Question

In young double slit interference experiment using two coherent waves of different amplitudes, the intensity ratio between bright and dark fringes is $3$. Then the value of the ratio of the amplitude of the waves that arrive there is:

• A. $\left[\frac{\sqrt{3}+1}{\sqrt{3}-1}\right]$
• B. $\left[\frac{\sqrt{3}-1}{\sqrt{3}+1}\right]$
• C. $\sqrt{3}:1$
• D. $1:\sqrt{3}$

Answer 1

Answer: (A), (B)

$\left[\frac{\sqrt{3}-1}{\sqrt{3}+1}\right]$
$\left[\frac{\sqrt{3}+1}{\sqrt{3}-1}\right]$

Given : $\frac{I_{max}}{I_{min}}=3$

Let the amplitude of the waves be $A_1$ and $A_2$ respectively.

Maximum intensity $I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2$ where $I_i\propto A_i^2$

$⟹$ $I_{max}=(A_1+A_2{)}^2$ ............(1)

Similarly, minimum intensity $I_{min}=(A_1-A_2{)}^2$

$\therefore$ $\frac{(A_1+A_2{)}^2}{(A_1-A_2{)}^2}=\frac{I_{max}}{I_{min}}=3$

OR $\frac{A_1+A_2}{A_1-A_2}=\sqrt{3}$ $⟹A_1=[\frac{\sqrt{3}+1}{\sqrt{3}-1}]A_2$

$⟹$ $\frac{A_1}{A_2}=[\frac{\sqrt{3}+1}{\sqrt{3}-1}]$ or, $\frac{A_2}{A_1}=[\frac{\sqrt{3}-1}{\sqrt{3}+1}]$