In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D = 0.2 m, the minimum distance between the slits S₁ and S₂ is _________ mm. Numerical answer type. Work it out, then open the answer.

Answer: 0.20. Path via S₂ (on the axis): 2D. Path via S₁: 2√(D² + d²). Δ = 2(√(D²+d²) - D) ≈ d²/D for d D. Minimum d for a dark fringe: d²/D = (λ)/(2) ⇒ d² = (λ D)/(2) = 400×10⁻⁹× 0.22 = 4×10⁻⁸ m². d = 2×10⁻⁴ m = 0.20 mm.

In a double slit experiment shown in figure, when light of wavelength… | JEE Main 2024 (29 Jan, Shift 1)

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YDSE: Film, Medium & Two Wavelengths

In a double slit experiment shown in figure, when light of wavelength $400\text{ nm}$ is used, dark fringe is observed at $P$ . If $D = 0.2\text{ m}$ , the minimum distance between the slits $S_1$ and $S_2$ is _________ $\text{mm}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 0.20

Path via $S_2$ (on the axis): $2D$ . Path via $S_1$ : $2\sqrt{D^2 + d^2}$ .
$\Delta = 2(\sqrt{D^2+d^2} - D) \approx \frac{d^2}{D}$ for $d \ll D$ .
Minimum $d$ for a dark fringe: $\frac{d^2}{D} = \frac{\lambda}{2} \Rightarrow d^2 = \frac{\lambda D}{2} = \frac{400\times10^{-9}\times 0.2}{2} = 4\times10^{-8}\ \text{m}^2$ .
$d = 2\times10^{-4}\ \text{m} = 0.20\ \text{mm}$ .

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