In Young's double-slit experiment, the fringe width will increase if ......

Light consisting of wavelengths $6000\,\mathring{A}$ and $5500\,\mathring{A}$ falls on the double slits in $YDSE$. The $n^{th}$ order bright fringe of $\lambda_1 = 6000\,\mathring{A}$ is found to coincide with the $m^{th}$ order bright fringe of $\lambda_2 = 5500\,\mathring{A}$. The smallest values of $n$ and $m$ are respectively:

In a Young's double-slit experiment, the angular width of a fringe is $1^\circ$ for light of wavelength $6000 \, \mathring{A}$. What is the distance between the slits in $mm$?

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}$)

Consider a Young's double slit experiment as shown in the figure. What should be the slit separation $d$ in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit $S_1$?

In Young's double slit experiment,the distance between the slits is $1 \,mm$ and the distance between the slit and the screen is $1 \,m$. If the $10^{th}$ fringe is $5 \,mm$ away from the central bright fringe,then the wavelength of the light used will be.....$\mathring A$.