Consider a Young's double slit experiment, What should be the slit separation `d` in terms of wavelength `lambda` such that the `5th` minimum occurs directly in front of the slit `(S_(1))`? plane of a slits and plane of screen have a separation of `3d`

Consider a Young's double slit experiment as shown in figure. What would be the slit separation d in terms of wavelength lamda such that the first minima occurs directly in front of the (S_(1)) ?

In Young.s double slit experiment the band width is minimum for the colour

In Young'double-slit interference experiment, if the slit separation is made threefold, the fringe width becomes

In Young's double slit experiment, if the distance between two slits is equal to the wavelength of used light. Then the maximum number of bright firnges obtained on the screen will be

In a Young’s double slit experiment, the ratio of the slit’s width is 4 : 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be:

In a Young's double slit experiment, (slit distance d) monochromatic light of wavelength lambda is used and the fringe pattern observed at a distance D from the slits. The angular position of the bright fringes are

In Young's double slit experiment, if the two slits are illuminated with separate sources, no interference pattern is observed because

White light is used to illuminate the two slits in a Young's double slit experiment. The separation between the slits is b and the screen is at a distance d(gtgtb) from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing. Some of these missing wavelength are

White light is used to illuminate the two slits in a Young's double slit experiment. The separation between the slits is b and theh screen is at a distance d (gt gtb) from the slits At a point on the screen directly in front of one of the slits, certain wavelengths are missing some of these missing wavelengths are