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

11th bright fringe is formed at P, as measured from Q. `therefore` 10th bright fringe is formed at P from the centre O.
`therefore` the path difference between the waves reaching at P
from `S_(1) and S_(2)`
`=S_(1)B=10lamda=10xx6000xx10^(-10)=6xx10^(-6)m`