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Visible light of wavelength 6000 × 10⁻⁸ cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60° from the central maximum. If the first minimum is produced at θ1, then θ1 is close to:

1. 20°
2. 30°
3. 25°
4. 45°

Correct answer: 25°

Solution

Second minimum: a*sin60 = 2*lambda. First minimum: a*sin(theta1) = lambda, so sin(theta1) = sin60/2 = 0.433, giving theta1 = 25.7 deg, closest to 25 deg.

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