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In a Lloyd's-mirror arrangement, a long narrow horizontal slit lies 1 mm above a plane mirror. The interference pattern formed by the slit and its image in the mirror is observed on a screen 1 m from the slit. The wavelength of the light is 600 nm. Find the distance of the first maximum above the mirror surface. Construct four plausible options including the computed value.

1. 0.15 mm
2. 0.075 mm
3. 0.30 mm
4. 0.60 mm

Correct answer: 0.15 mm

Solution

The two coherent sources are the slit and its mirror image, separated by d = 2 x 1 mm = 2 mm. Fringe width beta = lambda*D/d = (600e-9 * 1)/(2e-3) = 3e-4 m = 0.30 mm. In Lloyd's mirror the reflected beam undergoes a pi phase change, so the position of the mirror is a minimum (dark) and the first maximum lies half a fringe width above it: y1 = beta/2 = 0.15 mm.

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