In a double slit experiment shown in figure, when light of wavelength $400\mathrm{nm}$ is used, dark fringe is observed at $P$. If $D=0.2m$, the minimum distance between the slits ${S}_{1}$ and ${S}_{2}$ is $\alpha \mathrm{mm}$. Write the value of $10\alpha$ to the nearest integer. ![JEE Main 2024 Physics, Wave Optics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Wave_Optics/6682bd88ba0e94a3c4025d3c/question_1__q_6682bd88ba0e94a3c4025d3c__cdn-question-pool.getmarks.app__db11ceee-3588-4622-97d4-6541cebc69d4__94b12cc91e_final_ppt_sync.png)

JEE Main 2024 — Physics Optics

2024integerhard

In a double slit experiment shown in figure, when light of wavelength $400\mathrm{nm}$ is used, dark fringe is observed at $P$ . If $D=0.2m$ , the minimum distance between the slits ${S}_{1}$ and ${S}_{2}$ is $\alpha \mathrm{mm}$ . Write the value of $10\alpha$ to the nearest integer.

JEE Main 2024 Physics, Wave Optics — question figure

Official previous-year question

Held on 29 Jan 2024 · Verified 6 Jul 2026.

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Solution

JEE Main 2024 Physics, Wave Optics — solution figure

According to the given diagram, the path difference for minima at P is given by

$\Delta x=2\sqrt{{D}^{2}+{d}^{2}}-2D...(1)$

And, from the condition of minima, it follows that

$\Delta x=\frac{\lambda }{2}...(2)$

Equations (1) and (2) imply that

$2\sqrt{{D}^{2}+{d}^{2}}-2D=\frac{\lambda }{2}$

$\Rightarrow \sqrt{{D}^{2}+{d}^{2}}-D=\frac{\lambda }{4}$

$\Rightarrow \sqrt{{D}^{2}+{d}^{2}}=\frac{\lambda }{4}+D$

$\Rightarrow {D}^{2}+{d}^{2}={D}^{2}+\frac{{\lambda }^{2}}{16}+\frac{D\lambda }{2}$

$\Rightarrow {d}^{2}=\frac{D\lambda }{2}+\frac{{\lambda }^{2}}{16}$

$\Rightarrow {d}^{2}=\frac{0.2\times 400\times {10}^{-9}}{2}+\frac{16\times {10}^{-14}}{16}$

$\Rightarrow {d}^{2}\approx 400\times {10}^{-10}$

$\begin{matrix}\therefore d & = & 20\times {10}^{-5}m \\ & = & 0.20\mathrm{mm}\end{matrix}$

Therefore, $10\alpha =2$ .

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JEE Main 2024 — Physics Optics

2024integerhard

JEE Main 2024 Physics, Wave Optics — question figure

Official previous-year question

Held on 29 Jan 2024 · Verified 6 Jul 2026.

Did you get this right?

Solution

JEE Main 2024 Physics, Wave Optics — solution figure

According to the given diagram, the path difference for minima at P is given by

$\Delta x=2\sqrt{{D}^{2}+{d}^{2}}-2D...(1)$

And, from the condition of minima, it follows that

$\Delta x=\frac{\lambda }{2}...(2)$

Equations (1) and (2) imply that

$2\sqrt{{D}^{2}+{d}^{2}}-2D=\frac{\lambda }{2}$

$\Rightarrow \sqrt{{D}^{2}+{d}^{2}}-D=\frac{\lambda }{4}$

$\Rightarrow \sqrt{{D}^{2}+{d}^{2}}=\frac{\lambda }{4}+D$

$\Rightarrow {D}^{2}+{d}^{2}={D}^{2}+\frac{{\lambda }^{2}}{16}+\frac{D\lambda }{2}$

$\Rightarrow {d}^{2}=\frac{D\lambda }{2}+\frac{{\lambda }^{2}}{16}$

$\Rightarrow {d}^{2}=\frac{0.2\times 400\times {10}^{-9}}{2}+\frac{16\times {10}^{-14}}{16}$

$\Rightarrow {d}^{2}\approx 400\times {10}^{-10}$

$\begin{matrix}\therefore d & = & 20\times {10}^{-5}m \\ & = & 0.20\mathrm{mm}\end{matrix}$

Therefore, $10\alpha =2$ .

Your note

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Optics 2026 Optics 2025 Optics 2023 Optics 2022 Optics 2021 Optics 2020 Optics 2019 Optics 2018 Optics 2017 Optics 2016 Optics 2015 Optics 2014

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JEE Main 2024 — Physics Optics

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JEE Main 2024 — Physics Optics

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Solution

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JEE Main Optics in other years

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Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

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