The paint used on highway signs often contains small transparent spheres which provide nighttime illumination of the sign’s lettering by retro-reflecting vehicle headlight beams. Consider a light ray from air incident on one such sphere of radius r and index of refraction n. Let θ be its incident angle, and let the ray follow the path shown in Fig. 32–70, so that the ray exits the sphere in the direction exactly antiparallel to its incoming direction. Considering only rays for which sin θ can be approximated as θ, determine the required value for n.

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1. Identify the key concepts: The problem involves the phenomenon of retro-reflection, where light rays entering a transparent sphere are reflected back in the direction they came from. The sphere has a radius 'r' and an index of refraction 'n'. The incident angle is 'θ', and it is given that for small angles, sin θ can be approximated as θ.
2. Apply Snell's Law at the point of entry: Snell's Law, which relates the angles of incidence and refraction, is given by n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (for air), θ1 = θ, n2 = n (index of refraction of the sphere), and θ2 is the angle of refraction inside the sphere.
3. Consider the path of the light inside the sphere: The light ray must hit the opposite side of the sphere and reflect back along the same path. This requires that the angle of incidence inside the sphere equals the critical angle for total internal reflection at the interface between the sphere and air.
4. Determine the critical angle for total internal reflection: The critical angle θc can be found using sin(θc) = n1/n2, where n1 = 1 (air) and n2 = n. Since the ray must reflect back exactly, θ2 should be equal to or greater than θc.
5. Solve for the index of refraction 'n': Using the approximation sin θ ≈ θ for small angles, and the condition that θ2 should be equal to θc, set up the equation θ = sin(θc) = 1/n and solve for 'n'.