A beam of light consisting of two wavelengths `650 nm` and `520 nm` is used to obtain interference fringes in a Young's double slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength `650 nm`.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is `2 mm` and the distance between the plane of the slits and screen is `120 cm`.

Explanation

Given `lambda_1 = 650 nm = 65 xx 10^(-8)m`
`lambda_2 = 520 nm = 52 xx 10^(-8) m`
Let d be the separation between the slits and D be the distance of the screen from the slits.
Then, the linear distance of `n^(th)` bright fringe from the central maximum is given by
`y_n = (n lambda D)/d`
(a) Distance of third bright fringe (n = 3) from the central maximum for `lambda_1 = 650 nm` is
`lambda_3 = (3 lambda D)/d`
`= (3 xx 65 xx 10^(-8) xx D)/(d)`
Let `d = 2 mm = 2 xx 10^(-3) m and D = 1.2m`
Then, `y_3 = (3 xx 65 xx 10^(-8) xx 1.2)/(2 xx 1^(-3))`
`= 1.17 mm`
Note : The estimation of d and D has been provided in order to arrive at the final result as the information in question is incomplete.
(b) Let `n_1^(th)` bright fringe of wavelength `lambda_1` coincides with `n_2^(th)` bright fringe of wavelength `lambda_2`. Then
`y. = (n_1 lambda_1 D)/(d) = (n_2 lambda_2D)/(d)`
or `(n_1)/(n_2) = (lambda_2)/(lambda_1) = (52 xx 10^(-8))/(65 xx 10^(-8)) = 4/5`
As `n_1 and n_2` are integers , the minimum value of `n_1 and n_2` satisfying the above equation are 4 and 5, respectively.
The distance of `5^(th)` bright fringe for wavelength `52 xx 10^(-8) m` (or `4^(th)` bright fringe for wavelength `65 xx 10^(-8) m)` is
`:. y. = (n lambdaD)/(d)`
`=(4 xx 65 xx 10^(-8) xx 1.2)/(2 xx 10^(-3))`
`:. y. = 15.6 xx 10^(-4)m = 1.56 mm`