In a Young's double-slit experiment,the fringe pattern is observed on a screen placed at a distance $D$. The slits are separated by $d$ and are illuminated by light of wavelength $\lambda$. Find the distance from the central point where the intensity falls to half the maximum intensity.

In Young's double-slit experiment,we get $10$ fringes in the field of view using monochromatic light of wavelength $4000 \text{ } \mathring{A}$. If we use monochromatic light of wavelength $5000 \text{ } \mathring{A}$,then the number of fringes obtained in the same field of view is

In Young's double slit experiment,the wavelength of red light is $7800\,\mathring{A}$ and that of blue light is $5200\,\mathring{A}$. The value of $n$ for which the $n^{th}$ bright band due to red light coincides with the $(n + 1)^{th}$ bright band due to blue light is:

$A$ beam of light consisting of two wavelengths $7000 \; \mathring{A}$ and $5500 \; \mathring{A}$ is used to obtain an interference pattern in Young's double-slit experiment. The distance between the slits is $2.5 \; mm$ and the distance between the plane of the slits and the screen is $150 \; cm$. The least distance from the central fringe,where the bright fringes due to both the wavelengths coincide,is $n \times 10^{-5} \; m$. The value of $n$ is .............

$A$ Young's double slit interference arrangement with slits $S_1$ and $S_2$ is immersed in water (refractive index $\mu_w = 4/3$) as shown in the figure. The positions of maxima on the surface of water are given by $x^2 = p^2 m^2 \lambda^2 - d^2$,where $\lambda$ is the wavelength of light in air (refractive index $\mu_a = 1$),$2d$ is the separation between the slits,and $m$ is an integer. The value of $p$ is

In $YDSE$,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. Then,the fringe width: