Question

Ratio of amplitude of two waves is $1:16$. The ratio of maximum to minimum intensity when the waves interfere is

• A. $289:25$
• B. $1:256$
• C. $289:225$
• D. $3:1$

Answer 1

The correct option is C $289:225$

$\frac{A_1}{A_2}=\frac{1}{16}$
$\frac{\text{Maximum intensity}}{\text{Minimum intensity}}=\frac{(A_1+A_2{)}^2}{(A_1-A_2{)}^2}=\frac{(1+16{)}^2}{(1-16{)}^2}$
Dividing by $A_2^2$ in the numerator and denominator
$\frac{A_{max}}{A_{min}}=\frac{(17{)}^2}{(15{)}^2}=\frac{289}{225}=289:225$