Two doherent sources of intensity ratio `alpha` interfere in interference pattern `(I_(max)-I_(min))/(I_(max)+I_(min))` is equal to

Options

A. `(2alpha)/(1+alpha)`
B. `(2sqrt(alpha))/(1+alpha)`
C. `(2alpha)/(1+sqrt(alpha)`
D. `(1+alpha)/(2alpha)`

Answer

B

Explanation

`(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))`