NEET · Physics —

Wave Optics appeared 2 times across 1 year — 4.4% of Physics. This question is from Interference of Light Waves and Young's Experiment.

Year 2024 Total
Questions 2 2

In interference and diffraction, the light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. A. As there is no gain or loss of energy, these phenomena are consistent with the principle of conservation of energy. B. Diffraction and interference are characteristics exhibited only by light waves. Choose the correct answer from the options given below:

Solution & Explanation

Core Logic

Evaluate Statement A: In interference and diffraction, energy is neither created nor destroyed; it merely gets spatially redistributed from dark regions to bright regions. Thus, the average energy remains constant, validating the law of conservation of energy. Statement A is true.

Evaluate Statement B: Interference and diffraction are fundamental characteristics of all types of waves, not just light. Sound waves, water waves, and matter waves all exhibit interference and diffraction. Therefore, Statement B is false.

Step 1: Final Conclusion

Statement A is correct, but Statement B is incorrect.

Pattern Recognition

Whenever "only" is used in physics statements describing a general wave property (like diffraction/interference), the statement is almost always false. All waves interfere and diffract.

Chapter Mix

Class 12 Physics: Wave Optics

Reference Study Guides

More Wave Optics Previous-Year Questions

Q13 neet_2026_03_may_morning Interference of Light Waves and Young's Experiment
In Young's double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where the path difference is λ, is K units. The intensity of light at a point where the path difference is (λ)/(3) will be
  • A. (K)/(4)
  • B. K
  • C. (K)/(2)
  • D. 2K

Solution

Related Formula
I = I₀ ²((Δφ)/(2)) Δφ = (2π)/(λ) Δ x
Core Logic

Where Δ x = λ, the phase difference Δφ = 2π. Intensity I = I₀ ²((2π)/(2)) = I₀ ²(π) = I₀. We are given that at this point, Intensity = K. Thus, I₀ = K.

Step 1: Calculate Intensity at new path difference

Now evaluate at Δ x = (λ)/(3). The phase difference is:

Δφ = (2π)/(λ) × (λ)/(3) = (2π)/(3)

Substitute this into the intensity formula:

I₁ = I₀ ²((2π/3)/(2)) = I₀ ²((π)/(3))

Since ((π)/(3)) = (1)/(2):

I₁ = I₀ ((1)/(2))² = (I₀)/(4)

Since I₀ = K, we get:

I₁ = (K)/(4)
Pattern Recognition

In YDSE, phase angle is strictly mapped to path difference λ → 2π. The square of the cosine half-angle dictates the final intensity drop.

Chapter Mix

Class 12 Physics: Wave Optics

More Wave Optics Questions — neet_2026_03_may_morning

Practice all Wave Optics previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)