In the Young's double-slit experiment, the spacing between two slits is $0.1 \, mm$. If the screen is kept at a distance of $1.0 \, m$ from the slits and the wavelength of light is $5000 \, \mathring{A}$, then the fringe width is ........ $cm$.

In Young's double slit experiment, in an interference pattern, a minimum is observed exactly in front of one slit. The distance between the two coherent sources is $d$ and $D$ is the distance between the source and the screen. The possible wavelengths used are inversely proportional to

$A$ 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

Light consisting of wavelengths $6000\,\mathring{A}$ and $5500\,\mathring{A}$ falls on the double slits in $YDSE$. The $n^{th}$ order bright fringe of $\lambda_1 = 6000\,\mathring{A}$ is found to coincide with the $m^{th}$ order bright fringe of $\lambda_2 = 5500\,\mathring{A}$. The smallest values of $n$ and $m$ are respectively:

In a Young's double slit experiment,the source illuminating the slits is changed from blue to violet. The width of the fringes

In Young's double-slit experiment,the separation between the two slits is $3 \, mm$ and they are illuminated by light of wavelength $480 \, nm$. The screen is at a distance of $2 \, m$ from the plane of the slits. Find the separation between the $8^{th}$ bright fringe and the $3^{rd}$ dark fringe with respect to the central bright fringe.