Interference fringes were produced by Young's double slit method, the wavelength of light used being 6000Å. The separation between the two slits is 2 mm.The distance between the slits and screen is 10 cm. When a transparent plate of thickness 0.5 mm is placed over one of the slits. the fringe pattern is displaced by 5 min. Find the refractive index of the material of the plate.

In Young's double slit experiment, the slits are 3 mm apart. The wavelength of light used is 5000 Å and the distance between the slits and the screen is 90 cm. The fringe width in mm is

YDSE is conducted using light of wavelength 6000 Å to observe an interference pattern. The slits are 2mm apart. The fringes are observed on a screen 10cm away from the slits. When a film of some material 0.5mm thick was placed over one of the slits, the fringe pattern shifted by a 5mm . What is the refractive index of the material of the film ?

A two slit Young's interference experiment it done with monochromatic light of wavelength 6000 A . The slits are 2 mm apart. The fringes are observed on a screen placed 10 cm away from the slits. Now a transparent plate of thickness 0.5 mm is placed in front of one of the slits and it if found that the interference pattern shifts by 5 mm . The refractive index of the transparent plate is :

A double slit experiment is done with monochromatic light of wavelength 6000 Å . Slits are 2 mm apart and fringes are observed on a screen placed 10 cm away from the slits. If a transparent plate of thickness 0.5 mm is placed in front of one of the slit, interference pattern shifts by 5 mm. Then refractive index of transparent plate should be :

In Young's double slit experiement when wavelength used is 6000Å and the screen is 40cm from the slits, the fringes are 0.012cm wide. What is the distance between the slits?