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A semi-cylinder made of transparent plastic has refractive index mu = sqrt(2) and radius R. A narrow light ray enters perpendicular to the flat face of the semi-cylinder at a distance d from the axis of symmetry. What is the maximum value of d such that the light ray can still emerge from the curved surface?
- d = R / sqrt(2)
- d = R / (2*sqrt(2))
- d = R / 2
- None of these
Correct answer: d = R / sqrt(2)
Solution
The ray travels horizontally at height d inside the semi-cylinder. It hits the curved surface at a point where the outward normal is radial. The angle of incidence theta satisfies sin(theta) = d/R. Critical angle: sin(theta_c) = 1/mu = 1/sqrt(2), so theta_c = 45 deg. For emergence: sin(theta) <= 1/sqrt(2), meaning d <= R/sqrt(2). Maximum d = R/sqrt(2).
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