In $Y.D.S.E$, two slits are made $1 \, mm$ apart and the screen is placed $1 \, m$ away. The fringe separation when light of wavelength $500 \, nm$ is used is:

In a Young's double-slit experiment,white light is used. The distance between the two slits is $b$ and the distance between the slits and the screen is $d$ (where $d >> b$). At the point on the screen directly in front of one of the slits,certain wavelengths are missing. Some of the missing wavelengths are: $(1) \lambda = b^2/d$,$(2) \lambda = 2b^2/d$,$(3) \lambda = b^2/3d$,$(4) \lambda = 2b^2/3d$.

Interference fringes are observed on a screen by illuminating two thin slits $1 \, mm$ apart with a light source $(\lambda = 632.8 \, nm)$. The distance between the screen and the slits is $100 \, cm$. If a bright fringe is observed on a screen at a distance of $1.27 \, mm$ from the central bright fringe,then the path difference between the waves,which are reaching this point from the slits,is close to $.... \mu m$.

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

If the separation between slits in Young's double slit experiment is reduced to $\frac{1}{3}rd,$ the fringe width becomes $n$ times. The value of $n$ is

In a Young's double-slit experiment,the wavelength of light used is $5000 \, \mathring{A}$ and the fringe width obtained is $1 \, mm$. If the wavelength of light is changed to $6000 \, \mathring{A}$ while keeping the experimental setup unchanged,the new fringe width will be ........ $mm$.