Question

A beam of light consisting of two wavelengths 560 nm and 420 nm is used to obtain interference fringes in young's double slit experiment find the least distance from the central maximum where the bright fringes due to both the wavelengths coincide the distance between two slits is 4 m and the screen is at a distance of 1 m from the slits

Answer 1

Distance from the central maxima will be 420 nm

Explanation :

Let n fringes of 560 nm coincides with (n+1) fringes of 420 nm

Distance from the central maxima is given by,

X = nλ₁D/d = (n+1)λ₂D/d

=> (n+1)/n = 560/420 = 4/3

=> n = 3

Given,

D = distance of the screen = 1 m

d = distance between two slits = 4 m

Hence, Distance from the central maxima is given by,

X = nλ₁D/d

= 3 x 560 x 10⁻⁹ x 1/4

=> X = 420 nm

Answer 2

It'll be 420..........