In an interference experiment the ratio of amplitudes of coherent waves is `(a_(1))/(a_(2))=(1)/(3)`. The ratio of maximum and minimum intensities of fringes will be:

Answer

4

Explanation

`(I_(max))/(I_(min))=((a_2+a_1)/(a_2-a_1))^(2)`
Given `a_1/a_2=1/3`
`implies(a_2+a_1)/(a_(2)-a_1)=(3+1)/(3-1)=2 implies(I_(max))/(I_(min))=4`