Question

If the two wavelength of 800 NM and 600NM are used to get interference on screen 1.4 m away the distance of the two slit is 0.3 mm find the least distance from centre of fringe where the bright fringe of two will coincide

Answer 1

Dear Student,

Let the n1th maximum correspond to λ1 (800 nm) and n2th correspond to λ2 (600 nm).

Then,

n1λ1D/d = n2λ2D/d

n1 λ1 = n2 λ2

n1/n2 = 600/800 =3/4

Thus, 3rd maximum correspond to n1 and 4th to n2.

The minimum distance from the central maximum to the point where the maxima dye to both the wavelengths coincide is,

V1 = n1λ1D/d = 3×800×1.4 ×10^9/0.3 ×10^6 = 11200000 nm