Question

Two coherent waves of amplitude $A_1$ and $A_2$ superpose and the ratio of the maximum intensity to minimum intensity is $\alpha$. What is the value of $\frac{A_1}{A_2}$?

• A. $\frac{1+\alpha }{1-\alpha }$
  
• B. $\frac{1-\alpha }{1+\alpha }$
  
• C. $\frac{\sqrt{\alpha }-1}{\sqrt{\alpha }+1}$
  
• D. $\frac{\sqrt{\alpha }+1}{\sqrt{\alpha }-1}$

Answer 1

The correct option is D $\frac{\sqrt{\alpha }+1}{\sqrt{\alpha }-1}$

Let $I_{max}$ and $I_{min}$ are the maximum and minimum resultant intensities, respectively, then,
$\frac{I_{max}}{I_{min}}=\alpha$

We know that, $I\propto A^2$

$\Rightarrow \frac{A_{max}}{A_{min}}=\sqrt{\frac{I_{max}}{I_{min}}}=\sqrt{\alpha }$

We know that,
$A_{max}=A_1+A_2$ and $A_{min}=A_1-A_2$

$\Rightarrow \sqrt{\alpha }=\frac{A_1+A_2}{A_1-A_2}$

after doing componendo-dividendo on both sides, we get,

$\frac{A_1}{A_2}=\frac{\sqrt{\alpha }+1}{\sqrt{\alpha }-1}$

Hence, $\left(D\right)$ is the correct answer.