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ExamsJEE AdvancedPhysics

A light ray enters a flat surface of a glass semicylinder of refractive index n = sqrt(2) parallel to the flat face at a perpendicular distance d from the axis of the cylinder (radius R). What is the maximum value of d such that the light ray can still exit from the curved surface of the semicylinder (i.e., does not undergo total internal reflection)?

  1. (A) d = R / sqrt(2)
  2. (B) d = R / (2*sqrt(2))
  3. (C) d = R / 2
  4. (D) None of these

Correct answer: (A) d = R / sqrt(2)

Solution

For a ray entering the flat surface at distance d from the axis, it hits the curved surface where the local normal (radial direction) makes angle theta with the ray. Using geometry: sin(theta) = d/R. Critical angle theta_c = arcsin(1/n) = arcsin(1/sqrt(2)) = 45 deg. For the ray to exit, theta <= theta_c, so d/R <= 1/sqrt(2), giving d <= R/sqrt(2).

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