A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is 'x' nm. The value of 'x' to the nearest integer is ____________.

Solution

Physics sir · 4 steps

Fringe width is λ D/d, so flip it round for λ and keep every length in metres.

1. Fringe width formula. In Young's double slit experiment the fringe width is β = (λ D)/d, where D is the slit-to-screen distance and d is the slit separation. Rearranging gives the wavelength.

   λ = (β d)/D

2. Convert to metres. β = 6 mm = 6 × 10⁻³ m, d = 1 mm = 1 × 10⁻³ m and D = 10 m.
3. Put in numbers. λ = (6 × 10⁻³ × 1 × 10⁻³)/10 = 6 × 10⁻⁷ m. Since 1 nm = 10⁻⁹ m, this is 600 nm.
4. So. The wavelength is 600 nm, so x = 600.

The common slip: 0.6. Leaving β and d in millimetres gives λ = 6 × 1 / 10 = 0.6, which is in mm²/m and is not the wavelength in nm.

The idea to keep. Fringe width grows with wavelength and screen distance and shrinks with slit separation. Put all lengths in metres first, then convert the answer to nm at the very end.

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