Young's double -slit experiment is performed with a 600 nm light and a distance of 2.00 m between the slits and the screen. The tenth interference minimum is observed 7.5 ,, from the central maximum . Determine the spacing of the slits.

Explanation

In equation `d sin theta = (m + (1)/(2))lambda`,
The first minimum is described by `m = 0`
And the tenth by `m = 9 sin theta = (lambda)/(d) (9 + (1)/(2))`
Also, `tan theta = y//D`
But for small `theta, sin theta ~~ tan theta`
Thus, `d = (9.5 lambda)/(sin theta) = (9.5 lambda D)/(y)`
`D = (9.5(6000 xx 10^(-10) m) (2.00 m))/(7.5 xx 10^(-3) m)`
`= 1.52 xx 10^(-3) m = 1.52 mm`