Question

A beam of light, consisting of two wavelengths 4500 and 7500 is used to obtain intereference fringes in young’s double slit experiment. the separation between the slits is 2 mm and the distance between the plane of slits and the screen is 360cm. what is the minimum distance between two successive regions of complete darkness on the screen?

Answer 1

nλ2 = (n-1) λ1 Where, λ1 = 650 nm



n= 5





Wavelength of another light beam (λ2) = 520 nm, Distance of the slits from the screen = D , Distance between the two slits = d



(a)

Distance of the nth bright fringe on the screen from the central maximum is given by the relation,



x = n λ1(D/d), If third bright fringe. N=3



Then,



x = 3 x 650D/d = 1950(D/d) nm



(b)



Let the nth bright fringe due to wavelength λ2 and (n – 1)th bright fringe due to wavelength λ1 coincide on the screen. We can equate the conditions for bright fringes as:



nλ2 = (n-1) λ1

520n = 650n - 650

n = 5

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



x = λ2D/d

= 5 x 520 x D/d

= 260 D/d nm

Answer 2

2.7 mm will be the distance between two successive darkness