In Youngs Double Slit Experiment the incident radiation is of 590 nm wavelength and the fringe width is `3 xx 10^-3m`. If the incident wave. length is changed to 480 nm, What will be the new fringe width?

If a Young's double slit experiment were conducted inside water instead of air, the fringe width would

In Young's double slit experiment the distance between the two slits is d and wavelength of the light used is lambda . The angular width of the fringe

In Young's double slit experiment the magnitude of the fringe width is beta . If the whole set-up of the experiment is immersed in a liquid of refractive index mu then the magnitude of the fringe width will be

In Young's double slit experiment if distance between the two sources is increased how will the fringe width be changed?

In Young's double slit experiment the fringe width is 2.0 mm. Determine the separation between the 9th bright fringe and the 2nd dark fringe.

Using light of wavelength 600 mm in Young's double slit experiment 12 bands are found on one part of the screen. If the wavelength of light is changed to 400 nm , then what will be the number of bands on that part of the screen?

In Young's double slit experiment the separation d between the slits is 2 mm the wavelength lambda of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20^(@) . To increase the fringe angular width to 0.21^(@) (with same lambda and D ) the separation between the slits needs to be changed to

In Young's double slit experiment, the fringe width is beta . If the entire arrangement is placed in a liquid of refractive index n, the fringe width becomes :