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

In young double slit experiment, the intensity at a point where path difference is lamda/6 is l. IF l_0 denotes the maximum intensity l/l_0 .

In the Young's double-slit experiment,the intensity of light at a point on the screen where the path difference is lambda is K ( lambda being the wave length of light used).The intensity at a point where the path difference is lambda/4 , will be

The wavelength of the light used in Young's double slit experiment is lambda . The intensity at a point on the screen is I, where the path difference is (lambda)/(6) . If I_(0) denotes the maximum intensity, then the ratio of I and I_(0) is

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

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is lambda=l . The intensity of light at a point where the path difference becomes lambda//3 is

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 ?

In Young's double slit experiment the intensity of light at a point on the screen where the path difference lambda is K. The intensity of light at a point where the path difference is (lambda)/(6) [ lambda is the wavelength of light used] is