In Young's double slit experiment the two slits are 0.6 mm distance apart. Interference pattern is observed on a screen at a distance 80 cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be ____________ nm.

Solution

Physics sir · 4 steps

Opposite one slit means y = d/2 from the centre, so put that into the first dark-fringe condition.

1. Locate the point. The central bright fringe sits on the line midway between the slits. A point directly opposite one slit is therefore at y = d/2 = 0.3 mm from the centre.
2. First dark fringe. At height y the path difference is yd/D. The first dark fringe needs a path difference of λ/2.

   yd/D = λ/2 ⇒ ((d/2) d)/D = λ/2 ⇒ λ = d²/D

3. Put in numbers. λ = (0.6 × 10⁻³)²/0.8 = (3.6 × 10⁻⁷)/0.8 = 4.5 × 10⁻⁷ m.
4. So. The wavelength is 4.5 × 10⁻⁷ m, that is 450 nm.

The common slip: 900. Taking the point at y = d = 0.6 mm from the centre, instead of d/2, gives 900 nm.

The idea to keep. Fringe positions are measured from the centre line between the slits, not from a slit. A point in front of one slit is at y = d/2, and the first dark fringe needs a path difference of λ/2.

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