Two waves each of intensity $I$ interfere with a phase difference of $120^o$. The resultant intensity of the waves will be:

Two waves of amplitudes $A_1$ and $A_2$ respectively are superimposed. The ratio between the maximum and minimum intensities of the resultant waves is $9 : 4$. The value of $A_2 / A_1$ is [Assume $A_1 > A_2$]

Two waves of same frequency and intensity superimpose with each other in opposite phases,then after superposition the

Two interfering waves have the same wavelength,frequency,and amplitude. They are traveling in the same direction but are $90^o$ out of phase. Compared to the individual waves,the resultant wave will have the same.

When two sound waves having amplitude $3$ and $5$ units are superimposed,then the ratio of maximum to minimum intensity of the wave produced is:

The resultant amplitude due to the superposition of two waves $y_1 = 5 \sin (\omega t - kx)$ and $y_2 = -5 \cos (\omega t - kx - 150^{\circ})$ is: