NEET2026PhysicsChapterActual
In a Young's double slit experiment, a monochromatic light of wavelength 400 nm is used to illuminate the slits. The interference pattern is observed on a screen placed at a distance of 2.5 m from the plane of the slits. If the distance between two consecutive dark fringes is measured to be 2.0 mm , the separation between the two slits is
Options
- A2.5 10⁻⁴ m
- B5 10⁻⁴ m
- C1 10⁻³ m
- D5 10⁻⁵ m
Correct answer
B. 5 10⁻⁴ m
Step-by-step solution
Given = 400 nm = 400 10⁻⁹ m D = 2.5 m Fringe width = 2.0 mm = 2.0 10⁻³ m The distance between two consecutive dark fringes is equal to the fringe width . Using the formula for fringe width = D d d = D d = 400 10⁻⁹ 2.5 2.0 10⁻³ d = 1000 10⁻⁹ 2.0 10⁻³ d = 5 10⁻⁴ m Answer: 5 10⁻⁴ m