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

Let the amplitudes of the two sound waves be $A_1 = 3$ units and $A_2 = 5$ units. The maximum resultant amplitude occurs during constructive interference: $A_{\text{max}} = A_1 + A_2 = 8$ units. The minimum resultant amplitude occurs during destructive interference: $A_{\text{min}} = |A_1 - A_2| = 2$ units. Since intensity is proportional to the square of amplitude, the ratio is: $\frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{A_{\text{max}}}{A_{\text{min}}}\right)^2 = \left(\frac{8}{2}\right)^2 = 16$ The ratio of maximum to minimum intensity is 16:1.

The correct option is D.