In Young's double slit experiment, the distance between the two slits is $0.1 \, mm$ and the wavelength of light used is $4 \times 10^{-7} \, m$. If the width of the fringe on the screen is $4 \, mm$, the distance between the screen and the slit is:

In a Young's double-slit experiment,light of wavelength $5890 \ \mathring A$ is used,and the angular fringe width on the screen is $0.20^\circ$. If the entire apparatus is immersed in water,find the new angular fringe width. (Refractive index of water $\mu = 4/3$) (in $^\circ$)

In a Young's double slit experiment,the source $S$ and the two slits $A$ and $B$ are vertical,with slit $A$ above slit $B$. The fringes are observed on a vertical screen $K$. The optical path length from $S$ to $B$ is increased very slightly (by introducing a transparent material of higher refractive index) and the optical path length from $S$ to $A$ is not changed. As a result,the fringe system on $K$ moves:

In Young's experiment,the separation between the $5^{\text{th}}$ maxima and the $3^{\text{rd}}$ minima is how many times the fringe width?

In a double slit experiment,the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1\, m$ away. The wavelength of light used is $600\, nm$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water (in $^{\circ}$)? Take the refractive index of water to be $4/3$.

In Young's double-slit experiment,if the distance between the slits is twice the wavelength,the number of possible interference maxima is: