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Light of wavelength 550 nm falls normally on a slit of width 22.0×10⁻⁵ cm. The angular position of the second minima from the central maximum will be (in radians)

1. π/8
2. π/12
3. π/4
4. π/6

Correct answer: π/6

Solution

For single-slit minima, a sin(theta) = m*lambda. Slit width a = 22.0e-5 cm = 22.0e-7 m, lambda = 550 nm. For the 2nd minimum sin(theta) = 2*lambda/a = 2*(550e-9)/(22.0e-7) = 0.5, so theta = pi/6 rad.

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