The path difference between two interfering light waves meeting at a point on the screen is $\left(\frac{87}{2}\right) \lambda$. The band obtained at that point is

The intensity of each source in Young's double slit experiment is $I_0$. The distance between the slits is $d = 5\lambda$,where $\lambda$ is the wavelength of the monochromatic light used. What will be the intensity of light in front of one of the slits on a screen (where the slit and screen are at a distance $D = 10d$)?

Two light waves of wavelengths $800 \, nm$ and $600 \, nm$ are used in Young's double slit experiment to obtain interference fringes on a screen placed $7 \, m$ away from the plane of the slits. If the two slits are separated by $0.35 \, mm$,then the shortest distance from the central bright maximum to the point where the bright fringes of the two wavelengths coincide will be $............. \, mm$.

In a Young's double-slit experiment with wavelength $\lambda$,the fringe width is $\beta$. When two glass plates of thicknesses $t_1$ and $t_2$ $(t_1 > t_2)$ and refractive index $\mu$ are placed in the paths of the two light beams respectively,by what distance will the fringe pattern shift?

$A$ 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 . . . . . . $\times 10^{-5} \ m$.

Two slits are made $10 \ mm$ apart and the screen is placed $1.5 \ m$ away. What is the fringe separation when a wavelength of $7000 \ \mathring{A}$ is used (in $\mu m$)?