Question

a Beam of light consisting of two wavelengths 560nm and 420nm is used to obtain interference fringes in a young's double-slit experiment. Find the least distance from the central maxima where the bright fringes due to both wavelengths coincide. The distance between the two slits is 4mm and the screen is at a distance of 1mm from the slits.

Answer 1

The distance from the central maxima where the bright fringes due to both wavelengths coincide will be 420 nm

Explanation :

Given 1st wavelength = 560nm = 560 x 10⁻⁹m

2nd wavelength = 420 nm = 420 x 10⁻⁹m

D = distance of the screen and slit = 1 mm

d = distance between 2 slits = 4 mm

let n fringes of wavelength 560nm coincides with n+1 fringes of wavelength 420nm

we know that,

x = nλD/d

=> x = nλ₁D/d = (n+1)λ₂D/d

=> nλ₁ = (n+1)λ₂

=> n x 560 = (n+1) x 420

=> n x 4 = (n+1) x 3

=> 4n = 3n + 3

=> n = 3

Hence distance central maxima,

x = nλD/d

= 3 x 520 x 10⁻⁹ x 1/4

= 420 nm