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-

Write down the formula for the distance between two consecutive bright fringes or two consecutive dark fringes in Young's Double Slit Experiment.

In Young's double slit experiment,the separation $d$ between the slits is $2 \ mm$,the wavelength $\lambda$ of the light used is $5896 \ \mathring{A}$,and the distance $D$ between the screen and slits is $100 \ cm$. It is found that the angular width of the fringes is $0.20^\circ$. To increase the fringe angular width to $0.21^\circ$ (with the same $\lambda$ and $D$),the separation between the slits needs to be changed to ...... $mm$.

In a Young's double slit experiment, the slits are $2 \,mm$ apart and are illuminated with a mixture of two wavelengths ${\lambda _1} = 750 \,nm$ and ${\lambda _2} = 900 \,nm$. The minimum distance from the common central bright fringe on a screen $2 \,m$ from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is.....$mm$

In a Young's double-slit experiment,sources of equal intensities are used. The distance between the slits is $d$ and the wavelength of light used is $\lambda$ (where $\lambda \ll d$). Find the angular separation of the nearest points on either side of the central maximum where the intensity becomes half of the maximum value.

In Young's double slit experiment,if one of the slits is closed fully,then in the interference pattern: