Question

A beam of light, consisting of two wavelengths 560 nm and 420 nm, is used to obtain interference fringes in a Young’s double slit experiment. Find the least distance from the central maximum, where the bright frings, due to both the wavelengths coincide. The distance between the two slits is 4.0 mm and the screen is at a distance of 1.0 m from the slits.

Answer 1

Given that,

Wavelength of the first light beam, λa = 560 nm

Wavelength of second light beam, λb = 420 nm

Distance of the slits from the screen, D = 1m = 1000 mm

Distance between the two slits, d = 4.0 mm

Let the nth bright fringe due to wavelength, (λb) and (n − 1)th bright fringe due to wavelength coincide on the screen. We can equate the conditions for bright fringes as:

n(λb) = (n - 1) (λa)

420n = 560n - 560

560 = 140n

∴ n = 4

Hence, the least distance from the central maximum can be obtained by the relation:

x = nλb D/d = 4 x 420 D/d =  4 x 420 x 1000/4 = 420000 nm