In a Young's double slit experiment 15 fringes are observed on a small portion of the screen when light of wavelength 500 nm is used. Ten fringes are observed on the same section of the screen when another light source of wavelength λ is used. Then the value of λ is (in nm) __________.

Solution

Physics sir · 3 steps

Same patch of screen, so fringe count times fringe width stays fixed: fewer fringes, wider ones.

1. Fringe width. Each fringe has width β = (λ D)/d. Only the light changes here, so D and d stay the same and β is proportional to λ.
2. Same length. If n fringes fill a section of length L, then L = nβ. The section is the same both times, so nβ is the same, and so is nλ.

   n₁λ₁ = n₂λ₂ ⇒ 15 × 500 = 10 × λ

3. So. λ = 7500/10 = 750 nm, so the answer is 750. It makes sense: a longer wavelength gives wider fringes, so fewer of them fit.

The common slip: 333. Taking λ proportional to the number of fringes gives 500 × 10/15 ≈ 333 nm. More fringes in the same space means each one is narrower, so the proportion goes the other way.

The idea to keep. When the same stretch of screen holds a different number of fringes, use nλ = constant. The number of fringes and the wavelength are inversely proportional.

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