Two beams of light having intensities I and 4I interfere to produce a fringe pattern on the screen.Phase differenced between the beamsis (pi)/(2) atpoint A and pi at point B.Then difference between the resultant intensities at A and B is :

Two beams of light having intensities I and 4I interfere to produce a

1.	Two beams of light having intensities I and 4I interfere to produce a fringe pattern on the screen.Phase differenced between the beamsis (pi)/(2) atpoint A and pi at point B.Then difference between the resultant intensities at A and B is :
Answer» <html><body><p>3I<br/>4I<br/>5I<br/>6I</p>Solution :`I_(1) = I , I_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) = 4I` <br/> `theta_(1) = (<a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a>)/(2), theta_(2) = pi` <br/> Now, `I_(A) = I_(1) + I_(2) +2sqrt(I_(1)I_(2)) <a href="https://interviewquestions.tuteehub.com/tag/cos-935872" style="font-weight:bold;" target="_blank" title="Click to know more about COS">COS</a> (pi)/(2)` <br/> ` = I + 4I + 0 = 5I` <br/> `I_(<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) = I_(1) + I_(2) +2sqrt(I_(1)I_(2)) cos pi = 1` <br/> `therefore I_(A) - I_(B) = 5I - I = 4I`</body></html>

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Two beams of light having intensities I and 4I interfere to produce a fringe pattern on the screen.Phase differenced between the beamsis (pi)/(2) atpoint A and pi at point B.Then difference between the resultant intensities at A and B is :

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