A beam of light consisting of two wavelengths 7000 Aᵒ and 5500 Aᵒ is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is 2.5 mm and the distance between the plane of slits and the screen is 150 cm. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is n × 10⁻⁵ m. The value of n is __________.

Solution

Physics sir · 4 steps

Bright fringes coincide when n₁λ₁ = n₂λ₂, so find the smallest whole numbers that fit.

1. Condition for coinciding. The nth bright fringe sits at y = (nλ D)/d. Both wavelengths have a bright fringe at the same spot when n₁λ₁ = n₂λ₂.

   n₁/n₂ = λ₂/λ₁ = (550 nm)/(700 nm) = 11/14

2. Smallest whole numbers. The least distance comes from the smallest integers. So n₁ = 11 for 700 nm and n₂ = 14 for 550 nm. Check: 11 × 700 = 14 × 550 = 7700 nm.
3. Put in numbers. With D = 1.5 m and d = 2.5 × 10⁻³ m, D/d = 600. So y = 11 × 7 × 10⁻⁷ × 600 = 4.62 × 10⁻³ m.
4. So. y = 462 × 10⁻⁵ m, so n = 462.

The common slip: 588. Pairing order 14 with 700 nm instead of with 550 nm gives 588. The longer wavelength needs the smaller order.

The idea to keep. For two wavelengths, the first coincidence of bright fringes is at the LCM of their fringe widths. Reduce λ₂/λ₁ to lowest terms to get the fringe orders.

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