Q.16 Physics · Section A ⭐ 360° Option Elimination Active

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NEET 2024 Physics Question 16 Solution

Topic: Wave Optics · Official NTA Booklet Code T1

Question

If the monochromatic source in Young's double slit experiment is replaced by white light, then

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Verified Against NTA Final Answer Key

Option (1) : there will be a central bright white fringe surrounded by a few coloured fringes.

📚 Wave Optics Easy

📋 Given Parameters & Problem Setup

Light source White light (polychromatic containing continuous spectrum from 400 nm to 700 nm)

Experimental setup Standard Young’s Double Slit Experiment (YDSE)

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The 30-Second Exam Hall Shortcut: ⚡ Zero Path Difference Central Symmetry Hack

⏱️ Save ~40 seconds

At center, path difference Δx = 0 for ALL wavelengths -> All colors constructively interfere simultaneously -> Central WHITE fringe!

1 Central point on screen O has zero path difference: Δx = d sin θ = 0.

2 Zero path difference gives constructive interference for EVERY constituent wavelength of white light.

3 Mixing all colors in phase creates a bright WHITE central fringe.

4 Away from the center, fringe width β = λD/d varies with color: violet fringes form closest (smallest λ), red fringes farthest (largest λ).

5 This produces a few colored fringes near the center, after which overlapping creates uniform white illumination.

6 Matches Option (1) in under 5 seconds!

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Why You Can 100% Trust This Shortcut: NCERT Physics · Wave Optics (Interference with white light in Young’s double slit experiment).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Path Difference at the Central Point

In Young’s double slit experiment, the path difference between light waves emerging from slits S1 and S2 arriving at the central point O on the perpendicular bisector is identically zero:

\Delta x = S_2 O - S_1 O = 0

Step 2

Superposition at the Central Maximum

Since Δx = 0 satisfies the condition for constructive interference (Δx = nλ with n = 0) for every wavelength in the visible spectrum simultaneously, all color components (VIBGYOR) constructively combine with maximum intensity. Their additive combination yields a bright white central fringe.

\text{Phase difference } \Delta \phi = \frac{2\pi}{\lambda} (0) = 0 \implies I_{max} = 4 I_0 \text{ for all } \lambda

Step 3

Dispersion Away from the Center

For points away from the center (Δx ≠ 0), the fringe width β depends directly on wavelength:

\beta = \frac{\lambda D}{d}

Step 4

Color Fringing and Pattern Overlap

Because λ_violet ≈ 400 nm is smaller than λ_red ≈ 700 nm, the violet bright fringes form closest to the central white fringe, with red forming farther out. After a few colored fringes, the overlapping of different bright and dark bands of different colors produces uniform white illumination.

\beta_{violet} < \beta_{blue} < \dots < \beta_{red}

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360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 (1)

✓ 100% Correct

Verdict & Scientific Reason:

Standard textbook result: zero path difference ensures central white maximum.

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Why this is airtight: None.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

False; fringe width β = λD/d varies across different colors.

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Student Fallacy (Why candidates pick this trap): Assuming monochromatic fringe formula applies equally to all wavelengths.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

False; interference still occurs, clearly visible near the center.

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Student Fallacy (Why candidates pick this trap): Believing white light is completely incoherent and cannot produce any fringes.

4 (4)

✗ Incorrect

Verdict & Scientific Reason:

Directly contradicts the condition of zero path difference at the center (which is a maximum, not a minimum).

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Student Fallacy (Why candidates pick this trap): Confusing YDSE with Lloyd’s mirror where a 180° reflection phase shift occurs.

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NTA Trap Alert: Confusing Central Maximum with Destructive Interference

Students sometimes confuse thin film interference (reflected ray phase reversal) with YDSE and think the central fringe is dark.

🛡️ How to avoid the -1 mark: In standard YDSE with two in-phase slits, Δx = 0 is ALWAYS a bright maximum (white if source is white).

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Core NEET Syllabus Standard

Physics Core Wave Optics · Wave Optics Standard Benchmark

Core Scientific Principle Tested:

"When white light is used in YDSE, the central fringe is white because path difference is zero for all wavelengths; fringes on either side are colored with violet closest to center."

Syllabus Focus: Interference of Light Waves and Young’s Experiment

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Zero path difference blends all light: Central band is pure and WHITE!

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NEET 2024 Physics Question 16 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Zero Path Difference Central Symmetry Hack

Step-by-Step Rigorous Workflow & Mathematical Proof

Path Difference at the Central Point

Superposition at the Central Maximum

Dispersion Away from the Center

Color Fringing and Pattern Overlap

360° Option Elimination Engine Key Feature

NTA Trap Alert: Confusing Central Maximum with Destructive Interference

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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