Grade 12 Physics · Trapped light
The angle where the exit closes
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The angle where the exit closes

Push Snell's law and it breaks in an interesting way. Send a ray from a dense medium toward a rarer one and it bends away from the normal — further and further as you tilt it, until at one particular angle the refracted ray lies flat along the surface, θ2=90°\theta_2 = 90°. Past that, there is no refracted ray at all: the light is totally internally reflected, and the surface becomes a perfect mirror. Set sinθ2=1\sin\theta_2 = 1 in Snell's law and out drops sinθc=n2n1\sin\theta_c = \dfrac{n_2}{n_1}sine theta-c equals n-two over n-one. Here θc\theta_c is the critical angle in degrees, n1n_1 is the index of the medium the light is starting in, and n2n_2 the index of the one it is trying to reach. Same subscript convention as Snell's law, because it IS Snell's law.

The ratio n2/n1n_2/n_1 has to sit at or below 1 for an arcsin to exist, so the light must start in the denser medium. Dense to rare, always. That single fact rules out half the exam's scenarios before you compute anything — and it is what carries your voice down a fibre optic cable, bouncing off the core wall thousands of times a kilometre and never leaking out.

Refraction's second trick needs no angles at all. Look straight down into water and the bottom rises to meet you: d=dnd' = \dfrac{d}{n}, d-prime equals d over n, where dd is the real depth in metres, dd' (said “d-prime”) is the apparent depth in metres, and nn is the liquid's index. Water's 1.33 makes a 2.0 m pool look about 1.5 m deep. Every summer, somebody misjudges that. The formula is the small-angle child of Snell's law, which is why it only holds looking straight down.