $...... \ mm$.

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?

Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ and $5^{\text{th}}$ bright band. The wavelength of light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is:

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:

In Young's double slit experiment, the following figure shows that $Q$ is the position of the second bright fringe on the right side of point $O$. $P$ is the eleventh bright fringe on the other side measured from point $Q$. If the wavelength of light used is $6000 \text{ Å}$, then what will be the value of $S_1B$?