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 mm from the central maximum . Determine the spacing of the slits.

Answer

`1.52`

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 m m `