In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by 5 × 10⁻² m towards the slits, the change in fringe width is 3 × 10⁻³ cm. If the distance between the slits is 1 mm, then the wavelength of the light will be ____________ nm.

Solution

Physics sir · 4 steps

Fringe width is proportional to D, so the change in width depends only on how far the screen moved.

1. Fringe width. Fringe width is β = (λ D)/d. Here λ and d stay fixed and only D changes, so β changes in proportion to D.

   Δβ = (λ Δ D)/d ⇒ λ = (Δβ d)/(Δ D)

2. Convert to metres. Δβ = 3 × 10⁻³ cm = 3 × 10⁻⁵ m, d = 1 mm = 10⁻³ m and Δ D = 5 × 10⁻² m. You do not need the original screen distance.
3. Put in numbers. λ = (3 × 10⁻⁵ × 10⁻³)/(5 × 10⁻²) = 6 × 10⁻⁷ m.
4. So. 6 × 10⁻⁷ m = 600 × 10⁻⁹ m, so the wavelength is 600 nm.

The common slip: 60000. Using the change in fringe width as 3 × 10⁻³ m instead of converting it from cm gives 6 × 10⁻⁵ m, which is 60000 nm, not 600.

The idea to keep. When the screen moves, Δβ = λ Δ D/d. The unknown original distance drops out. Put every length in metres before you divide.

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