Two slits are made $1\, mm$ apart and the screen is placed $1\, m$ away. The fringe separation when monochromatic light of wavelength $500\, nm$ is used is

At two points $P$ and $Q$ on the screen in Young's double-slit experiment,waves from slits $S_1$ and $S_2$ have a path difference of $0$ and $\frac{\lambda}{4}$ respectively. The ratio of intensities at $P$ and $Q$ will be

In a Young's double-slit experiment,interference fringes are obtained on a screen placed at a certain distance from the slits using monochromatic light. If the screen is moved towards the slits by $5 \times 10^{-2} \, m$,the fringe width changes by $3 \times 10^{-5} \, m$. If the distance between the slits is $10^{-3} \, m$,find the wavelength of the light.

In a double-slit interference experiment, the fringe width obtained with light of wavelength $5900 \ \mathring{A}$ was $1.2 \ \text{mm}$ for parallel narrow slits placed $2 \ \text{mm}$ apart. In this arrangement, if the slit separation is increased by one-and-a-half times the previous value, then the fringe width is: (in $\text{mm}$)

Two coherent light sources $S_1$ and $S_2$ $(\lambda = 6000 \ \mathring{A})$ are $1 \, mm$ apart from each other. The screen is placed at a distance of $25 \, cm$ from the sources. The width of the fringes on the screen should be (in $, cm$)

In Young's double slit experiment,the intensity on the screen at a point where path difference is $\lambda$ is $K.$ What will be the intensity at the point where path difference is $\lambda /4$?