Question

In interference, I_(max)/I_(min) = alpha, find. (a) A_(max)/A_(min) (b) A_1/A_2 (c) I_1/I_2

Answer 1

1. A_(max)/ A_(min)= ∝
2. A_1/A_2= $\frac{\sqrt{ \alpha} +1}{\sqrt{ \alpha} -1}$
3. I_1/ I_2=$(\frac{\sqrt{ \alpha} +1}{\sqrt{ \alpha} -1})^{2}$

1. We know that for interference, I is directly proportional to A² .

I ∝ A²

A ∝ √I

∴ A_max/A_min = $\sqrt{\frac{I_{max}}{I_{min} }$

∴ A_max/A_min= $\sqrt{\alpha }$  

2. A_1/A_2 can be found from the above relation in 1.

$\frac{I_{max} }{I_{min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}$

$\frac{(\sqrt{\frac{I_1}{I_2}) } + 1)^2}{(\sqrt{\frac{I_1}{I_2}} - 1)^2 }$  

   $\alpha = \frac{(({\frac{A_1}{A_2}) } + 1)^2}{(({\frac{A_1}{A_2}}) - 1)^2 }$  

On simplifying we get ,

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

3. For I_1/ I_2,

$\frac{I_1}{I_2} =( \frac{A_1}{A_2} )^{2}$

in 2, we have already found the value of A_1/A_2.

Substituting that value we get,

I_1/I_2 = $(\frac{\sqrt{ \alpha} +1}{\sqrt{ \alpha} -1})^{2}$