In a Young's double slit experiment, a laser light of 560 nm produces an interference pattern with consecutive bright fringes' separation of 7.2 mm. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of 8.1 mm. The wavelength of second light is __________ nm.

Solution

Physics sir · 3 steps

Same slits, same screen, so fringe width just scales with wavelength. Set up one ratio and you are done.

1. Fringe width. The gap between consecutive bright fringes is β = (λ D)/d. Only the light changes here, so D and d stay the same and β ∝ λ.

   λ₂/λ₁ = β₂/β₁

2. Put in numbers. λ₂ = 560 nm × 8.1/7.2 = 560 × 1.125 nm. Both fringe widths are in mm, so the units cancel and nothing needs converting.
3. So. λ₂ = 630 nm, so the answer is 630.

The common slip: 497.8. Writing the ratio upside down, λ₂ = 560 × 7.2/8.1, gives about 498, not 630. A wider fringe must come from a longer wavelength.

The idea to keep. When only one quantity in β = (λ D)/d changes, do not hunt for D or d. Write a ratio and let everything else cancel.

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