In a Young's double slit experiment, two slits are illuminated with a light of wavelength 800 nm. The line joining is perpendicular to as shown in the figure. If the first minimum is detected at , the value of slits separation '' will be:
The distance of screen from slits D = 5 cm
The distance of screen from slits D = 5 cm
- A.0.2 mm
- B.0.4 mm
- C.0.5 mm
- D.0.1 mm
Show answer and solution
Correct option: A
Solution
The geometry shows A₁P is perpendicular to A₁A₂, meaning P is directly opposite slit A₁. The path difference at P is the distance A₂P - A₁P. Since A₁P ⊥ A₁A₂, we have A₁P = D = 5 cm and A₁A₂ = a, so A₂P = √(D² + a²).
Path difference = √(D² + a²) - D.
For first minimum, path difference = λ/2:
√(D² + a²) - D = λ/2
√(D² + a²) = D + λ/2
D² + a² = D² + Dλ + λ²/4
a² = Dλ + λ²/4 ≈ Dλ (since λ << D)
a = √(Dλ) = √(5×10⁻² × 800×10⁻⁹) = √(4×10⁻⁸) = 2×10⁻⁴ m = 0.2 mm.
Path difference = √(D² + a²) - D.
For first minimum, path difference = λ/2:
√(D² + a²) - D = λ/2
√(D² + a²) = D + λ/2
D² + a² = D² + Dλ + λ²/4
a² = Dλ + λ²/4 ≈ Dλ (since λ << D)
a = √(Dλ) = √(5×10⁻² × 800×10⁻⁹) = √(4×10⁻⁸) = 2×10⁻⁴ m = 0.2 mm.
[JEE Main 2023 · 29 Jan · Shift 1]