Using Young's double slit experiment, a monochromatic light of wavelength 5000 Aᵒ produces fringes of fringe width 0.5 mm. If another monochromatic light of wavelength 6000 Aᵒ is used and the separation between the slits is doubled, then the new fringe width will be :

Solution

Physics sir · 3 steps

Fringe width goes up with wavelength and down with slit gap, so scale 0.5 mm by both ratios.

1. Fringe width formula. In Young's double slit experiment the fringe width is β = (λ D)/d. Here D is the screen distance and d is the slit separation. D does not change, so only λ and d matter.

   β₂/β₁ = λ₂/λ₁ × d₁/d₂

2. Put in numbers. The wavelength ratio is 6000/5000 = 1.2. The slit gap doubles, so d₁/d₂ = 1/2. Then β₂ = 0.5 mm × 1.2 × 1/2 = 0.3 mm.
3. So. The new fringe width is 0.3 mm, option (d).

The idea to keep. Since β = λ D/d, never recompute from scratch. Multiply the old fringe width by the ratio of each quantity that changed, upstairs or downstairs as the formula says.

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