Two light waves of wavelengths 800 and 600 nm are used in Young's double slit experiment to obtain interference fringes on a screen placed 7 m away from plane of slits. If the two slits are separated by 0.35 mm, then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be ______ mm.

Solution

Physics sir · 4 steps

Bright fringes of both colours meet where both fringe widths fit a whole number of times: take their LCM.

1. Fringe widths. The nth bright fringe sits at y = nβ, where β = (λ D)/d. With D = 7 m and d = 0.35 × 10⁻³ m, β₁ = (800 × 10⁻⁹ × 7)/(0.35 × 10⁻³) = 16 mm and β₂ = (600 × 10⁻⁹ × 7)/(0.35 × 10⁻³) = 12 mm.
2. Coincidence condition. Two bright fringes overlap when they are at the same distance from the centre. That needs whole numbers n₁ and n₂ with the same y.

   n₁ λ₁ = n₂ λ₂ ⇒ n₁/n₂ = 600/800 = 3/4

3. Smallest orders. The smallest whole numbers are n₁ = 3 for 800 nm and n₂ = 4 for 600 nm. Check: 3 × 16 = 48 mm and 4 × 12 = 48 mm. This is the LCM of 16 and 12.
4. So. The bright fringes first coincide at 48 mm from the central maximum.

The common slip: 64. Putting the 4th order on the 800 nm wave instead of the 3rd gives 4 × 16 = 64 mm. There the 600 nm fringes are at 60 mm and 72 mm, so they do not overlap.

The idea to keep. For two wavelengths in YDSE, the first point where the bright fringes overlap is at the LCM of the two fringe widths. The longer wavelength always needs the smaller order there.

Answer from the NTA final answer key. Solution by PrepWiser.