A single‑slit of width 0.02 cm is illuminated by light of wavelength 500 nm. Find the smallest angle θ (in degrees) at which a minimum in the diffraction pattern occurs.
Answer
0.143° (approximately)
The correct answer is 0.143° (approximately) — Find θ: θ ≈ sin⁻¹(2.5×10⁻³) ≈ 0.143°.
Step-by-step solution
- Write condition for minima: a sinθ = λ.
- Convert slit width to metres: 0.02 cm = 2×10⁻⁴ m.
- Convert wavelength to metres: 500 nm = 5×10⁻⁷ m.
- Solve for sinθ: sinθ = λ/a = (5×10⁻⁷)/(2×10⁻⁴) = 2.5×10⁻³.
- Find θ: θ ≈ sin⁻¹(2.5×10⁻³) ≈ 0.143°.
Common mistake: Neglecting to convert units to metres before applying the formula
Concept tested: Single‑slit diffraction minima
Published · How this solution was prepared and checked
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