InterviewSolution
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InterviewSolution
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Two doherent sources of intensity ratio `alpha` interfere in interference pattern `(I_(max)-I_(min))/(I_(max)+I_(min))` is equal toA. `(2alpha)/(1+alpha)`B. `(2sqrt(alpha))/(1+alpha)`C. `(2alpha)/(1+sqrt(alpha)`D. `(1+alpha)/(2alpha)` |
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Answer» Correct Answer - B `(I_(max)-I_(min))/(I_(max)+I_(min))=((a_(1)+a_(2))^(2)-(a_(1)-a_(2))^(2))/((a_(1)+a_(2))^(2)+(a_(1)-a_(2))^(2))` `[becauseI_(max)=(a_(1)+a_(2))^(2),I_(min)=(a_(1)-a_(2))^(2)` where a=amplitude] `=(4a_(1)a_(2))/(2(a_(1)^(2)+a_(2)^(2)))=(2a_(1)a_(2))/((a_(1)^(2)+a_(2)^(2)))` Now, dividing the numerator and denominator by `a_(1)a_(2)` we get `(I_(max)-I_(min))/(I_(max)+I_(min))=2/((a_(1)/a_(2)+a_(2)/a_(1)))(because I_(1)/I_(2)=alpharArra_(1)/a_(2)=sqrt(alpha))` `rArr(I_(max)-I_(min))/(I_(max)+I_(min))=2/((sqrt(alpha+1/sqrt(alpha))))=(2sqrt(alpha))/((alpha+1))` |
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