Consider an interference experiment using eight equally spaced slits. Determine the smallest phase different in the waves from adjacent slits such that the resultant wave has zero amplitude.

In Young's double slit experiment, the phase difference between the two waves reaching at the location of the third dark fringe is

In Young's double slit experiment, the phase difference between the two waves reaching at the location of the third dark fringe is

In a two-slit interference pattern, the maximum intensity is l_0 . (a) At a point in the pattern where the phase difference between the waves from the two slits is 60^@ , what is the intensity? (b) What is the path difference for 480nm light from the two slits at a point where the phase angle is 60^@ ?

In a two-slit interference pattern, the maximum intensity is l_0 . (a) At a point in the pattern where the phase difference between the waves from the two slits is 60^@ , what is the intensity? (b) What is the path difference for 480nm light from the two slits at a point where the phase angle is 60^@ ?

In Young's double slit interference experiment the wavelength of light used is 5000A .If the phase difference between waves reaching a point p on the screen is 6 pi ,then at the point p is ?

In Young's double slit interference experiment, using two coherent waves of different amplitudes, the intensities ratio between bright and dark fringes is 3 Then, the value of the wave amplitudes ratio that arrive there is

In Young's double slit interference experiment, using two coherent waves of different amplitudes, the intensities ratio between bright and dark fringes is 3 Then, the value of the wave amplitudes ratio that arrive there is

Do you think that interference pattern in Young's double slit experiment is the consequence of the superposition of the diffracted waves from the two slits?

In Youngs double slit experiment the slits are of different lengths and widths. The amplitude of the light waves is directly proportional to the :