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 $120 \, cm$. What is the minimum distance from the central maximum where the bright fringes of both wavelengths coincide (in $, cm$)?

Assertion: In Young's double slit experiment,the two slits are at a distance $d$ apart. An interference pattern is observed on a screen at a distance $D$ from the slits. At a point on the screen directly opposite to one of the slits,a dark fringe is observed. Then,the wavelength of the wave is proportional to the square of the distance between the two slits.
Reason: For a dark fringe,the intensity is zero.

Young's double slit experiment is first performed in air and then in a medium other than air. It is found that the $8^{th}$ bright fringe in the medium lies where the $5^{th}$ dark fringe lies in air. The refractive index of the medium is nearly:

In $YDSE$ setup,light of wavelength $640 \, nm$ is used with $d = 0.8 \, mm$ and $D = 1 \, m$. If intensity at central maximum is $I_0$ and its position is $y = 0$,then:

Two sources of light of wavelengths $2500 \,\mathring{A}$ and $3500 \,\mathring{A}$ are used in Young's double slit experiment simultaneously. Which orders of fringes of the two wavelength patterns coincide?

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900 \text{ Å}$ 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-half times the previous value, then the fringe width is (in $ \text{ mm}$)