Wave Optics — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Wave Optics" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Wave Optics" — 7 important questions with detailed answers for CBSE board exam prepara…
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Key Questions Covered:
- Explain Huygens' principle and show how it accounts for reflection and refrac…
- What is interference? Derive the condition for constructive and destructive i…
- In Young's double slit experiment, the slits are 1 mm apart and screen is 1 m…
- + 4 more questions in the full chapter
Solutions Summary:
| Question | Status |
|---|---|
| Explain Huygens' principle and show how it accounts for r… | ✓ Solved |
| What is interference? Derive the condition for constructi… | ✓ Solved |
| In Young's double slit experiment, the slits are 1 mm apa… | ✓ Solved |
Showing 3 of 7 questions
Q1: Explain Huygens' principle and show how it accounts for reflection and refraction.
Huygens' Principle: Every point on a wavefront acts as a source of secondary wavelets, and the new wavefront is the envelope of these secondary wavelets.
Reflection using Huygens' Principle:
- Incident plane waves AB fall on plane surface MM'
- Each point on AB generates secondary wavelets
- At tim...
Q2: What is interference? Derive the condition for constructive and destructive interference in Young's double slit experiment.
Interference: Superposition of two coherent waves resulting in maximum or minimum intensity.
Young's Double Slit Experiment:
Two slits S₁ and S₂ separated by distance d
Distance to screen: D >> d
Wavelength: λ
Path difference at point P on screen:
Δ = S₂P - S₁P ≈ d sin θ ≈ d × y/D
Where y = ...
Q3: In Young's double slit experiment, the slits are 1 mm apart and screen is 1 m away. Light of wavelength 500 nm is used. Find fringe width and position of 3rd bright fringe.
Given:
Slit separation: d = 1 mm = 10⁻³ m
Distance to screen: D = 1 m
Wavelength: λ = 500 nm = 500 × 10⁻⁹ m = 5 × 10⁻⁷ m
Fringe width:
β = λD/d
β = (5 × 10⁻⁷ × 1)/(10⁻³)
β = 5 × 10⁻⁴ m = 0.5 mm
Position of 3rd bright fringe (n = 3, counting from center at n = 0):
For bright fringes: y_n = nλD/d
Fo...
Showing 3 of 7 questions. Visit the full page for complete solutions.