If the slit widths of $YDSE$ are in the ratio $1:4$,then the ratio of maximum to minimum intensity on the screen is: (in $:1$)

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 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 Young's double slit experiment,the distance between the sources is $1 \ mm$ and the distance between the screen and the source is $1 \ m$. If the fringe width on the screen is $0.06 \ cm$,then $\lambda = \dots \mathring{A}$.

In Young's double slit experiment,the fringes are displaced by a distance $x$ when a glass plate of refractive index $1.5$ is introduced in the path of one of the beams. When this plate is replaced by another plate of the same thickness,the shift of fringes is $(3/2)x$. The refractive index of the second plate is

In an interference experiment, the third bright fringe is obtained at a point on the screen with light of wavelength $700 \, nm$. What should be the wavelength of the light source in order to obtain the $5^{th}$ bright fringe at the same point (in $nm$)?