In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\angle S_1 C S_2 = \theta$ and $\lambda$ is the wavelength,the fringe width is

In a two-slit experiment with monochromatic light,fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by $5 \times 10^{-2} \ m$ towards the slits,the change in fringe width is $3 \times 10^{-5} \ m$. If the separation between the slits is $10^{-3} \ m$,the wavelength of light used is ....... $\mathring{A}$.

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-

In a Young's double-slit experiment,the slits are separated by $0.28 \; mm$ and the screen is placed $1.4 \; m$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2 \; cm$. Determine the wavelength of light used in the experiment (in $nm$).

At two points $P$ and $Q$ on the screen in Young's double-slit experiment,waves from slits $S_1$ and $S_2$ have a path difference of $0$ and $\frac{\lambda}{4}$ respectively. The ratio of intensities at $P$ and $Q$ will be

$A$ double slit interference experiment performed with a light of wavelength $600 \ nm$ forms an interference fringe pattern on a screen with $10^{\text{th}}$ bright fringe having its centre at a distance of $10 \ mm$ from the central maximum. The distance of the centre of the same $10^{\text{th}}$ bright fringe from the central maximum when the source of light is replaced by another source of wavelength $660 \ nm$ would be . . . . . . $mm$.