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The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio

$\frac{{I_{\max } - I_{\min } }} {{I_{\max } + I_{\min } }}$

will be

Option 1)
$\frac{{\sqrt n }} {{n + 1}}$

Option 2)
$\frac{{2\sqrt n }} {{n + 1}}$

Option 3)
$\frac{{\sqrt n }} {{\left( {n + 1} \right)^2 }}$

Option 4)
$\frac{{2\sqrt n }} {{\left( {n + 1} \right)^2 }}$

Answers (1)

best_answer

As we learnt in

Minimum amplitude & Intensity -

$heta = left ( 2n+1 ight )pi$

- wherein

$A_{min }= A_{1}-A_{2}$

$I_{min }= left ( sqrt{I_{1}}-sqrt{I_{2}} ight )^{2}$

Maximum amplitude & Intensity -

When $heta = 0,2pi ---2npi$

- wherein

$A_{max }= A_{1}+A_{2}$

$I_{max }= left ( sqrt{I_{1}}+sqrt{I_{2}} ight )^{2}$

$\frac{I_{1}}{I_{2}}=n$

$I_{max}=\left ( \sqrt{I_{1}} +\sqrt{I_{2}}\right )^{2}=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}$

$=nI_{2}+I_{2}+2\sqrt{X}I_{2}$

$I_{min}=\left ( \sqrt{I_{1}} \sqrt{I_{2}}\right )^{2}$

$=I_{2}+nI_{2}-2\sqrt{x}I_{2}$

$\frac{I_{max}-I_{min}}{I_{max}+I_{min}}=\frac{4\sqrt{x}I_{2}}{2\left ( 1+n \right )I_{2}}=\frac{2\sqrt{n}}{n+1}$

Option 1)

$\frac{{\sqrt n }} {{n + 1}}$

This is incorrect option

Option 2)

$\frac{{2\sqrt n }} {{n + 1}}$

This is correct option

Option 3)

$\frac{{\sqrt n }} {{\left( {n + 1} \right)^2 }}$

This is incorrect option

Option 4)

$\frac{{2\sqrt n }} {{\left( {n + 1} \right)^2 }}$

This is incorrect option

Posted by

Aadil

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