In YDSE path difference at a point on screen is `lambda/8` . Find ratio of intensity at this point with maximum intensity.

In an interference experiment using waves of same amplitude, path difference between the waves at a point on the screen is lambda//4 . The ratio of intensity at this point with that at the central bright fringe is

The path of difference between two interfering waves at a point on the screen is (lambda)/(8) . The ratio of intensity at this point and that at the central fringe will be :

The path difference between two interfering waves of equal intensities at a point on the screen is lambda//4 . The ratio of intensity at this point and that at the central fringe will be

The path difference between two interfering waves of equal intensities at a point on the screen is lambda//4 . The ratio of intensity at this point and that at the central fringe will be

The path difference between two interfering waves at a point on the screen is lambda // 6 , The ratio of intensity at this point and that at the central bright fringe will be (assume that intensity due to each slit is same)

The path difference between two interfering waves at a point on the screen is lambda // 6 , The ratio of intensity at this point and that at the central bright fringe will be (assume that intensity due to each slit is same)

The path difference between two interfering waves at a point on the screen is lambda // 6 , The ratio of intensity at this point and that at the central bright fringe will be (assume that intensity due to each slit is same)

In Young’s double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is frac{7 lambda}{4} . The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is :