(a) In Young's double slit experiment, derive the condition for (i) constructive interference and (ii) destructive interference at a point on the screen. (b) A beam of light consisting of two wavelengths, 800nm and 600nm is used to obtain the interference fringes in a Young's double slit experiment on a screen placed 1.4m away. If the two slits are separated by 0.28nm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.

Young double slit experiment: Let S1 and S2 are the two narrow rectangular slits placed perpendicular to the plane of paper. Screen is placed on the perpendicular bisector of S1S2 and is illuminated with monochromatic light. The slits are separated by a small distance d. A screen is placed at a distance D from S1S2. Consider a point P on the screen at a distance x from O. The path difference between the waves reaching P from S1 and S2 is P = S2P - S1P, draw a perpendicular S1N on S2P P = S2P - S1P = S2P - NP = S2N From the triangle S2NS2S1 = sinθ ∴P = S2N = S2S1sinθ = dsinθ when θ is small, sinθ ≈ θ = tanθ = xD ∴P = xdD For constructive interference xdD = nλ; n = 0, 1, 2. Position of nth bright fringe Xn = nDλd when n = 0 Xn = 0, central bright fringe is formed at O. For destructive interference , xdD = (2n - 1)λ2 Xn = (2n + 1)λD2 Thus alternative bright fringe and dark fringe are formed on the screen (b) Let the nth1 maximum corresponds to n1 and nth2 maximum corresponds to n2, then n1λ1Dd = n2λ2Dd ➝n1n2 = 600800 = 34 ymin = N1λ1Dd = 3 × 800 × 10⁻⁹ × 1.40.28 × 10⁻³ = 12nm