Question

In an interference pattern, $\frac{I_{max}}{I_{min}}=4$, then $\frac{A_1}{A_2}$ is

Answer 1

Given,

$\frac{I_{max}}{I_{min}}=4\left[\text{Let us say}\right(\alpha \left)\right]$

And we know that,

$\frac{I_{max}}{I_{min}}=\frac{({\sqrt{I}}_1+{\sqrt{I}}_2{)}^2}{({\sqrt{I}}_1-{\sqrt{I}}_2{)}^2}$

Since we know that,

$I\propto A^2\Rightarrow I=k^2{A}^2\left(k\text{is constant}\right)$

$\therefore \sqrt{I}=kA$

So, $\frac{I_{max}}{I_{min}}=\frac{(k{A}_1+k{A}_2{)}^2}{(k{A}_1-k{A}_2{)}^2}=\frac{(A_1+A_2{)}^2}{(A_1-A_2{)}^2}$

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

$\Rightarrow \sqrt{\alpha }{A}_1-\sqrt{\alpha }{A}_2=A_1+A_2$

$\Rightarrow A_1(\sqrt{\alpha }-1)=A_2(\sqrt{\alpha }+1)$

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

Taking $\alpha =4$

we get, $\frac{A_1}{A_2}=3$