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`.

Answer

A, B, C

Explanation

(i) The distance of the nth bright fringe from the central maxima is given by the expression
`y_(n)=(nlamdaD)/(d)`, For 3rd bright fringe `n=3`
`:'``y=(3xx6500xx10^(-10)xx120xx10^(-2))/(2xx10^(-3))=1.17xx10^(-3)m`
(ii) Let nth bright fringe of wavelength `6500 A` coincide with mth bright fringe of wavelength `5200 A `. Their distance will be same from the central bright. Therefore,
`(nlamda_(1)D)/(d)=(mlamda_(2)D)/(d)` `:'` `(n)/(m)=(5200)/(6500)=(4)/(5)`
i.e., at the least distance 4th bright fringe of `6500A` will coincide with 5th bright fringe of `5200A`. Its distance from the central maxima will be
`y_(n)=(4xx6500xx10^(-10)xx120xx10^(-2))/(2xx10^(3))=1.56xx10^(-3)`m`