Physics

Physics

Problem

When light travels from medium 1 with refractive index n1n_1 to medium 2 with refractive index n2n_2 (n1>n2n_1>n_2), the critical angle ici_c is correctly given by a) sin⁡ic=1n1\sin i_c = \dfrac{1}{n_1} b) sin⁡ic=n1−n2n1\sin i_c = \dfrac{n_1-n_2}{n_1} c) sin⁡ic=n2n1\sin i_c = \dfrac{n_2}{n_1} d) sin⁡ic=n2n1\sin i_c = \sqrt{\dfrac{n_2}{n_1}}
Intuitive explanation

When light travels from a denser to a less dense medium, it bends away from the normal—and at some angle, it bends so much that the refracted ray grazes along the boundary. That's the critical angle, where refraction becomes impossible and total internal reflection takes over. Use Snell's law at this special moment: the refracted angle is exactly 90°, so sin⁡(90°)=1\sin(90°) = 1, which lets you solve directly for the incident angle ici_c in terms of the two refractive indices.

Intuitive explanation

When light travels from a denser to a less dense medium, it bends away from the normal—and at some angle, it bends so much that the refracted ray grazes along the boundary. That's the critical angle, where refraction becomes impossible and total internal reflection takes over. Use Snell's law at this special moment: the refracted angle is exactly 90°, so sin⁡(90°)=1\sin(90°) = 1, which lets you solve directly for the incident angle ici_c in terms of the two refractive indices.

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