In a Young's double-slit experiment,the distance between the two sources is $0.1 / \pi \, mm$. The distance between the source and the screen is $25 \, cm$. The wavelength of light is $5000 \, \mathring{A}$. The angular position of the first dark fringe is ........$^o$.

Two wavelengths $\lambda_1 = 450 \ nm$ and $\lambda_2 = 650 \ nm$ are used in Young's double slit experiment. The minimum order of fringe produced by $\lambda_2$ which overlaps with a fringe produced by $\lambda_1$ is $n$. The value of $n$ is . . . . . . .

In a Young's double slit experiment,the ratio of the amplitude of light coming from the slits is $2:1$. The ratio of the maximum to minimum intensity in the interference pattern is:

$A$ beam of light consisting of two wavelengths $6500 \mathring{A}$ and $5200 \mathring{A}$ is used to obtain interference fringes in Young's double-slit experiment. The distance between slits is $2 \text{ mm}$ and the distance of the screen from the slits is $120 \text{ cm}$. What is the least distance from the central maximum where the bright fringes due to both wavelengths coincide (in $\text{ cm}$)?

In Young's double-slit experiment,the amplitudes of two sources are $3a$ and $a$ respectively. The ratio of intensities of bright and dark fringes will be: (in $: 1$)

Write the formula for fringe width.