JEE Main 2023 · 29 January · morning · Single correct · +4 / −1
In a Young's double slit experiment, two slits are illuminated with a light of wavelength 800 nm. The line joining A₁ P is perpendicular to A₁ A₂ as shown in the figure. If the first minimum is detected at P, the value of slits separation 'a' will be:
The distance of screen from slits D = 5 cm
- (a) 0.2 mm
- (b) 0.5 mm
- (c) 0.4 mm
- (d) 0.1 mm
Answer and solution
Answer (a) 0.2 mm
Solution
Physics sir · 4 steps
P sits right in front of A₁, so it is a/2 off the centre. That makes the path difference a²/2D.
- Locate P. A₁P is perpendicular to the slit plane, so P is level with A₁. The central line passes midway between the slits, so P is at height y = a/2 above the centre of the screen.
- Path difference. For a point at height y, the path difference is Δ = (a y)/D. With y = a/2 this becomes Δ = a²/2D. You get the same result from √(D² + a²) - D ≈ a²/2D, because a ≪ D.
a²/2D = λ/2 ⇒ a² = λ D
- Put in numbers. The first minimum needs Δ = λ/2. So a² = (800 × 10⁻⁹ m) × (5 × 10⁻² m) = 4 × 10⁻⁸ m², which gives a = 2 × 10⁻⁴ m.
- So. The slit separation is a = 0.2 mm, which is option (a).
The idea to keep. In YDSE, first find where the point is measured from the central line, not from a slit. The path difference is always a y / D with y measured from the centre.
Answer from the NTA final answer key. Solution by PrepWiser.