In a Young’s double slit experiment, 16 fringes are observed in a certain segment of the screen when light of a wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be

Solution

Physics sir · 3 steps

The segment length is fixed, so number of fringes times fringe width stays the same.

1. Fringe width. Fringe width is β = (λ D)/d. With D and d unchanged, β is proportional to λ.
2. Same segment. The segment length equals the number of fringes times the fringe width. That length does not change, so n₁ β₁ = n₂ β₂, which means n₁ λ₁ = n₂ λ₂.

   n₂ = (n₁ λ₁)/λ₂ = (16 × 700)/400 = 28

3. So. A shorter wavelength gives narrower fringes, so more of them fit. The same segment now holds 28 fringes, option (a).

The idea to keep. Fringe width scales with wavelength. For a fixed stretch of screen, the number of fringes goes as 1/λ, so nλ stays constant.

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