JEE Main 2021 · 18 March · morning · Single correct · +4 / −1
In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890 Aᵒ is :-
- (a) 1178 × 10⁻⁹ m
- (b) 1178 × 10⁻⁶ m
- (c) 1178 × 10⁻¹² m
- (d) 5890 × 10⁻⁷ m
Answer and solution
Answer (b) 1178 × 10⁻⁶ m
Solution
Physics sir · 4 steps
First to third bright fringe is two fringe widths apart, so find β once and double it.
- Count the gaps. Neighbouring bright fringes are one fringe width β apart. From the 1st to the 3rd bright fringe there are two gaps, so the distance is 2β.
β = (λ D)/d
- Convert to SI. λ = 5890 Å = 5890 × 10⁻¹⁰ m, D = 0.5 m and d = 0.5 mm = 0.5 × 10⁻³ m.
- Put in numbers. β = (5890 × 10⁻¹⁰ × 0.5)/(0.5 × 10⁻³) = 5890 × 10⁻⁷ m = 5.89 × 10⁻⁴ m.
- So. The distance is 2β = 2 × 5.89 × 10⁻⁴ = 1.178 × 10⁻³ m = 1178 × 10⁻⁶ m, which is option (b).
The idea to keep. The distance from the mth to the nth bright fringe is (n - m)β. Count the gaps between the fringes, not the fringes, and put every length in metres before dividing.
Answer from the NTA final answer key. Solution by PrepWiser.