White light is passed through a double slit and interference is observed on a screen 1.5 m away. The separation between the slits is 0.3 mm. The first violet and red fringes are formed 2.0 mm and 3.5 mm away from the central white fringes. the difference in wavelengths of red and violet light is ................ nm.

Solution

Physics sir · 4 steps

Fringe position is λ D/d, so the gap between the two fringes gives the gap in wavelengths.

1. First bright fringe. In YDSE the n-th bright fringe is at yₙ = (nλ D)/d. For the first fringe, n = 1, so the position is proportional to the wavelength.

   λᵣ - λᵥ = d/D(yᵣ - yᵥ)

2. Put in numbers. yᵣ - yᵥ = 3.5 - 2.0 = 1.5 mm = 1.5 × 10⁻³ m, d = 0.3 × 10⁻³ m and D = 1.5 m. So Δλ = (0.3 × 10⁻³ × 1.5 × 10⁻³)/1.5 = 3 × 10⁻⁷ m.
3. Check. Taken separately, λᵥ = 400 nm and λᵣ = 700 nm. Both are sensible values for violet and red light.
4. So. Δλ = 3 × 10⁻⁷ m, that is 300 nm.

The common slip: 600. Treating the first fringes as dark fringes, y = λ D/2d, doubles each wavelength and gives 600 nm, not 300.

The idea to keep. Fringe position grows linearly with wavelength, y = nλ D/d. A difference in positions converts straight into a difference in wavelengths. Put everything in metres first.

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