JEE Main202325 Jan 2023Morning ShiftPhysicsWave OpticsActual
In Young's double slits experiment, the position of 5 th bright fringe from the central maximum is 5 cm . The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm . The separation between the slits is:
Options
- A60   μm
- B48   μm
- C12   μm
- D36   μm
Correct answer
A. 60   μm
Step-by-step solution
Given here, D = 1   m , λ = 600 × 10 - 9   m and n = 5 Distance of nth bright fringe is given by, y n = n λ D d , where, d is separation between the slits. ⇒ 5 × 600 × 10 - 9 × 1 d = 5 × 10 - 2 ⇒ d = 5 × 600 × 10 - 9 × 1 5 × 10 - 2 = 60 × 10 - 6   m ⇒ d = 60   μ m .