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 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 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, 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.