In a Young's double-slit experiment,`lambda=500nm`. `d=1nm` and `D=1m` .The minimum distance from the central maximum,at which the intensity is half of the maximum intensity,is `x times10^(-4)m` .The value of `x is(---)`.

In a Young's double slit experiment lamda= 500nm, d=1.0 mm andD=1.0m . Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.

In Young's double-slit experiment lambda = 500 nm, d = 1 mm , and D = 4 m . The minimum distance from the central maximum for which the intensity is half of the maximum intensity is '**' xx 10^(-4) m . What is the value of '**' ?

In a YDSE experiment, d = 1mm, lambda = 6000Å and D= 1m. The minimum distance between two points on screen having 75% intensity of the maximum intensity will be

In a double experiment D=1m, d=0.2 cm and lambda = 6000 Å . The distance of the point from the central maximum where intensity is 75% of that at the centre will be:

In a Young's double slit experiment, I_0 is the intensity at the central maximum and beta is the fringe width. The intensity at a point P distant x from the centre will be

In a young's double slit experiment, two wavelength of 500 nm and 700 nm were used. What is the minimum distance from the central maximum where their maximas coincide again ? Take D//d = 10^(3) . Symbols have their usual meanings.

In a Young's double slit experiment, slits are seperated by a distance d and screen is at distance D, (D gt gt d) from slit plane. If light source of wavelength lambda(lambda ltlt d) is used. The minimum distance from central point on the screen where intensity is one fourth of the maximum is (D lambda)/(nd) . Find the value of n.

In Young's double slit experiment, the intensity of central maximum is I . What will be the intensity at the same place if one slit is closed ?