In Young's double slit experiment two disturbances arriving at a point P have phase difference fo `(pi)/(3)`. The intensity of this point expressed as a fraction of maximum intensity `I_(0)` is

In Young's double slit experiment,the intensity at a point where the path difference is

In Young's double slit experiment, the constant phase difference between two source is (pi)/(2) . The intensity at a point equidistant from theslits in terms of maximum intensity I_(0) is

In a Young's double slit experiment the intensity at a point where the path difference is (lambda)/(6)(lambda being the wavelength of the light used) is I. If I_(0) denotes the maximum intensity then (I)/(I_(0))=…

In a Young's double slit experiment the intensity at a point where tha path difference is (lamda)/(6) ( lamda being the wavelength of light used) is I. If I_0 denotes the maximum intensity, (I)/(I_0) is equal to

In a Young's double slit experiment the intensity at a point where tha path difference is (lamda)/(6) ( lamda being the wavelength of light used) is I. If I_0 denotes the maximum intensity, (I)/(I_0) is equal to

In a Young's double slit experiment the intensity at a point where tha path difference is (lamda)/(6) ( lamda being the wavelength of light used) is I. If I_0 denotes the maximum intensity, (I)/(I_0) is equal to

In a Young's double slit experiment, the intensity at a point where the path difference is (lamda)/6 ( lamda - wavelength of the light) is 1. If I_(0) denotes the maximum intensity, then I/(I_(0)) is equal to

In Young's double slit experiment, the intensity on the screen at a pont where path difference lamda is K. what will be the intensity at the point where the path difference is lamda/4 ?