In a two slit experiment, light of wavelength 600 nm is used. The slit spacing i

Question

In a two slit experiment, light of wavelength 600 nm is used. The slit spacing is 60000 nm. The interference pattern is observed on a screen 300 cm from the slits. (You may assume that the small angle approximation is valid and that the slits are so narrow that diffraction effects are negligible.) The m = 1 maximum is 30 mm from the center of the interference pattern. At a point which is one-third this distance from the center of the interference pattern, what is the phase difference between the two waves?

Please answer with step-by-step instructions, showing all work, including all calculations and formulas used.

Explanation / Answer

y = 30/3 = 10 mm = 10e-3 m

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sin = y/L = 10e-3/(300e-2) = 0.0033333

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path difference = = d sin = 60000 * 0.0033333 = 200 nm

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= (2/) = (2*3.14/600)*200 = 2.0933 radian = 120 degrees

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