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 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 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 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 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 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 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 Young's double slit experiment,two slits are 1.5 mm apart and a screen in placed at a distance 1.2 m from the plane of the slits .The slits are illuminated with light of wavelength 6,000 overset @A .Find the inimum distance from the central maximum for which the intensity is half of the maximum intensity.

In a Young's double slit experiment, D equals the distance of screen and d is the separation between the slits. The distance of the nearest point to the central maximum where the intensity is same as that due to a single slit is equla to