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Help me please, - Optics - NEET

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)
97 Views

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

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