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QJEEM24Phc62b: JEE Main 2024 Physics - Class 12 JEEM24Phc62b Solution: In a double slit experiment shown in figure, when light of…

In a double slit experiment shown in figure, when light of wavelength 400 nm400 \mathrm{~nm} is used, dark fringe is observed at PP . If D=0.2 m\mathrm{D}=0.2 \mathrm{~m} , the minimum distance between the slits S1S_1 and S2S_2 is _________ mm\mathrm{mm} .

JEE Main 2024 (Online) 29th January Morning Shift Physics - Wave Optics Question 26 English

Step-by-Step Verified Solution

Path difference for minima at P\mathrm{P}

amp;2D2+d2−2D=λ2amp;∴D2+d2−D=λ4amp;∴D2+d2=λ4+Damp;⇒D2+d2=D2+λ216+Dλ2amp;⇒d2=Dλ2+λ216amp;⇒d2=0.2×400×10−92+4×10−144amp;⇒d2≈400×10−10amp;∴d=20×10−5amp;⇒d=0.20 mm\begin{aligned}& 2 \sqrt{\mathrm{D}^2+\mathrm{d}^2}-2 \mathrm{D}=\frac{\lambda}{2} \\& \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}-\mathrm{D}=\frac{\lambda}{4} \\& \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}=\frac{\lambda}{4}+\mathrm{D} \\& \Rightarrow \mathrm{D}^2+\mathrm{d}^2=\mathrm{D}^2+\frac{\lambda^2}{16}+\frac{\mathrm{D} \lambda}{2} \\& \Rightarrow \mathrm{d}^2=\frac{\mathrm{D} \lambda}{2}+\frac{\lambda^2}{16} \\& \Rightarrow \mathrm{d}^2=\frac{0.2 \times 400 \times 10^{-9}}{2}+\frac{4 \times 10^{-14}}{4} \\& \Rightarrow \mathrm{d}^2 \approx 400 \times 10^{-10} \\& \therefore \mathrm{d}=20 \times 10^{-5} \\& \Rightarrow \mathrm{d}=0.20 \mathrm{~mm}\end{aligned}

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