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Light of wavelength 6000 angstrom falls on a thin glass plate of refractive index 1.5. The angle of refraction inside the plate is 60 deg. Find the smallest thickness of the plate that makes it appear dark in reflected light.

1. 4000 angstrom
2. 2000 angstrom
3. 6000 angstrom
4. 8000 angstrom

Correct answer: 4000 angstrom

Solution

Reflection from a denser medium adds a pi phase shift at the top surface, so the dark (destructive) condition in reflected light is 2*mu*t*cos(r) = m*lambda. For the smallest thickness take m = 1: t = lambda/(2*mu*cos r) = 6000/(2*1.5*cos60deg) = 6000/(3*0.5) = 6000/1.5 = 4000 angstrom.

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