Question

Ratio of maximum to minimum intensity during interference of two waves is $25:1$. The amplitude ratio of these two waves is

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\frac{4}{5}$
• D. $\frac{5}{4}$

Answer 1

The correct option is B $\frac{3}{2}$

Given that,

$\frac{I_{max}}{I_{min}}=\frac{25}{1}$

We know that,

Maximum intensity of the resultant wave
$I_{max}\propto (A_1+A_2{)}^2$

Minimum intensity of the resultant wave
$I_{min}\propto (A_1-A_2{)}^2$

So, $\frac{I_{max}}{I_{min}}={\left(\frac{A_1+A_2}{A_1-A_2}\right)}^2=\frac{25}{1}$

$\Rightarrow \frac{A_1+A_2}{A_1-A_2}=\frac{5}{1}$

$\Rightarrow A_1+A_2=5(A_1-A_2)\Rightarrow 4A_1=6A_2$

$\Rightarrow \frac{A_1}{A_2}=\frac{3}{2}$

Thus, option (b) is the correct answer.