In Young's double-slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $k$ units; $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\lambda/4$ will be:

Two sources of light are $0.6 \, mm$ apart and the screen is placed at a distance of $1.2 \, m$ from them. $A$ light of wavelength $6000 \, \mathring{A}$ is used. The phase difference between the two light waves interfering on the screen at a point at a distance of $3 \, mm$ from the central bright band is:

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 a double-slit experiment,the angular width of the fringes is $0.20^o$ for sodium light $(\lambda = 5890 \ \mathring{A})$. In order to increase the angular width of the fringes by $10\%$,the necessary change in the wavelength is:

In Young's experiment,obtain the distance between two consecutive bright fringes and two consecutive dark fringes.

In a double-slit experiment performed in air,the angular width of a fringe is found to be $0.15^{\circ}$ on a screen placed $80 \ cm$ away. The wavelength of light used is $490 \ nm$. What is the angular width of the fringe if the entire apparatus is immersed in a medium of refractive index $\frac{5}{3}$ (in $^{\circ}$)?