If the ratio of amplitudes of two interfering waves is $4: 3$, then the ratio of maximum and minimum…
If the ratio of amplitudes of two interfering waves is $4: 3$, then the ratio of maximum and minimum intensity is
$16: 18$
$18: 16$
$49: 1$
$94: 1$
Solution
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.
$\frac{I_{\max }}{I_{\min }}=\left(\frac{\frac{A_1}{A_2}+1}{\frac{A_1}{A_2}-1}\right)^2=\left(\frac{\frac{4}{3}+1}{\frac{4}{3}-1}\right)^2=\left(\frac{7}{1}\right)^2=\frac{49}{1}$