Physics

Physics

Problem

The x−zx-z plane separates two media AA and BB of refractive indices μ1=1.5\mu_1 = 1.5 and μ2=2\mu_2 = 2. A ray of light travels from AA to BB. Its directions in the two media are given by unit vectors u⃗1=ai^+bj^\vec{u}_1 = a\hat{i} + b\hat{j} and u⃗2=ci^+dj^\vec{u}_2 = c\hat{i} + d\hat{j}. Then: (a) ac=43\frac{a}{c} = \frac{4}{3} (b) ac=34\frac{a}{c} = \frac{3}{4} (c) bd=43\frac{b}{d} = \frac{4}{3} (d) bd=34\frac{b}{d} = \frac{3}{4}
Intuitive explanation

Refraction at a boundary is governed by Snell's law, which relates the angles of incidence and refraction to the refractive indices—but the key insight is that the component of the wave vector parallel to the interface is conserved. Since the xx-zz plane is the boundary, the i^\hat{i} and k^\hat{k} components of the direction vectors don't "bend," only the perpendicular (j^\hat{j}) component does. This means you can set up a relationship between the parallel components using μ1sin⁡θ1=μ2sin⁡θ2\mu_1 \sin\theta_1 = \mu_2 \sin\theta_2, where the sines come directly from the i^\hat{i} (or k^\hat{k}) parts of your unit vectors.

Intuitive explanation

Refraction at a boundary is governed by Snell's law, which relates the angles of incidence and refraction to the refractive indices—but the key insight is that the component of the wave vector parallel to the interface is conserved. Since the xx-zz plane is the boundary, the i^\hat{i} and k^\hat{k} components of the direction vectors don't "bend," only the perpendicular (j^\hat{j}) component does. This means you can set up a relationship between the parallel components using μ1sin⁡θ1=μ2sin⁡θ2\mu_1 \sin\theta_1 = \mu_2 \sin\theta_2, where the sines come directly from the i^\hat{i} (or k^\hat{k}) parts of your unit vectors.

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