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

In Young's double-slit experiment,the angular width of the fringes formed on the screen is $\pi / 200$. If light of wavelength $4800 \, \mathring{A}$ is used,find the distance between the slits.

In Young's double slit experiment, for the $n^{th}$ dark fringe $(n = 1, 2, 3, \dots)$ the phase difference of the interfering waves in radian will be

Two ideal slits $S_1$ and $S_2$ are at a distance $d$ apart and are illuminated by light of wavelength $\lambda$ passing through an ideal source slit $S$ placed on the line through $S_2$ as shown. The distance between the plane of the slits and the source slit is $D$. $A$ screen is held at a distance $D$ from the plane of the slits. The minimum value of $d$ for which there is darkness at $O$ is

In Young's double-slit experiment,the distance between the two slits is $1 \, mm$ and the distance between the slits and the screen is $1 \, m$. If the wavelength of the light used is $500 \, nm$,then the distance of the third minimum from the central maximum is ... $mm$.

In a Young's double slit experiment,$I_0$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ at a distance $x$ from the central maximum will be