In $Y.D.S.E.$,how many maxima can be obtained on the screen if the wavelength of light used is $200 \ nm$ and $d = 700 \ nm$?

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}$)

In a $YDSE$,light of two different wavelengths $(\lambda_1, \lambda_2)$ is incident normal to the plane of slits. The $n^{th}$ maxima of $\lambda_1$ coincides with the $m^{th}$ maxima of $\lambda_2$ exactly in front of one of the slits. Given $D = 1.5 \ m$,$d = 3 \ mm$,and $4500 \ \mathring{A} < \lambda_1, \lambda_2 < 7000 \ \mathring{A}$,then $n, m$ and $\lambda_1$ are:

In a double-slit experiment,for light of which colour will the fringe width be minimum?

Two parallel slits $0.6 \, mm$ apart are illuminated by a light source of wavelength $6000 \, \mathring{A}$. The distance between two consecutive dark fringes on a screen $1 \, m$ away from the slits is ........ $mm$.

The width of the fringe is $2\,mm$ on the screen in a double-slit experiment for light of wavelength $400\,nm$. The width of the fringe for light of wavelength $600\,nm$ will be $..............\,mm$.