Two sound waves each of wavelength `lamda` and having the same amplitude A form two sources `S_(1) and S_(2)` interfere at a point P. If the path difference `S_(2)P - S_(1)P = (lamda)/(3)`, then the amplitude of the resultant were at P will be

Two waves of equal amplitude a from two coherent sources (S_1 & S_2) interfere at a point R such that S_2R-S_1R =1.5 lambda The intensity at point R Would be

In the case of light waves from two coherent sources S_(1) and S_(2) , there will be constructive interference at an arbitrary point P, the path difference S_(1)P - S_(2)P is

Two waves of the same frequency have amplitudes 1.60 and 2.20. They interfere at a point where their phase difference is 60.0^(@) . What is the resultant amplitude?

A monochromatic light source of wavelength lamda is placed at S. three slits S_(1),S_(2) and S_(3) are equidistant from the source S and the point P on the screen S_(1)P-S_(2)P=lamda//6 and S_(1)P-S_(3)P=2lamda//3 . if I be the intensity at P when only one slit is open the intensity at P when all the three slits are open is-

Two S.H.M.’s have zero phase difference and equal amplitudes A. The resultant amplitude on their composition will be

At two point P and Q on screen in Young's double slit experiment, waves from slits S_(1) and S_(2) have a path difference of 0 and (lamda)/(4) respectively. The ratio of intensities at P and Q will be:

The wavelength of the waves arrving at P from two coherent sources S_1 and S_2 is 4m, while intensity of each wave is I_0 . The resultant intensity at P is 2I_0 . Find the minimum valus of S_2P