In an interference experiment,the spacing between successive maxima or minima is

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 $\lambda /4$?

In a Young's double-slit experiment,two slits are separated by a distance of $3 \, mm$ and are illuminated by light of wavelength $480 \, nm$. The screen is at a distance of $2 \, m$ from the plane of the slits. Find the separation between the $8^{th}$ bright fringe and the $3^{rd}$ dark fringe with respect to the central fringe.

In a $YDSE$,bichromatic light of wavelengths $\lambda_1 = 400\, nm$ and $\lambda_2 = 560\, nm$ is used. The distance between the slits is $d = 0.1\, mm$ and the distance between the plane of the slits and the screen is $D = 1\, m$. The minimum distance between two successive regions of complete darkness is.....$mm$.

In a Young's double-slit experiment,the light beam consists of two wavelengths $6500 \, \mathring A$ and $5200 \, \mathring A$. The distance between the slits is $2 \, mm$ and the distance between the plane of the slits and the screen is $120 \, cm$. Find the distance of the third bright fringe from the central maximum for the wavelength $6500 \, \mathring A$ in $mm$.

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?