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Light of wavelength 520 nm passes through a double slit, producing an interference pattern of relative intensity versus angular position theta. From the graph, the first interference minima occur at theta = +/- 0.5 degrees (so the angular fringe spacing/relevant angle can be read off). Determine the slit separation d.

1. about 30 micrometre
2. about 60 micrometre
3. about 15 micrometre
4. about 120 micrometre

Correct answer: about 30 micrometre

Solution

Reading the first minimum at theta = 0.5 degrees, the double-slit minimum condition d sin(theta) = lambda/2 (m=0) gives d = lambda/(2 sin(theta)). With theta approx 0.5 deg = 0.00873 rad and lambda = 520 nm, d comes out near 30 micrometre. (Note: the original graph supplies the angle; reconstructed here for a self-consistent numerical answer.)

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