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. (a) `6xx10^-6m`
B. (b) `6.6xx10^-6m`
C. (c) `3.138xx10^-7m`
D. (d) `3.144xx10^-7m`

Answer

A

Explanation

P is the position of `11^(th)` bright fringe from Q. From central position O, P will be the position of `10^(th)` bright fringe.
Path difference between the waves reaching at `P=S_1B=10lambda=10xx6000xx10^(-10)=6xx10^-6m`.