The distance between two slits is $3 \ mm$ and the screen is placed at $2 \ m$ distance. When blue-green light of wavelength $500 \ nm$ is used, then the distance between two consecutive fringes will be (in $mm$)?

In Young's double slit experiment,if the wavelength of the light used is doubled and the distance between the two slits is halved,the resultant fringe width becomes $n$ times the initial fringe width. Find the value of $n$.

In a Young's double slit experiment,the intensities at two points,for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits,then $\frac{I_1 + I_2}{I_0} = \dots$

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 $... \mathring{A}$.

$A$ beam of light consisting of wavelengths $650 \ nm$ and $550 \ nm$ illuminates the Young's double slits with separation of $2 \ mm$ such that the interference fringes are formed on a screen, placed at a distance of $1.2 \ m$ from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is . . . . . . $\times 10^{-5} \ m$.

$A$ Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is