In Young's double slit experiment the slits are 0.5 mm apart and interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that the `9^(th)` bright fringe is at a distance of 8.835 mm from the second dark fringe from the centre of the fringe pattern. Find the wavelength of light used.

Explanation

Given : `d = 0.5 mm = 0.5 xx 10^(-3) m`
`D = 100 cm = 1 m`,
`(X_(9))_(B) - (X_(2))_(D) = 8.835 mm`
`= 88.35 xx10^(-3) m`,
As we know,
The distance of `n^(th)` bright fringe from central fringe is given by :
`(X_(n))B = (nlambdaD)/(d)`
`:. ` For 9th bright fringe
`(X_(9))_(B) = (9lambdaD)/(d)`
Similarly, the distance of `n^(th)` dark fringe from the central fringe is given by
`(X_(n))d = (2n-1). (lambda)/(2). (D)/(d)`
`:.` For the `2^(nd)` band,
`(X_(2))d = (2xx2-1). (lambda)/(2). (D)/(d)`
`(X_(2))d = 3/2. (lambdaD)/(d)`
Now, `(X_(9))_(B) - (X_(2)) = 8.835 xx 10^(-3)`
` 9 - (3)/(2). (lambdaD)/(d) =8.835 xx 10^(-3)`
`((18-3)/(2)). (lambdaD)/(d) = 8.835 xx 10^(-3)`
`(15)/(2). (lambdaD)/(d) = 8.835xx 10^(-3)`
`lambda = (8.835 xx 10^(-3) xx 2 xx d)/(15 xx D)`
`:. lambda = (8.835 xx 10^(-3) xx 2 xx 0.5 xx 10^(-3))/(15 xx 1)`
`lambda = (8.835)/(15) xx 10^(-6)`
`lambda = 5.890 xx 10^(-7) m`
`lambda = 5.890 xx 10^(-7) m`
`:. lambda = 5890 xx 10^(-10) Å`
`:. lambda = 5890 Å`