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Physics · Wave Optics

JEE Main 2024 — 29 January, Shift 1 — Question 54

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\mathrm{D}=0.2 \mathrm{~m}. the minimum distance between the slits S1S_{1} and S2S_{2} is _______\_\_\_\_\_\_\_ mm .

figure

Answer: 0.2

Numerical answer — enter this value.

Step-by-step solution

Path difference for minima at P 2D2+d2−2D=λ22 \sqrt{\mathrm{D}^{2}+\mathrm{d}^{2}}-2 \mathrm{D}=\frac{\lambda}{2} ∴D2+d2−D=λ4\therefore \sqrt{\mathrm{D}^{2}+\mathrm{d}^{2}}-\mathrm{D}=\frac{\lambda}{4} ∴D2+d2=λ4+D\therefore \sqrt{\mathrm{D}^{2}+\mathrm{d}^{2}}=\frac{\lambda}{4}+\mathrm{D} ⇒D2+d2=D2+λ216+Dλ2\Rightarrow D^{2}+d^{2}=D^{2}+\frac{\lambda^{2}}{16}+\frac{D \lambda}{2} ⇒d2=Dλ2+λ216\Rightarrow \mathrm{d}^{2}=\frac{\mathrm{D} \lambda}{2}+\frac{\lambda^{2}}{16} ⇒d2=0.2×400×10−92+4×10−144\Rightarrow \mathrm{d}^{2}=\frac{0.2 \times 400 \times 10^{-9}}{2}+\frac{4 \times 10^{-14}}{4} ⇒d2≈400×10−10\Rightarrow \mathrm{d}^{2} \approx 400 \times 10^{-10}

∴d=20×10−5\therefore \mathrm{d}=20 \times 10^{-5}

⇒d=0.20 mm\Rightarrow \mathrm{d}=0.20 \mathrm{~mm}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 1
Subject
Physics
Chapter
Wave Optics
Topic
Young's Double Slit Experiment and Its Modifications