In the young's double slit experiment the fringe width of the interference pattern is found to be $$3.2 \times 10^{-4} \mathrm{~m}$$, when the light of wave length $$6400^{\circ} \mathrm{A}$$ is used. What will be change in fringe width if the light is replaced with a light of wave length $$4800^{\circ} \mathrm{A}$$

A light having wavelength $$6400^{\circ} \mathrm{A}$$ is incident normally on a slit of width $$2 \mathrm{~mm}$$. Then the linear width of the central maximum on the screen kept $$2 \mathrm{~m}$$ from the slit is :

In Young's double slit experiment, the two slits are separated by 0.2 mm and they are 1 m from the screen. The wavelength of the light used is 500 nm. The distance between 6th maxima and 10th minima on the screen is closest to

In Young's double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in a liquid of refractive index $$\frac{4}{3}$$ without changing geometrical arrangement, the new fringe width will be