Two sources produce an interference pattern which is observed on a screen,at a distance $D$ from the sources. The fringe width is $2w$. If the distance $D$ is now doubled,the 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}$)

If the distance between the first maxima and the fifth minima of a double slit pattern is $7\,mm$ and the slits are separated by $0.15\,mm$ with the screen $50\,cm$ from the slits,then find the wavelength of the light used in $nm$.

In a double slit experiment,the distance between slits is increased $10$ times whereas their distance from the screen is halved,then what is the fringe width?

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 $4/3$.

Which of the following statements is incorrect regarding interference fringes?