In Young's double slit experiment, carried out with light of wavelength 5000 Aᵒ, the distance between the slits is 0.3 mm and the screen is at 200 cm from the slits. The central maximum is at x=0 cm. The value of x for third maxima is __________ mm.

Solution

Physics sir · 3 steps

The nth bright fringe sits n fringe widths from the centre, so put n = 3 into nλD/d.

1. Bright fringe position. The nth maximum forms where the path difference is nλ. On the screen that point lies at xₙ = (nλ D)/d from the central maximum.
2. Convert to SI. λ = 5000 Å = 5 × 10⁻⁷ m, d = 0.3 mm = 3 × 10⁻⁴ m and D = 200 cm = 2 m.

   x₃ = (3 × (5 × 10⁻⁷) × 2)/(3 × 10⁻⁴) = 1 × 10⁻² m

3. So. x₃ = 0.01 m = 10 mm, so the answer is 10.

The common slip: 3.33. Stopping at the fringe width λ D/d gives 3.33 mm, which is only the first maximum. Multiply by 3 for the third.

The idea to keep. Bright fringes sit at whole multiples of the fringe width β = λ D/d. Put every length in metres first, then convert the answer to the unit asked for.

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