If the ratio of amplitudes of two interfering waves is $4: 3$, then the ratio of maximum and minimum intensity is

Given, the ratio of amplitudes of two interfering waves is $4: 3$ i.e $\frac{A_1}{A_2}=\frac{4}{3}$ ...(i) We know that, intensity of wave is proportional to square of amplitude. $\therefore \frac{I_{\max }}{I_{\min }}=\left(\frac{A_{\max }}{A_{\min }}\right)^2$, where maximum and minimum amplitudes are $A_{\text {max }}=A_1+A_2$ and $\quad A_{\min }=A_1-A_2$ Substituting the values, we get $\frac{I_{\max }}{I_{\min }}=\left(\frac{A_1+A_2}{A_1-A_2}\right)^2$ Multiplying and dividing by $A_2^2$ in RHS, we get. $\frac{I_{\max }}{I_{\min }}=\left(\frac{\frac{A_1}{A_2}+1}{\frac{A_1}{A_2}-1}\right)^2=\left(\frac{\frac{4}{3}+1}{\frac{4}{3}-1}\right)^2=\left(\frac{7}{1}\right)^2=\frac{49}{1}$