In Young's double slit experiment, the slits are horizontal. The intensity at a point P as shown in figure is `(3)/(4) I_(0)`,where`I_(0)` is the maximum intensity.
Then the value of `theta` is,
(Given the distance between the two slits `S_(1)` and `S_(2)` is `2 lambda`)

In Young.s double slit experiment intensity at a point is (1"/"4) of the maximum intensity. Angular position of this points is

In a Young's double slit experiment the amplitudes of the two waves incident on the two slits are A and 2A . If I_(0) is the maximum intensity, then the intensity at a spot on the screen where the phase difference between the two interfering waves is phi .

In Young's double-slit experiment, the intensity at a point P on the screen is half the maximum intensity in the interference pattern. If the wavelength of light used is lambda and d is the distance between the slits, the angular separation between point P and the center of the screen is

In Young's double-slit experiment, the intensity at a point P on the screen is half the maximum intensity in the interference pattern. If the wavelength of light used is lambda and d is the distance between the slits, the angular separation between point P and the center of the screen is

In Young's double-slit experiment, the intensity of light in front of one of the slits on a screen is I_(0)//2 where I_(0) is the maximum intensity. The distance between the slits is 5lambda where lambda is the wavelength of monochromatic light. How far away is the screen from the slit?

In Young's double slit experiment, the maximum intensity is I_(0) . What is the intensity at a point on the screen where the path difference between the interfering waves is lamda/4 ?

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