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

`P` is `11^(th)` fringe from `Q` i.e. `10^(th)` fringe from `O`.
For `10^(th)` bright fringe, `Deltax=nlambda=10lambda=S_(1)B`
`S_(1)B=10lambda=10xx600xx10^(-9)=6xx10^(-6)m`