In Young's double slit experiment, carried out with light of wavelength $\lambda = 5000\;\mathring{A}$, the distance between the slits is $d = 0.2\;mm$ and the screen is at $D = 200\;cm$ from the slits. The central maximum is at $x = 0$. The third maximum (taking the central maximum as the zeroth maximum) will be at $x$ equal to......$cm$.

In a Young's double-slit experiment,the light beam consists of two wavelengths $6500 \, \mathring{A}$ and $5200 \, \mathring{A}$. The distance between the slits is $2 \, mm$ and the distance between the plane of the slits and the screen is $120 \, cm$. What is the minimum distance from the central maximum where the bright fringes of both wavelengths coincide (in $, cm$)?

In Young's double-slit experiment,the maximum intensity is $I_0$. If one slit is closed,the new maximum intensity is:

In Young's double slit interference experiment,the slit separation is made $3$ fold. The fringe width becomes

In a Young's double slit experiment,the width of one of the slits is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

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 $... \mathring{A}$.