In a Young's double slit experiment,if there is no initial phase difference between the light from the two slits,what is the path difference at a point on the screen corresponding to the fifth minimum?

In a double slit experiment,the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1\, m$ away. The wavelength of light used is $600\, nm$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water (in $^{\circ}$)? Take the refractive index of water to be $4/3$.

If the sodium light in Young's double slit experiment is replaced by red light, the fringe width will

In Young's double-slit experiment,the distance between the two slits is $3 \, cm$,the distance from the slits to the screen is $7 \, cm$,and the wavelength of light used is $1000 \, \mathring{A}$. Calculate the fringe width.

The width of one of the two slits in a Young's double slit experiment is $4$ times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: (in $: 1$)

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 of equal amplitude $A$ and wavelength $\lambda$ but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is