JEE Main 2025 · 24 January · morning · Single correct · +4 / −1
The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is
- (a) 8
- (b) 4
- (c) 5
- (d) 6
Answer and solution
Answer (c) 5
Solution
Physics sir · 4 steps
Two bright fringes coincide when they sit at the same spot, so set n₁λ₁ = n₂λ₂.
- Bright fringe position. The nth bright fringe sits at yₙ = (nλ D)/d. Both wavelengths pass through the same slits onto the same screen, so D and d are the same for both.
- Coincidence condition. Two bright fringes overlap when their positions are equal. D/d cancels and leaves a condition on orders and wavelengths alone.
n₁ × 480 = n₂ × 600 ⇒ n₁/n₂ = 600/480 = 5/4
- Smallest whole numbers. The orders must be whole numbers, so the smallest choice is n₁ = 5 and n₂ = 4. Check: 5 × 480 = 4 × 600 = 2400 nm.
- So. The 5th bright fringe of 480 nm light falls on the 4th bright fringe of 600 nm light. The answer is 5.
The idea to keep. For overlapping bright fringes of two colours, equate nλ. The shorter wavelength always needs the higher order. Take the LCM of the wavelengths and divide by each one.
Answer from the NTA final answer key. Solution by PrepWiser.