The path difference between two interfering waves at a point on the screen is `lambda // 6`, The ratio of intensity at this point and that at the central bright fringe will be (assume that intensity due to each slit is same)

Options

A. 0.853
B. 8.53
C. 0.75
D. 7.5

Answer

C

Explanation

At path difference `(lambda)/6`, phase difference is `(pi)/3`
`I=I_(O)+I_(O)+2I_(O) cos .(pi)/(3) =3I_(O) I_(max)=4I_(O)`
So the required ratio is `(3I_(O))/(4I_(0))=0.75`