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JEE Main20268 April 2026Evening ShiftPhysicsRay OpticsActual

Light ray incident along a vector AO ( AO = 2 i -3 j ) emerges out along vector OB ( OB = C i -4 j ) as shown in the figure below. The value of C is ________.

Options

  1. A1.6
  2. B0.16
  3. C11.6
  4. D16

Correct answer

A. 1.6

Step-by-step solution

From the given figure, the interface is horizontal and the normal is along the y-axis. The incident ray is along the vector AO = 2 i - 3 j . The angle of incidence is the angle between the incident ray and the normal. = 2 2^2 + (-3)^2 = 2 13 The refracted ray is along the vector OB = C i - 4 j . The angle of refraction is the angle between the refracted ray and the normal. = C C^2 + (-4)^2 = C C^2 + 16 Applying Snell's law at the interface: ₁ = ₂ 1 2 13 = 1.5 C C^2 + 16 2 13 = 3 2 C C^2 + 16 Squaring both sides: 4

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