NTA Abhyas JEE Main2020PhysicsWave OpticsPractice
In Young’s double-slit experiment, the slit separation is 1 mm and the screen is at a distance of 1 m from the slit. For a monochromatic light of wavelength 500 nm , the distance of 3 rd minima from the central maxima is
Options
- A0.50   mm
- B1.25   mm
- C1.50   mm
- D1.75   mm
Correct answer
B. 1.25   mm
Step-by-step solution
Distance of n th minima from central bright fringe x n = 2 n - 1 λ D 2 d For n = 3 i.e., 3 rd minima x 3 = 2 × 3 - 1 × 5 0 0 × 1 0 - 9 × 1 2 × 1 × 1 0 - 3 = 5 × 5 0 0 × 1 0 - 6 2 = 1.25 × 1 0 - 3   m = 1.25   mm