Two slits are made $10 \ mm$ apart and the screen is placed $1.5 \ m$ away. What is the fringe separation when a wavelength of $7000 \ \mathring{A}$ is used (in $\mu m$)?

In Young's double slit experiment,the two slits act as coherent sources of equal amplitude $A$ and wavelength $\lambda$. In another experiment with the same setup,the two slits are sources of equal amplitude $A$ and wavelength $\lambda$ but are incoherent. The ratio of the intensity of light at the midpoint of the screen in the first case to that in the second case is:

If the source of light used in a Young's double slit experiment is changed from red to violet:

In a Young's double slit experiment,the fringe width is found to be $0.4 \, mm$. If the whole apparatus is immersed in water of refractive index $4/3$ without disturbing the geometrical arrangement,the new fringe width will be

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}$)

Laser light $(630 \, nm)$ incident on a pair of slits produces an interference pattern in which the bright fringes are separated by $8.3 \, mm$. $A$ second light produces an interference pattern in which the bright fringes are separated by $7.6 \, mm$. What is the wavelength of this second light in $nm$?