In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment, a source of light of wavelengh 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is :

Solution

Physics sir · 4 steps

Fringe width is just λ D/d. Put everything in metres and the answer falls out.

1. Fringe width. The gap between two successive bright fringes is the fringe width β. It depends only on the wavelength λ, the slit-to-screen distance D and the slit separation d.

   β = (λ D)/d

2. Convert to SI. λ = 589 nm = 589 × 10⁻⁹ m, D = 1.5 m, and d = 0.15 mm = 0.15 × 10⁻³ m.
3. Put in numbers. D/d = 1.5/(0.15 × 10⁻³) = 10⁴. So β = 589 × 10⁻⁹ × 10⁴ = 5.89 × 10⁻³ m.
4. So. β = 5.89 mm ≈ 5.9 mm, which is option (b).

The idea to keep. Successive bright fringes, or successive dark fringes, are always one fringe width β = λ D/d apart. Convert nm and mm to metres before you divide.

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