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Light of wavelength 600 nm illuminates a single slit of width 4*10⁻⁴ m. On a screen 2 m away a single-slit diffraction pattern forms. Find the distance 's' from the central maximum to the first diffraction minimum.

1. 0.002 m
2. 0.003 m
3. 0.004 m
4. 0.006 m

Correct answer: 0.003 m

Solution

For a single slit, the first minimum occurs at a*sin(theta) = lambda. Using small angle, s = lambda*D/a = (600e-9 * 2)/(4e-4) = 1.2e-6/4e-4 = 3e-3 m = 0.003 m.

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