In Young's double-slit experiment,the intensity at a point is $1/4^{th}$ of the maximum intensity. The angular position of this point is

$A$ beam of light consisting of two wavelengths, $650\; nm$ and $520\; nm$, is used to obtain interference fringes in a Young's double-slit experiment.
$(a)$ Find the distance of the third bright fringe on the screen from the central maximum for wavelength $650\; nm$.
$(b)$ What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Electrons accelerated from rest by an electrostatic potential are collimated and sent through a Young's double slit experiment. The fringe width is $\omega$. If the accelerating potential is doubled,then the width is now close to ............. $\omega$.

The distance between two slits in a double slit experiment is $1 \ mm$. The distance between the slits and the screen is $1 \ m$. If the distance of the $10^{th}$ fringe from the central fringe is $5 \ mm$, then the wavelength of light is $... \mathring{A}$.

$A$ Young's double slit experiment apparatus has slits separated by $0.2 \ mm$ and a screen $60 \ cm$ away from the slits. The whole apparatus is immersed in a liquid medium of refractive index $\mu = \frac{11}{9}$ and the slits are illuminated with green light $(\lambda = 550 \ nm$ in vacuum$)$. Find the fringe width of the pattern formed on the screen. (in $mm$)

In Young's double-slit experiment, one slit is wider than the other, such that the amplitude of the light wave from one slit is twice that of the other. If the maximum intensity is $I_m$, then the resultant intensity $I$ when they interfere with a phase difference of $\phi$ is: