Question

The ratio of maximum intensity and minimum intensity obtained in the interference of waves emitted by two coherent sources is 121 : 81. The ratio of amplitude of two coherent sources will be

• A. 1 : 10
• B. 10 : 1
• C. 81 : 121
• D. 121 : 81

Answer 1

The correct option is B 10 : 1

$\frac{I_{max}}{I_{min}}={\left(\frac{\sqrt{\frac{I_1}{I_2}}+1}{\sqrt{\frac{I_1}{I_2}}-1}\right)}^2$
$={\left(\frac{\frac{A_1}{A_2}+1}{\frac{A_1}{A_2}-1}\right)}^2=\frac{121}{81}$
$\Rightarrow \frac{A_1}{A_2}=\frac{10}{1}$