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If the ratio of the intensity of two coherent sources is 4 then the visibility \frac{\left ( I_{max}-I_{min} \right )}{\left ( I_{max}+I_{min} \right )}  of the fringes is

Option: 1 4

Option: 2 0.8

Option: 3 0.6

Option: 4 9

Answers (1)

best_answer

 

Resultant Intensity of two wave -

I= I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos \theta

- wherein

I_{1}= Intencity of wave 1

I_{2}= Intencity of wave 2

\theta = Phase difference

 

 

 

Maximum amplitude & Intensity -

When \theta = 0,2\pi ---2n\pi
 

- wherein

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

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

 

 

Minimum amplitude & Intensity -

\theta = \left ( 2n+1 \right )\pi
 

- wherein

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

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

 

 

\frac{I_{max}- I_}{I_{max} + I_{min}} = \frac{\left ( \sqrt{I}_{1}+ \sqrt{I}_{2} \right )^{2} -\left ( \sqrt{I}_{1} -\sqrt{I}_{2}\right )^{2} }{\left ( \sqrt{I}_{1}+ \sqrt{I}_{2} \right )^{2}+\left ( \sqrt{I}_{1}+ \sqrt{I}_{2}\right )^{2}} -

= \frac{I_{1}}{I_{2}}\times \frac{\left [ 1+\sqrt{\frac{I_{2}}{I_{1}}} \right ]^{2}- \left [ 1-\sqrt{\frac{I_{2}}{I_{1}}} \right ]^{2}}{\left [ 1+\sqrt{\frac{I_{2}}{I_{1}}} \right ]^{2}+\left [ 1-\sqrt{\frac{I_{2}}{I_{1}}} \right ]^{2}}

= \frac{\left ( 1+2 \right )^{2} -\left ( 1-2 \right )^{2}}{(1+2)^{2}+(1+2)^{2}} = \frac{8}{10} = \frac{4}{5}

Posted by

Sanket Gandhi

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