In the figure is shown Young's double slit experiment. Q is the position of the first bright fringe on the right side of O. P is the `11^(th)` bright fringe on the other side, as measured from Q. If the wavelength of the light used is `600 nm`. Then `S_(1)B` will be equal to

Options

A. `6xx10^(-6)m`
B. `6.6xx10^(-6)m`
C. `3.138xx10^(-7)m`
D. `3.144xx10^(-7)m`

Answer

A

Explanation

Path difference, `S_(1)B=Deltax=nlamda`
As P is the position of `11^(th)` fringe from Q, so from O it will be 10.
`therefore" "Deltax=nlamda=10lamda`
= `10xx6000xx10^(-10)=6xx10^(-6)m`