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)

In a double slit experiment… - JEE Main 2024 Physics PYQ

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.

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

JEE Main Optics in other years

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

More JEE Main Optics questions

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index $n$ of the prism satisfies.

2026medium

A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. Refractive index of glass is 1.5. The rays converge to a point at a distance $x \mathrm{~cm}$ from the centre of the curvature of the spherical surface. The value of $x$ is $\_\_\_\_$ cm.

2026medium

A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification $m_{1}$ when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away, which lead to a change in magnification of total lens system to $m_{2}$. The value of $\left|\frac{m_{1}}{m_{2}}\right|$ is $\_\_\_\_$.

2026medium

As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface. ![JEE Main 2026 Physics, Ray Optics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Ray_Optics/6983c1828bfdbb243563959e/question_1__q_6983c1828bfdbb243563959e__cdn-question-pool.getmarks.app__21S2_q_34_1__70267be84d_final_ppt_sync.png) If refractive index of the material of prism is $\sqrt{2}$, the angle $\theta$ of prism is.

2026medium

The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is $\_\_\_\_$ cm.

2026medium

View all Optics questions

Keep practicing Optics

Work through every JEE Main Optics PYQ, year by year.

Browse all Physics topics

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

JEE Main Optics in other years

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

Keep practicing Optics

Work through every JEE Main Optics PYQ, year by year.

Browse all Physics topics

Did you get this right?

Solution

Your note

JEE Main Optics in other years

More JEE Main Optics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

Keep practicing Optics

Did you get this right?

Solution

Your note

JEE Main Optics in other years

More JEE Main Optics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

Keep practicing Optics