When the distance between the slits is increased $4$ times in $YDSE$,the fringe width will

In the Young's double slit experiment,the ratio of intensities of bright and dark fringes is $9$. This means that

In a Young's double slit experiment with slit separation $0.1\, mm$,one observes a bright fringe at angle $\frac{1}{40}\, rad$ by using light of wavelength $\lambda_1$. When the light of wavelength $\lambda_2$ is used,a bright fringe is seen at the same angle in the same setup. Given that $\lambda_1$ and $\lambda_2$ are in the visible range ($380\, nm$ to $740\, nm$),their values are:

Minimum thickness of a mica sheet having $\mu = 3/2$ which should be placed in front of one of the slits in $YDSE$ is required to reduce the intensity at the centre of the screen to half of the maximum intensity is-

The distance between two coherent sources is $0.1 \, mm$. The fringe-width on a screen $1.2 \, m$ away from the source is $6.0 \, mm$. The wavelength of light used is ......... $\mathring{A}$.

In $YDSE$ experiment,the $4^{\text{th}}$ dark band is formed opposite to one of the slits. The wavelength of light used is ($d=$ distance between the slits,$D=$ distance between source and the screen)