In Young's double slits experiment, the position of 5ᵗʰ bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is :

Solution

Physics sir · 3 steps

The nth bright fringe sits n fringe widths from the centre, so put n = 5 and solve for d.

1. Bright fringe position. The nth bright fringe is at yₙ = (nλ D)/d from the central maximum. That is n fringe widths, each of size (λ D)/d.

   d = (nλ D)/yₙ

2. Put in numbers. Use SI units: n = 5, λ = 600 × 10⁻⁹ m, D = 1 m, y₅ = 0.05 m. Then d = (5 × 600 × 10⁻⁹ × 1)/0.05 = 6 × 10⁻⁵ m.
3. So. d = 6 × 10⁻⁵ m = 60 μm, which is option (a).

The idea to keep. Bright fringes sit at yₙ = nλ D/d, counting the central maximum as n = 0. Convert cm and nm to metres before you divide.

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